Notes

Method 2: using sum, pair sum and product of roots

Method 1:

B1: States the other complex root. May be implied by later working.

M1: Uses the area of the triangle to determine at least one value for the real root. You can award for sight of one correct value of \(z_3\)
(Note: \(h = 3 \Rightarrow z_3 = 5\) or \(-1\))

dM1: Dependent on the previous method mark. Deduces a correct expression for one of their real roots. Award for \(\big(z - (2 - 4\mathrm{i})\big)\big(z - (2 + 4\mathrm{i})\big)(z - z_3)\) or equivalent e.g. \(\left(z^2 - 4z + 20\right)(z - z_3)\)
\(z_3\) must be real.

ddM1: Dependent on previous method marks. Multiplies out to find values for the constants \(a\), \(b\) and \(c\). Award for the sight of at least 2 correct values of the constants \(a\), \(b\) and \(c\) in any one of their attempts at the function.

A1: Deduces the correct two functions.

Method 2:

B1: States the other complex root. May be implied by later working.

M1: Uses the area of the triangle to determine at least one value for the real root, You can award for sight of one correct value of \(z_3\) (Note: \(h = 3 \Rightarrow z_3 = 5\) or \(-1\))

dM1: Dependent on the previous method mark. Attempts sum, pair sum and product for at least one of their real roots.

ddM1: Dependent on previous method marks. Multiplies out to find values for the constants \(a\), \(b\) and \(c\)
Award for the sight of at least 2 correct values of the constants \(a\), \(b\) and \(c\) in any one of their attempts at the function.

A1: Deduces the correct two functions.

Note: There are several methods that can be used to obtain the first method mark:

• Uses vertices \((2,\ 4)\), \((2,\ -4)\) and \((x_3,\ 0)\) to determine at least one value for the real root. Alternatively uses \((x_3,\ y_3)\). This may come from a diagram. Do not be concerned about the variables they use.
  e.g. \(12 = \dfrac{1}{2} \times 8 \times \lvert x_3 - 2\rvert \Rightarrow z_3 =\) or \(x_3 =\)
• Uses the formula for the area of a triangle with vertices \((x_1,\ y_1),\ (x_2,\ y_2),\ (x_3,\ y_3)\) to determine at least one value for the real root. Do not be concerned about the variables they use.
  \(\left[\dfrac{1}{2}\left\lvert x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right\rvert\right]\)
  \(12 = \dfrac{1}{2}\left\lvert 2(-4 - y_3) + 2(y_3 - 4) + x_3(4 - {-4})\right\rvert\)
  \(24 = \lvert -16 + 8x_3\rvert\)
  \(\Rightarrow z_3 =\) or \(x_3 =\)
• Uses the shoelace method with no sign errors.
  e.g. Area \(= \dfrac{1}{2}\begin{vmatrix}2 & 2 & x_3 & 2\\ 4 & -4 & 0 & 4\end{vmatrix}\)
  \(12 = \dfrac{1}{2}\left\lvert (2 \times {-4} + 4x_3) - (2 \times 4 - 4x_3)\right\rvert\)
  \(24 = \lvert -16 + 8x_3\rvert\)
  \(\Rightarrow z_3 =\) or \(x_3 =\)

If you are uncertain in how to apply a method, then please send to review.

Notes

M1: Recalls that the locus represents a circle

A1: Circle with centre (1, 0) and radius 1, ignore any tangents drawn

Notes

(Corrected from the printed mark scheme: in Alternative 1 the term \(2x\tan^2\theta\) is printed as \(2x\tan^2 x\).)

M1: Complete method to find one possible value of \(\theta\)

Alternative 1: condone slips with algebra as long as intention is clear

A1: One correct value \(\theta\)

A1: Deduces the other correct value of \(\theta\) and no other incorrect angles

Notes

M1: A correct method to find a complex number using one of their values of \(\theta\)

A1: One correct complex number check special case for coordinates only

A1ft: Deduces the other correct complex number. Follow through on \(z = a + b\mathrm{i}\) so scored for \(z = a - b\mathrm{i}\) as long as the method mark is scored

Special case:
If leave as two correct coordinates or give values for \(x\) and \(y\) then SC M1 A1 A0

Alternative

Scheme	Marks	AO
Using equilateral triangles to find \(z = \ldots\) \[z = 1\left(\cos\left(\pm\frac{\pi}{3}\right) + \mathrm{i}\sin\left(\pm\frac{\pi}{3}\right)\right)\]	M1	3.1a
\(z = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}\mathrm{i}\)	A1	1.1b
\(z = \dfrac{1}{2} - \dfrac{\sqrt{3}}{2}\mathrm{i}\)	A1ft	2.2a

Notes

B1: States either \(b + d = 0\) or \(c - a = 2\) (oe)

M1: Uses the modulus information to write down two more equations (need not be squared). Accept if they forget to square one of the moduli but the sum of squares must be correct.

M1: Uses their equations to solve simultaneously to find at least one value. If they assume \(a = 2\), you may allow this mark for finding at least one of \(b\) or \(d\).

ddM1: Dependent on both previous method marks. Uses their equations to find values for all of the constants. This mark is not available if \(a = 2\) is assumed. All must be found from algebra.

A1: Correct values from correct work.

Notes

B1: One circle drawn with correct intercepts OR both circles drawn in correct position, not overlapping, but with no intercepts shown. Be tolerant on the shape but must be a clear attempt at a circle – a closed loop with no obvious kinks. Labels on imaginary axis may be just numbers or i and 9i.

B1: Both circles drawn with correct intercepts and the circles does not intersect each other. Again be tolerant as per first B mark.

Notes

M1: Finds the distance between the centres of the circles.

dM1: Dependent on previous method mark. Full method to find the nearest and furthest points of circles. E.g. Uses their distance +/- sum of radii. If their distance between circles is less than 11, there must a check seen to show this has been considered. If by error they show the circles touch (not cross) then note that finding twice the sum of radii is a valid method for the greatest distance.

A1: Correct answer. Allow if the centres were on the negative axes.

Alt: There may be attempts via finding the line through the centres.

M1: Full method to find the equation of the line through the two centres:
\(m = \dfrac{0 - 5}{12 - 0} = -\dfrac{5}{12} \Rightarrow y = -\dfrac{5}{12}(x - 12)\) (oe)

dM1: Full method to find the nearest and furthest points of circles, so finds the intersection points of this line with the circles and finds distances between the relevant points.
\(x^2 + \left(-\dfrac{5}{12}x + 5 - 5\right)^2 = 16 \Rightarrow \dfrac{169}{144}x^2 = 16 \Rightarrow x = \pm\dfrac{48}{13} \Rightarrow y = \dfrac{45}{13},\ \dfrac{85}{13}\)
\((x - 12)^2 + \left(-\dfrac{5}{12}x + 5\right)^2 = 49 \Rightarrow \dfrac{169}{144}x^2 - \dfrac{169}{6}x + 120 = 0 \Rightarrow x = \dfrac{72}{13},\ \dfrac{240}{13} \Rightarrow y = \pm\dfrac{35}{13}\)
nearest \(= \sqrt{\left(\dfrac{72}{13} - \dfrac{48}{13}\right)^2 + \left(\dfrac{35}{13} - \dfrac{45}{13}\right)^2} = 2\), furthest \(= \sqrt{\left(\dfrac{240}{13} + \dfrac{48}{13}\right)^2 + \left(-\dfrac{35}{13} - \dfrac{85}{13}\right)^2} = 24\)

A1: Correct answer.

Notes

M1: Finds an expression for \(\dfrac{z}{z^*}\) by rationalising the denominator and collects real parts, eliminating \(\mathrm{i}^2\) terms.
e.g. Accept \(\dfrac{a^2 - b^2 + 2ab\mathrm{i}}{a^2 + b^2}\) or \(\dfrac{a^2 - b^2}{a^2 + b^2} + \dfrac{2ab\mathrm{i}}{a^2 + b^2}\)
but \(\dfrac{a^2 + 2ab\mathrm{i} - b^2}{a^2 + b^2}\) is insufficient unless they then go on to identify the real parts
Allow sign slips but must be a correct use of rationalising the denominator.

A1*: Sets real part equal to zero and then achieves the correct equation, with no errors seen.
\(\dfrac{z}{z^*} = \dfrac{a^2 - b^2 + 2ab\mathrm{i}}{a^2 + b^2} \Rightarrow a^2 = b^2\) is insufficient, as they have not set or identified their real part as zero.

Alternative

M1: Substitutes \(z = a + b\mathrm{i}\) and \(z^* = a - b\mathrm{i}\), sets equal to an imaginary number such as \(k\mathrm{i}\) but not \(\mathrm{i}\). Then expands brackets, eliminating \(\mathrm{i}^2\) terms, achieving an equation equivalent to \(a + b\mathrm{i} = ka\mathrm{i} + kb\)

A1*: Equates real parts and imaginary parts and uses their pair of equations to eliminate \(k\) and then achieve the correct equation, with no errors seen.

If you see a method that involves using exponentials, or modulus argument form, or working backwards, or any other method which may be worthy of credit, please send to review.

Notes

M1: Uses the information \(zz^* = 50\) to form another equation for \(a\) and \(b\), with no imaginary terms present. Then solves simultaneously to find a value for \(a\) or \(b\)
If they obtain \(a^2 - b^2 = 50\) this will be M0

A1: At least two correct complex numbers. Accept e.g. \(\pm(5 + 5\mathrm{i})\)

A1: Deduces all 4 correct complex numbers. Accept e.g. \(\pm 5 \pm 5\mathrm{i}\) or \(\pm(5 \pm 5\mathrm{i})\)

Note: Correct answers with no workings can achieve M1A1A1

Notes

B1: Correct modulus or correct argument may be correct to 3 s.f for this mark

B1: Fully correct must be exact values for this mark, angle must be in radians

Notes

M1: If no coordinates given then it must be in the correct quadrant and either z below the line \(y = -x\) or i\(z\) below the line \(y = x\) or the correct coordinate shown. If incorrect coordinates shown then M0

A1: Both \(z\) and i\(z\) drawn correctly on an Argand diagram. with i\(z\) below the line \(y = x\) and z below the line \(y = -x\) or correct coordinates indicated

If there are no coordinates and the points line close to the lines \(y = x\) and \(y = -x\) send to review

Notes

(c) Note must be a single transformation, there is no follow through

B1: Deduces rotation

B1: Correct angle (radians or degrees), direction and centre

Alternative

Alternatives must have scored M1A1 in (b)

Alternative: must be from a correct (b)

B1: Deduces reflection or matrix or translation

B1: Correct line or matrix or vector

Notes

B1: Correct values for \(\lvert z\rvert\) and \(\lvert w\rvert\). No need for method, may see modulus-argument forms (polar or exponential) used. Alternatively, correctly identifies \(z = -2w\)

M1: Finds \(\dfrac{z}{w}\) and then \(\left\lvert\dfrac{z}{w}\right\rvert\). Look for multiplication of numerator and denominator by the conjugate of the denominator for finding \(\dfrac{z}{w}\). In the look for alternative factoring out the \(-2\) and cancelling \(z\) oe. Allow this mark if modulus-argument forms (polar or exponential) are used to deduce the modulus.

A1: Correctly shows that \(\dfrac{\lvert z\rvert}{\lvert w\rvert} = 2\) and \(\left\lvert\dfrac{z}{w}\right\rvert = 2\) hence state that \(\left\lvert\dfrac{z}{w}\right\rvert = \dfrac{\lvert z\rvert}{\lvert w\rvert}\) cso. If the result is not stated at the end, we require both \(\left\lvert\dfrac{z}{w}\right\rvert\) and \(\dfrac{\lvert z\rvert}{\lvert w\rvert}\) to have explicitly been seen at some stage in their work and a minimal conclusion (e.g. “LHS=RHS”) to be given. Note \(\dfrac{z}{w} = \ldots = -2\) followed by \(\lvert -2\rvert = 2\) without ever seeing \(\left\lvert\dfrac{z}{w}\right\rvert\) will be A0.
If modulus-argument forms (polar or exponential) were used all working must have been correct.

Notes

B1: Correct values for \(\arg(z)\) and \(\arg(w)\). Accept degrees equivalents here and throughout.
Alternatively, correct values for \(\arg(w)\) and \(\arg\left(w^2\right)\).

M1: Finds \(zw\) and then \(\arg(zw)\). Do not allow attempts via modulus-argument forms (polar or exponential) that use the sum of arguments to prove the sum of arguments of the product.
Alternatively, finds both \(\arg(z)\) and \(\arg(zw)\) in terms of \(\arg(w)\) using \(\arg(-p) = \arg(p) \pm \pi\).

A1: Shows that \(\arg(z) + \arg(w) = \dfrac{\pi}{3}\) and following \(\arg(zw) = \dfrac{\pi}{3}\) hence that \(\arg(zw) = \arg(z) + \arg(w)\) cso If the result is not stated at the end, we require both sides of the result to have been seen at some stage in their work and a minimal conclusion (e.g. “LHS=RHS”) to be given.
Alternatively, correctly establishes the result using \(\arg(zw) = \arg\left(w^2\right) \pm \pi = 2\arg(w) \pm \pi\) and \(\arg z + \arg w = \arg w \pm \pi + \arg w = 2\arg w \pm \pi\) with \(\arg w^2 = 2\arg w\) verified.

Notes

M1: For sight of \(\mathrm{e}^{\frac{k\pi}{3}\mathrm{i}}\)
Accept any value for \(k\)

A1: All six roots fully defined as shown or listed separately with their values of \(\theta\) within the given range with no incorrect or extra values. Ensure i and \(\pi\) are present in each term.

Note: Roots if listed are \(\mathrm{e}^0, \mathrm{e}^{\frac{\pi}{3}\mathrm{i}}, \mathrm{e}^{\frac{2\pi}{3}\mathrm{i}}, \mathrm{e}^{\pi\mathrm{i}}, \mathrm{e}^{\frac{4\pi}{3}\mathrm{i}}, \mathrm{e}^{\frac{5\pi}{3}\mathrm{i}}\), condone 1 for \(\mathrm{e}^0\) and/or \(-1\) for \(\mathrm{e}^{\pi\mathrm{i}}\)

Notes

B1: Plots 6 points that form a hexagon, with a point on the positive real axis and a point on the negative real axis, and one point in each quadrant.
Do not be concerned about the position of each point from the centre, however the sketch must convey a hexagon.

dB1: The points form a hexagon, centre the origin (see diagram), axes need not be labelled. Look for the axes acting as lines of symmetry.
(Drawing line/vectors to each point is acceptable but not necessary for either mark)

Examples

B1dB1

B1dB1

B1dB1

B1dB1

B0dB0

B0dB0

Notes

M1: Converts \(\sqrt{3} + \mathrm{i}\) to polar form to obtain \(r\mathrm{e}^{\mathrm{i}\theta}\) with at least \(r = 2\) or \(\theta = \dfrac{\pi}{6}\) and applies the power of 6 correctly to obtain \(r^6\mathrm{e}^{6\theta\mathrm{i}}\)

A1*: Obtains the given answer with sufficient working shown.
As a minimum need to see \(2^6\mathrm{e}^{\frac{6\pi i}{6}} = -64\) or \(2^6\mathrm{e}^{\pi\mathrm{i}} = -64\)
If \(r = -2\) is seen in their workings withhold this mark.

OR

M1: Converts \(\sqrt{3} + \mathrm{i}\) to modulus-argument form \(r(\cos\theta + \mathrm{i}\sin\theta)\) with at least \(r = 2\) or \(\theta = \dfrac{\pi}{6}\) and applies the power of 6 correctly to obtain \(r^6(\cos 6\theta + \mathrm{i}\sin 6\theta)\)

A1*: Obtains the given answer with sufficient working shown.

OR

M1: Attempts to expand \(\left(\sqrt{3} + \mathrm{i}\right)^6\) fully using an attempt at the binomial expansion. Must have 7 terms for \((a + b)^n\) and correct binomial coefficients with \(a = \sqrt{3}\), \(b = \mathrm{i}\) and \(n = 6\)

A1*: Obtains the given answer with at least one intermediate line.

OR

M1: Attempts the full expansion of \(\left(\sqrt{3} + \mathrm{i}\right)^6 = \left(\sqrt{3} + \mathrm{i}\right)\left(\sqrt{3} + \mathrm{i}\right)\left(\sqrt{3} + \mathrm{i}\right)\ldots\left(\sqrt{3} + \mathrm{i}\right) =\)
There must be no brackets, no irrational numbers and no terms in i in their simplified answer.

A1*: Obtains the given answer with sufficient working shown including correct full expansion, with at least one intermediate line.

Notes

B1: Deduces \(r = 2\) (only)

M1: Obtains at least one value of \(z\) in the form \(r\mathrm{e}^{\mathrm{i}\theta}\) with their consistent value of \(r\), and \(\theta\) taking one of \(\left\{\dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \dfrac{7\pi}{6}, \dfrac{3\pi}{2}, \dfrac{11\pi}{6}\right\}\)

A1: For \(2\mathrm{e}^{\frac{\pi}{6}\mathrm{i}}, 2\mathrm{e}^{\frac{\pi}{2}\mathrm{i}}, 2\mathrm{e}^{\frac{5\pi}{6}\mathrm{i}}, 2\mathrm{e}^{\frac{7\pi}{6}\mathrm{i}}, 2\mathrm{e}^{\frac{3\pi}{2}\mathrm{i}}, 2\mathrm{e}^{\frac{11\pi}{6}\mathrm{i}}\) with no incorrect or extra values. Accept unsimplified arguments such as having a solution of \(2\mathrm{e}^{\frac{9\pi}{6}\mathrm{i}}\). Ensure i and \(\pi\) are present in each term.
Accept \(2\mathrm{e}^{\frac{\pi}{2}\mathrm{i}}\) as \(2\mathrm{i}\) and \(2\mathrm{e}^{\frac{3\pi}{2}\mathrm{i}}\) as \(-2\mathrm{i}\)

Notes

(a)(i) B1: Correct centre, condone \((-5, 12\mathrm{i})\)

(a)(ii) B1: Correct radius

Notes

B1ft: A circle drawn with the inside shaded. Follow through their centre and radius.
The centre must be in the correct quadrant and intercept the axes as appropriate.
If they have the correct centre and radius then the centre must be in the second quadrant and the circle must only intercept the imaginary-axis. If diagram is correct consider this a restart B1.

Notes

M1: Calculates the distance from \(O\) to the centre of their circle.

M1: Fully correct strategy for the maximum. E.g. Finds distance from \(O\) to centre of their circle and adds their radius.

A1: Correct answer.
Correct answer with no working and following a correct centre and radius scores M1M1A1

Alternative

M1: Finds the equation of the line from the origin to centre and the Cartesian equation of the circle. Solves simultaneously to find the \(x\) values (or \(y\)) where the line intersects the circle.

M1: Selects the \(x\) coordinate to give the largest distance, find the corresponding \(y\) value and then the distance from the origin.
Selects the \(x\) coordinate to give the smallest distance, find the corresponding \(y\) value and then adds on 2 times the radius.

A1: Correct answer

Notes

M1: Recognises that \(\left\{z : 0 \leqslant \arg(z + 5 - 20\mathrm{i}) \leqslant \pi\right\}\) represents the line \(y = 20\) and uses this with the circle in an attempt to find the angle or half angle at the centre.

A1: Correct value for the angle or half angle at the centre.

M1: Fully correct strategy for the area of the segment using their values.

A1: Awrt 16.4

Note: finding the area of the major segment using \(100\pi - 16.4 = \ldots\) scores M1A1M1A0

Notes

M1: Circle drawn with centre on the real axis
Look for real axis acting as a line of symmetry of the circle.

A1: Circle in the correct position with the imaginary axis as a tangent
Centre need not be labelled for either mark.

M1: Half line starting at the origin, must be in the first quadrant. Do not award if their line continues into the third quadrant

A1: Fully correct diagram that requires

• A circle in the correct position
• A half line intersecting the circle at the origin and in the first quadrant
• The \(x\) coordinate of the intersection in the first quadrant must be to the left of the centre of the circle

Q5 (a) and (b) examples

a) M1A1M1A1 b) B1

a) M1A1M1A1 b) B0

a) M0A0M1A0 b) B1

a) M1A0M1A0 b) B1

Notes

B1ft: Shades the region in their circle above the real axis and below the half line
For this mark to be awarded their line must intersect the circle.

See the examples in the notes for (a).

Notes

Alternative 1 for part (c)

Alternative 2 for part (c)

Alternative 3 for part (c) Polar coordinates

Notes

This is not for finding their shaded area of their diagram in part (a) but is for a correct process for finding the correct required area.

M1: Correct strategy for identifying both coordinates of the point of intersection.
Award for substituting a line of the form \(y = kx\) into the equation of a circle \((x - 4)^2 + y^2 = 16\) then proceeds to find a value for \(x\) from a quadratic, where \(x \neq 0\) and then find their \(y\) coordinate.

A1: Correct coordinates, either written separately, or as \(\left(2, 2\sqrt{3}\right)\) or \(2 + 2\sqrt{3}\,\mathrm{i}\)

dM1: Fully correct strategy for the area, must be consistent with a half line making an angle of \(\dfrac{\pi}{3}\) with the real axis.
Can be found by subtracting the area of a segment from the area of a semicircle or by adding the area of a sector to the area of a triangle.
The candidate may do combinations of semicircles, triangles and sectors so look carefully for the method.

A1: Correct answer in the required form.

Alternative 1
M1: Deduces that since the half line makes an angle of \(\dfrac{\pi}{3}\) with the real axis, the horizontal and vertical distances from the origin to the point of intersection are \(4\cos\dfrac{\pi}{3}\) and \(4\sin\dfrac{\pi}{3}\).

A1: \(4\cos\dfrac{\pi}{3}\) and \(4\sin\dfrac{\pi}{3}\) oe are seen or used in their workings

dM1: Fully correct strategy for the area, must be consistent with a half line making an angle of \(\dfrac{\pi}{3}\) with the real axis.
Can be found by subtracting the area of a segment from the area of a semicircle or by adding the area of a sector to the area of a triangle.
The candidate may do combinations of semicircles, triangles and sectors so look carefully for the method.

A1: Correct answer in the required form.

Alternative 2
M1: Alternatively deduces that the angle implies that there is an equilateral triangle of radius 4

A1: 4 and \(\sin\dfrac{\pi}{3}\) are seen or used in their workings

dM1: Fully correct strategy for the area, must be consistent with a half line making an angle of \(\dfrac{\pi}{3}\) with the real axis.
Can be found by subtracting the area of a segment from the area of a semicircle or by adding the area of a sector to the area of a triangle.
The candidate may do combinations of semicircles, triangles and sectors so look carefully for the method.

A1: Correct answer in the required form.

Alternative 3: using polar coordinates
M1: Achieves a polar equation of the form \(r = k\cos\theta\) and uses \(\left(\dfrac{1}{2}\right)\displaystyle\int r^2\,\mathrm{d}\theta\) to obtain \(k\displaystyle\int \cos^2\theta\,\mathrm{d}\theta\)

A1: Obtains \(\displaystyle\int (16 + 16\cos 2\theta)\,\mathrm{d}\theta\) oe, \(\mathrm{d}\theta\) may be missing

dM1: Integrates to obtain an expression of the form \(a\theta + b\sin 2\theta\), substitutes in limits of 0 and \(\dfrac{\pi}{3}\) and subtracts. If they reach a correct answer with no integration seen then withhold this mark.

A1: Correct answer in the form required

e.g. \((x - 4)^2 + y^2 = 16,\ \Rightarrow r^2 = 8r\cos\theta \Rightarrow r = 8\cos\theta\)
Area \(= \displaystyle\int \frac{1}{2}r^2\,\mathrm{d}\theta = \int \frac{1}{2}.64\cos^2\theta\,\mathrm{d}\theta = \int \frac{32(1 + \cos 2\theta)}{2}\,\mathrm{d}\theta = \int (16 + 16\cos 2\theta)\,\mathrm{d}\theta = \left[16\theta + 8\sin 2\theta\right]\)
\(\left[16\theta + 8\sin 2\theta\right]_0^{\frac{\pi}{3}} = \left[16\left(\dfrac{\pi}{3}\right) + 8\sin 2\left(\dfrac{\pi}{3}\right)\right] - [0] = \dfrac{16}{3}\pi + 4\sqrt{3}\)

Notes

Note: if there are multiple attempts or attempts which start in main scheme before switching to an Alt, then mark the most complete attempt.

B1: Identifies the correct complex conjugate as another root.

M1: Formulates a correct strategy – main scheme. Uses the conjugate pair and a correct method to find a quadratic factor.

A1: Correct quadratic.

M1: Uses the given quartic and their quadratic to establish the other quadratic factor – by inspection or long division etc. resulting in a 3TQ

A1: Deduces the correct second quadratic.

M1: Solves their second quadratic.

A1: Correct second conjugate pair (isw if they try to write in a factorised equation).

Alt 1 (a)

B1: Identifies the correct complex conjugate as another root

M1: Formulates a correct strategy – sum and product approach. Attempts formulae for sum and product of the two know roots and all four roots – allow sign errors, \(\alpha + \beta = 4,\ \alpha\beta = 2^2 + 5^2 = \ldots\) \(\alpha + \beta + \gamma + \delta = \pm 6,\ \alpha\beta\gamma\delta = \pm 145\)

A1: Correct equations \(\alpha + \beta = 4,\ \alpha\beta = 29\) \(\alpha + \beta + \gamma + \delta = 6,\ \alpha\beta\gamma\delta = 145\) (seen or implied).

M1: Uses the sum and product of known roots to reduce to equations in just the unknown roots and attempts to solve simultaneously. Note allow if they assume complex roots and use \(\lambda \pm \mu\mathrm{i}\) for the two unknown roots.

A1: Deduces the correct quadratic for remaining roots.

M1: Solves their quadratic

A1: Correct second conjugate pair

Alt 2 (a)

B1: Identifies the correct complex conjugate as another root- seen anywhere

M1: Formulates a correct strategy – factor theorem approach. Applies the factor theorem with either complex root and equates real and imaginary terms to form simultaneous equations in \(a\) and \(b\)

A1: Correct equations need not be simplified.

M1: Solves the equations to find values for \(a\) and \(b\). Do not be concerned with the algebra.

A1: Correct values

M1: Solves the resulting quartic – may be by calculator. Note that if roots are stated by calculator they must correspond to their found \(a\) and \(b\), but allow for rounding to nearest integer.

A1: Correct second conjugate pair from fully correct work.

Notes

B1ft: A pair of complex roots plotted correctly, either both \(2 \pm 5\mathrm{i}\) or follow through their second pair of complex roots. Just the points are needed. Allow if there is no labelling for this mark, as long as there is a pair of roots symmetric in the real axis.

B1: Fully correct and labelled sketch with both sets of roots shown in approximately the correct locations. Just the points are needed. Labelling may be by coordinates or labels on axes, and accept tick marks as unit increments for labels. The \((1,\ \pm 2)\) roots must be within the sector spanned by \((2,\ \pm 5)\) (as in the diagram).

Notes

B1: One correct line described

B1: Two correct lines described

Special case: If candidate states that "any line is possible" score B1B1 (as they may be considering a cubic with complex coefficients).

Notes

B1: Interprets the conclusion from (a) in context by identifying the correct two roots.

Notes

B1: Deduces the real root is the one in common.

M1: Sets product of roots = – 12 using their \(\dfrac{3}{2} \times -4 \times \alpha = -12\). Alternatively forms an equation for \(\mathrm{g}(z)\) using the roots

M1: Solves their equation to find the third root, condone use of 12 for this mark. Alternatively multiply their constant and sets = 12 to find the third root. Condone a sign slip for this mark

A1: Correct third root

Notes

M1: Uses their roots of \(\mathrm{f}(z)\) to form a cubic expression for \(\mathrm{f}(z)\), and expands to at least a linear term times a quadratic with real coefficients (which may be seen later). May just expand the complex brackets.
They must have the factor 8 for this mark
Alternative uses \(b = -8(\text{sum of their roots})\) \(c = 8(\text{pair sum of their roots})\) and \(d = -8(\text{product of their roots})\)
Note: \(\mathrm{f}(z) = 8z^3 - 36z^2 + 72z - 54\)

M1: Sets their expressions equal and factorises out or cancels the common term to achieve a quadratic expression in \(z\). Allow if \(\mathrm{f}(z)\) is not yet expanded or factor of 8 is missing.
Alternatively finds the expression for \(\mathrm{g}(z)\) by multiplying out bracket or uses P = - sum roots and Q = pair sum. Sets their \(\mathrm{f}(z)\) = their \(\mathrm{g}(z)\) both must be cubic

M1: Expands, gathers terms and solves the resulting quadratic. Allow this mark if the \(z = \dfrac{3}{2}\) solution is not given.
Alternatively simplifies for form a cubic = 0 and solves using calculator to find complex roots

A1: All three correct solutions given. Note decimals \(\dfrac{13}{7} \pm \mathrm{i}1.68\ldots\) is A0

Special case If the candidate just forgets the factor of 8 for \(\mathrm{f}(z)\) this scores M0M1M0A0

Notes

M1: Draws a single straight line through both axes with a negative gradient. Ignore any line joining (3, 0) and (0, −6)

A1: Draws a single straight line through both axes with a negative gradient which has a negative \(y\) intercept. Ignore any intercept marked on the axes. Ignore any line joining (3, 0) and (0, −6)

B1: Shades the area above their straight line (not a bounded region such as a triangle bounded by the axes and the line)

Notes

M1: Finds the Cartesian equations for both loci by using the gradient as tan(argument) and correct coordinate. Must be an attempt at both equations but one correct equation scores this mark

A1: One equation correct, need not be simplified
A1: Both equations correct, need not be simplified
M1: Solve simultaneously to find either the real or imaginary component.
M1: Finds the other component to complete the process of finding \(w\).
A1: Correct exact answer

Note: If leaves the answer as a coordinate this is A0. If defines \(w = a + b\mathrm{i}\) and then states \(a = \dfrac{7}{2}\) and \(b = \dfrac{3\sqrt{3}}{2}\) this is A1

Note: If candidates use decimal instead of exact values throughout allow the method marks \(y = 1.73x - 3.46\) and \(y = 0.58x + 0.58\)

Alternative

Scheme	Marks	AO
\(\tan\left(\dfrac{\pi}{6}\right) = \dfrac{\sqrt{3}}{3} = \dfrac{y}{x_{-1}}\) and \(\tan\left(\dfrac{\pi}{3}\right) = \sqrt{3} = \dfrac{y}{x_2}\)	M1	1.1b
	A1 A1	1.1b 1.1b
\(y\sqrt{3} = y\dfrac{\sqrt{3}}{3} + 3 \Rightarrow y = \ldots\)	M1	3.1a
Uses \(x = y\sqrt{3} - 1\) or \(x = \dfrac{\sqrt{3}}{3}y + 2\) with their value of \(y\) leading to a value for \(x\)	M1	1.1b
\((w =)\ \dfrac{7}{2} + \dfrac{3\sqrt{3}}{2}\mathrm{i}\)	A1	2.1
	(6)

M1: Use both arguments to form equations involving \(x\) and \(y\)
A1: (One correct triangle) value for \(x\) in terms of \(y\)
A1: (Two correct triangles), values for \(x\) in terms of \(y\)

M1: Forms and solves an equation \(y\sqrt{3} = y\dfrac{\sqrt{3}}{3} + 3 \Rightarrow y = \ldots\) must be come from \(x_2 = x_{-1} + 3\)

M1: Uses their \(y\) value and \(x = y\sqrt{3} - 1\) or \(x = \dfrac{\sqrt{3}}{3}y + 2\) to find a value for \(x\)
A1: Correct exact answer

Q7(ii) Two alternatives seen

Alternative 3

M1: Uses correct geometry to form equations involving \(a\) and \(c\)
A1: One correct equation
A1: Two correct equations
M1: Finds the imaginary component
M1: Uses 2 + their \(c\) to find the real component
A1: Correct exact answer

Alternative 4

(Corrected from the printed mark scheme: \(AC = \dfrac{7}{2}\). \(AC = 3\sqrt{3}\sin 60 = \dfrac{9}{2}\); the real part of \(w\) is then \(\dfrac{9}{2} - 1 = \dfrac{7}{2}\).)

M1: Uses the sine rule to find the length \(AB\)
A1: Correct length \(AB\)
M1: Uses trigonometry to find either the real or imaginary component
A1: Correct real or imaginary component
M1: Uses trigonometry to find the other component
A1: Correct exact answer

Notes

M1: Attempts the midpoint of \(z_1\) and \(z_2\)

A1*: Correct point.

Notes

M1: Substitutes into the equation with \(z_1\) and \(\alpha\) and \(\beta\), simplifies and expands and applies \(\mathrm{i}^2 = -1\), this may be implied by their working.

A1*: Completes the proof to find the correct answer with no errors seen, all necessary brackets as required

Notes

Alternative 1

Alternative 2

Notes

Mark (c) as one

(c) (i)
B1: Correct roots, accept all 6 listed or given in general form as in the scheme. Need not show the 1.

(c)(ii)
M1: Realises the need to set the roots of unity equal to \(\beta(z - \alpha)\) and solve for \(z\). Must be attempted at least once with any of their roots.
M1: Finds the Cartesian form for their equation for at least one of the roots other than \(z_1\) and \(z_2\)
A1: At least two correct other roots than \(z_1\) and \(z_2\) in Cartesian form.
A1: Deduces all four correct in Cartesian form and no extra solutions

Alternative 1
M1: Finds the modulus and argument of \((z - \alpha)^6\)
M1: Finds the Cartesian form for one of their modulus and arguments
A1A1: same as above

Alternative 2
M1: Finds the rotation matrix and subtracts the centre from \(z_1\) or \(z_2\). Or finds the exponential from for \(z_1 - \alpha\) or \(z_2 - \alpha\)
M1: Finds the Cartesian form by multiplying by the rotation matrix and adding the centre. Or multiplies by \(\mathrm{e}^{\frac{\pi}{3}\mathrm{i}}\) write in Cartesian form and adds on the centre
A1A1: same as above

Note all four correct decimal answers or written as coordinates score A1A0
46.7 + 18.7i   – 40.7 – 4.7i   14.7 + 50.7i   – 8.7 – 36.7i
Note: Correct answers implies the method marks

Notes

M1: Selects the process \(\dfrac{2 + 3\mathrm{i}}{5 + \mathrm{i}} \times \dfrac{5 - \mathrm{i}}{5 - \mathrm{i}}\)

dM1: Evidence of multiplying out brackets

A1: Achieves \(\dfrac{1}{2}(1 + \mathrm{i})\) or \(\dfrac{13}{26}(1 + \mathrm{i})\) with no errors cso, isw.

Note: Correct answer from no working score no marks

Note: Going from \(\dfrac{13 + 13\mathrm{i}}{26}\) and then stating \(k = \dfrac{1}{2}\) is A0, they have not shown the form asked for

Alternative
M1: Multiplies across by \((5 + \mathrm{i})\) and expands the brackets
dM1: Collects terms
A1: Achieves \(2 + 3\mathrm{i} = k(4 + 6\mathrm{i})\) and draws the conclusion that therefore \(\dfrac{2 + 3\mathrm{i}}{5 + \mathrm{i}} = k(1 + \mathrm{i})\) where \(k = \dfrac{1}{2}\)

Notes

B1: Deduces \(n = 4\) only

Notes

Note: Send to review any attempts where they are finding additional solutions such as arguments of \(\boldsymbol{z}\) is \(\dfrac{(6k - 5)\pi}{30}\) For example correctly uses \(\arg(z) = \dfrac{\pi}{30}\)

B1 (M1 on ePen): \(|z| = 3\) can be implied by \(a^2 + b^2 = 9\) isw

M1: Uses \(\arg(z_1z_2) = \arg(z_1) + \arg(z_2)\) to find \(\arg(z) = -\dfrac{5\pi}{3} \div 10\) or \(\arg(z) = \dfrac{\pi}{3} \div 10\)

M1: Uses \(z = \text{their } |z|\left(\cos(\text{their arg}) + \mathrm{i}\sin(\text{their arg})\right)\) to find the complex number \(z\) or values for \(a\) or \(b\). As long as the modulus has changed.

A1: Correct complex number or values for \(a\) and \(b\).

Alternative

(Corrected from the printed mark scheme: the printed scheme has \(\dfrac{b}{a} = \arctan\left(-\dfrac{\pi}{6}\right)\) and \(\dfrac{b}{a} = \arctan\dfrac{\pi}{30} \Rightarrow b = 0.104\ldots a\); it should be \(\tan\), giving \(b = 0.105\ldots a\).)

B1: \(a^2 + b^2 = 9\) isw

M1: Uses \(\arg(z_1z_2) = \arg(z_1) + \arg(z_2)\) to find \(\arg(z) = -\dfrac{5\pi}{3} \div 10\)

M1: Uses the argument of \(z\) find an equation in \(a\) and \(b\). Then solve simultaneously to find a value for \(a\) or \(b\).
As long as \(\boldsymbol{\sqrt{a^2 + b^2} \neq 59049}\)

A1: Correct complex number or values for \(a\) and \(b\).

Note there are other correct answers

Notes

(a) Correct answer with no working scores both marks in (a)

M1: Any correct expression for \(|z_1|\) or \(\arg z_1\) e.g. \(|z_1| = \sqrt{(-4)^2 + 4^2}\) or \(\arg z_1 = \pi - \dfrac{\pi}{4}\)

A1: Correct expression. The "\(z_1 =\)" is not required.
This mark is not for correct modulus and correct argument it is for the complex number written in the required form. Condone the missing closing bracket e.g. \((z_1 =)\sqrt{32}\left(\cos\dfrac{3\pi}{4} + \mathrm{i}\sin\dfrac{3\pi}{4}\right.\)

Notes

(b)(i) Correct answer with no working scores no marks in (b)(i)

M1: Employs a correct method to find the quotient. E.g.

• uses modulus argument form and divides moduli and subtracts arguments the right way round
• uses exponential form and divides moduli and subtracts arguments the right way round
• converts \(z_2\) to Cartesian form and multiplies numerator and denominator by the complex conjugate of the denominator. Allow if the “3” is missing for this method. Allow with decimals for this method e.g. \(\dfrac{z_1}{z_2} = \dfrac{-4 + 4\mathrm{i}}{-0.258\ldots - 0.965\ldots\mathrm{i}} \times \dfrac{-0.258\ldots + 0.965\ldots\mathrm{i}}{-0.258\ldots + 0.965\ldots\mathrm{i}} = \ldots\)

If they convert \(z_2\) to Cartesian form it must be correct as shown or correct decimals.

A1: Correct exact answer in the required form.
Do not allow e.g. \(-\dfrac{2}{3}\left(\sqrt{2} + \sqrt{6}\mathrm{i}\right)\) or \(\dfrac{-2\sqrt{2} - 2\sqrt{6}\mathrm{i}}{3}\) unless a correct form is seen previously then apply isw.

Provided a correct method is shown as above, allow to go from the forms in the main scheme to the correct exact answer with no intermediate step.

(b)(ii) Correct answer with no working scores no marks in (b)(ii)

M1: Applies De Moivre’s theorem correctly to \(z_2\). E.g. uses polar form or exponential form and calculates the modulus as \(3^4\) and the argument as \(4 \times \dfrac{17\pi}{12}\)
For attempts at \(z_2^4 = \left\{3\left(\left(\dfrac{\sqrt{2} - \sqrt{6}}{4}\right) - \mathrm{i}\left(\dfrac{\sqrt{2} + \sqrt{6}}{4}\right)\right)\right\}^4\) you would need to see:

• the correct exact form used
• a clear and convincing attempt to expand the brackets e.g. by using a full binomial expansion or a complete attempt to multiply all 4 brackets together but you are not expected to check every detail
• a final answer in the required form with no obvious errors seen

So \(z_2^4 = \left\{3\left(\left(\dfrac{\sqrt{2} - \sqrt{6}}{4}\right) - \mathrm{i}\left(\dfrac{\sqrt{2} + \sqrt{6}}{4}\right)\right)\right\}^4 = \dfrac{81}{2} - \dfrac{81\sqrt{3}}{2}\mathrm{i}\) scores no marks.
Similar guidance applies if they attempt to expand \(\left\{3\left(\cos\dfrac{17\pi}{12} + \mathrm{i}\sin\dfrac{17\pi}{12}\right)\right\}^4\)

A1: Correct exact answer in the required form.
Do not allow e.g. \(\dfrac{81}{2}\left(1 - \dfrac{81\sqrt{3}}{2}i\right)\) or \(\dfrac{81 - 81\sqrt{3}\mathrm{i}}{2}\) unless a correct form is seen previously then apply isw.

Provided a correct method is shown as above, allow to go from the forms in the main scheme to the correct exact answer with no intermediate step.

Notes

M1: Uses the given root and its complex conjugate to form a quadratic equation. Uses the quadratic equation to write \(\mathrm{f}(z)\) in the form \(\left(z^2 + pz + q\right)(z + r)\) where \(p\), \(q\) and \(r\) are real values

A1: Correct expression for \(\mathrm{f}(z) = \left(z^2 + 6z + 25\right)(z + 7)\)

M1: Multiplies out and simplifies to find the \(z^2\) or \(z\) term.
A1: Correct values for \(a\) and \(b\) or cubic

Alternative 1

M1: Uses the complex conjugate and product of roots = −175 to find the third root.
A1: Correct third root
M1: A complete method to find the values of \(a\) or \(b\). Either uses the sum and pairs sum or multiplies out three brackets \((z - (-3 + 4\mathrm{i}))(z - (-3 - 4\mathrm{i}))(z - \text{their third root})\) to find the \(z^2\) or \(z\) term.
A1: Correct values for \(a\) and \(b\) or cubic

Alternative 2

M1: Substitutes \(-3 + 4\mathrm{i}\) or \(-3 - 4\mathrm{i}\) into \(\mathrm{f}(z)\), sets the real and imaginary parts = 0 to form two simultaneous equations in \(a\) and \(b\).
A1: Correct, unsimplified equations.
M1: Solves simultaneous equations to find values for \(a\) or \(b\) following an attempt at \(\mathrm{f}(-3 + 4\mathrm{i}) = 0\) or \(\mathrm{f}(-3 - 4\mathrm{i}) = 0\). Allow this mark for seeing a value for \(a\) or \(b\) following simultaneous equation, you do not need to check.
A1: Correct values for \(a\) and \(b\).

Notes

B1: Correctly plotting \(-3 + 4\mathrm{i},\ -3 - 4\mathrm{i}\)

B1: Correctly plotting −7

Notes

B1ft: \(-5 + 4\mathrm{i},\ -5 - 4\mathrm{i}\) and subtracts 2 from their real root shown on their Argand diagram

Notes

M1: States or implies \(z^* = a - b\mathrm{i}\) and finds an expression for \(zz^*\)

A1: Achieves \(zz^* = a^2 + b^2\) and draws the conclusion that \(zz^*\) is a real number. Accept \(\in \mathbb{R}\) as conclusion, but not just “no imaginary part”.

Notes

M1: Starts the process of solving by using the conjugate to form an equation with real denominators, and without \(z^*\) or \(\mathrm{i}^2\) in the equation. Accept as shown in scheme, or may multiply through by \(a - b\mathrm{i}\) and expand and gather terms. May be implied by correct extraction of equation(s).

M1: Uses the given information to form two equations involving \(a\) and \(b\) at least one of which includes both. It must involve equating real or imaginary parts of \(\dfrac{z}{z^*} = \dfrac{7}{9} + \dfrac{4\sqrt{2}\mathrm{i}}{9}\)

A1: Any two correct equations arising from use of both given facts. (Note: if multiplying through by \(a - b\mathrm{i}\) then equating real and imaginary terms gives the same equation.)

dM1: Dependent on previous method mark, solves the equations to find a value for either \(a\) or \(b\).

A1: Deduces the correct complex numbers and no extras. Do not accept \(\pm 4 \pm \sqrt{2}\mathrm{i}\)

Note: it is possible to solve via polar coordinates, but unlikely to succeed. If you see responses you think are worthy of credit but are unsure how to mark, use review. Example solutions shown below.

Alt (b)

Notes

M1: A complete method to find both \(z_1\) and \(z_2\) in the form \(a + b\mathrm{i}\) and adds them together.

dM1: Dependent on previous method mark, finds the modulus and argument of \(z_1 + z_2\). They must show their method, just stating modulus = 12 and argument \(= \dfrac{2\pi}{3}\) is not sufficient as this is a show question.

Alternative 1: Factorises out 12 and find the argument

Alternative 2: uses \(12\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}} = 12\left(\cos\dfrac{2\pi}{3} + \mathrm{i}\sin\dfrac{2\pi}{3}\right) = \ldots\)

A1*: Achieves the correct answer following no errors or omissions.
Alternatively shows that \(12\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}} = -6 + 6\sqrt{3}\mathrm{i}\) and concludes therefore \(z_1 + z_2 = 12\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}}\) *

Alternative 1 and Alternative 2 (for the dM1 and A1*)

Alternative 3

M1: Factorises out 12 and writes in the form \(12\left[\ldots\cos\left(\dfrac{\pi}{3}\right) + \ldots\mathrm{i}\sin\left(\dfrac{\pi}{3}\right) + \ldots\cos\left(\dfrac{5\pi}{6}\right) + \ldots\mathrm{i}\sin\left(\dfrac{5\pi}{6}\right)\right]\)

dM1: Dependent on previous mark. Writes in the form \(12(a + b\mathrm{i})\) leading to the form \(12(\cos\theta + \mathrm{i}\sin\theta)\)

A1*: Achieves the correct answer following no errors or omissions.

Alternative 4

M1: Factorises out 6 and writes in the form \(6\mathrm{e}^{\frac{\pi}{3}\mathrm{i}}\left(1 + \sqrt{3}\,\mathrm{e}^{\frac{\pi}{2}\mathrm{i}}\right) = 6\mathrm{e}^{\frac{\pi}{3}\mathrm{i}}(1 + a\mathrm{i})\)

dM1: Dependent on previous method mark, finds the modulus and argument of \((1 + a\mathrm{i})\) or \(12(a + b\mathrm{i})\) leading to the form \(12(\cos\theta + \mathrm{i}\sin\theta)\)

A1*: Achieves the correct answer following no errors or omissions.

Alternative 5

Scheme	Marks	AO
Uses geometry to show that \(z_1\), \(z_2\) and \(z_1 + z_2\) form a right-angled triangle	M1	3.1a
\(\arg(z_1 + z_2) = \dfrac{\pi}{3} + \tan^{-1}\left(\dfrac{6\sqrt{3}}{6}\right) = \ldots\left\{\dfrac{2\pi}{3}\right\}\) \(|z_1 + z_2| = \sqrt{(6)^2 + \left(6\sqrt{3}\right)^2} = \ldots\{12\}\)	dM1	1.1b
\(z_1 + z_2 = 12\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}}\) *	A1*	1.1b
	(3)

(In the printed diagram the axis labels Re and Im are the wrong way round: the real axis is the horizontal one.)

M1: Draws a diagram to show that \(z_1\), \(z_2\) and \(z_1 + z_2\) form a right-angled triangle.

dM1: Dependent on previous method mark, finds the modulus and argument of \(z_1 + z_2\)

A1*: Achieves the correct answer following no errors or omissions.

Note: Writing \(\arg(z_1 + z_2) = \arctan\left(\dfrac{6\sqrt{3}}{-6}\right) = -\dfrac{\pi}{3}\) therefore \(\arg(z_1 + z_2) = \pi - \dfrac{\pi}{3} = \dfrac{2\pi}{3}\) with no diagram or finding \(z_1 + z_2\) is M0dM0A0

Notes

(In the printed diagram the axis labels Re and Im are the wrong way round: the real axis is the horizontal one.)

M1: Draws a diagram and recognises that the shortest distance will form a right-angled triangle.

M1: Uses trigonometry to find the shortest length.

A1: Correct exact value.

Alternative 1

M1: Finds the equation of the half-line by attempting \(m = -\tan\left(\dfrac{\pi}{3}\right)\) \(c = 5\tan\left(\dfrac{\pi}{3}\right)\). Finds \(x^2 + y^2\) in terms of \(x\), differentiates, sets \(= 0\) and finds the value of \(x\).

M1: Uses their value of \(x\) to find the minimum value of \(\sqrt{x^2 + y^2}\)

A1: Correct exact value.

Alternative 2

M1: Finds the equation of the half-line by attempting \(m = -\tan\left(\dfrac{\pi}{3}\right)\) \(c = 5\tan\left(\dfrac{\pi}{3}\right)\). Finds the equation of the line perpendicular which passes through the origin. Finds the point of intersection of the lines

M1: Finds the distance from the origin to their point of intersection

A1: Correct exact value.

Notes

B1: Correct modulus

M1: Attempts the argument. Allow for \(\arctan\left(\dfrac{\pm 4}{\pm 4\sqrt{3}}\right)\) or equivalents using the modulus (may be in wrong quadrant for this mark).

A1: Correct argument \(-\dfrac{\pi}{6}\) (must be in fourth quadrant but accept \(\dfrac{11\pi}{6}\) or other difference of \(2\pi\) for this mark).

A1: Correct expression found for \(w\), in the correct form, must have positive \(r = 8\) and \(\theta = -\dfrac{\pi}{6}\).

Note: using degrees B1 M1 A0 A0

Notes

(b)(i)&(ii)
B1: \(w\) plotted in correct quadrant with either the correct coordinate clearly seen or above the line \(y = -x\)

M1: Half line drawn starting on the imaginary axis away from \(O\) with positive gradient (need not be labelled)

A1: Sketch on one diagram – both previous marks must have been scored and the half line should pass between \(O\) and \(w\) starting from a point on the imaginary axis below \(w\). (You may assume it starts at \(-10\mathrm{i}\) unless otherwise stated by the candidate)

Note: If candidates draw the loci on separate diagrams the maximum they can score is B1 M1 A0

Notes

M1: Formulates a correct strategy to find the shortest distance, e.g. uses right angle \(OXA\) where \(X\) is where the lines meet and proceeds at least as far as \(OX\).

M1: Full method to achieve the shortest distance, e.g. for \(WX = OW - OX\).

A1: cao shortest distance is 3

Alternative 1:
M1: Uses a correct method to find the equation of the line from \(O\) to \(w\), \(y = -\dfrac{1}{\sqrt{3}}x\) and the equation of the half line \(y = \sqrt{3}x - 10\), solves to find the point of intersection \(X\left(\dfrac{5\sqrt{3}}{2}, -\dfrac{5}{2}\right)\)
If the incorrect gradient(s) is used with no valid method seen this is M0

M1: Finds the length \(WX = \sqrt{\left(\text{their }\dfrac{5\sqrt{3}}{2} - 4\sqrt{3}\right)^2 + \left(\text{their } -\dfrac{5}{2} - -4\right)^2} = \ldots\) condone a sign slip in the brackets.

A1: cao shortest distance is 3

Alternative 2:
M1: Uses a correct method to find the length \(AW\) and a correct method to find the angle between the horizontal and the line \(AW\)

M1: Finds the length of \(WX = \text{their }\sqrt{84} \times \sin\left(\dfrac{\pi}{3} - \text{their } 0.7137\right) = \ldots\)

A1: cao shortest distance is 3

Alternative 3
M1: Finds the vector equation of the half line, then \(XW\).
Then either: Sets dot product \(XW\) and the line = 0 and solves for \(\lambda\). Substitutes their \(\lambda\) into the equation of the half line to find the point of intersection.
Or finds the length of \(XW\) and differentiates, set = 0 and solve for \(\lambda\)

M1: Finds the length \(WX = \sqrt{\left(\text{their }\dfrac{5\sqrt{3}}{2} - 4\sqrt{3}\right)^2 + \left(\text{their } -\dfrac{5}{2} - -4\right)^2} = \ldots\) condone a sign slip in the brackets.
Or substitutes their value for \(\lambda\) into the length of (d)

A1: cao shortest distance is 3

Notes

B1: \(2 + 3\mathrm{i}\)

Notes

(i)

M1: A complete method to find the third root. E.g. forms the quadratic factor and uses this to find the linear factor leading to roots. Alternatively uses sum of roots \(= 0\) or product of roots \(= \pm 52\) (condone sign error) with their complex roots to find the third. Note they may have used the factor theorem to find \(a\) first, which is fine. If they have found \(a\) first, then the correct third root seen implies this mark. The method may be implied by the third root seen on the diagram.

A1: Correct roots, all three must be clearly stated somewhere in (b), not just seen on a diagram in part (c).

(ii)

M1: Complete method to find a value for \(a\) e.g. multiplies out their quadratic and linear factors to find the coefficient of \(z\), or uses pair sum, or uses factor theorem with one of the roots (may be done before finding the third root) but must reach a value for \(a\).

A1: Deduces the correct value of \(a\). May be seen as the \(z\) coefficient in the cubic (need not be extracted, but if it is it must be correct).

Notes

B1ft: Correctly plots all three roots following through their third root in part (b). Must be labelled with the \(\text{``}{-4}\text{''}\) further from \(O\) than 2, but don’t be concerned about \(x\) and \(y\) scale. If correct look for one root on the negative real axis, with the other two symmetric about real axis in quadrants 1 and 4, but follow through their real root if positive. Accept \((0, -4)\) labelled on the real axis in correct place as a label.

Notes

(i) B1: See scheme, Condone \(-45\)
Any incorrect arguments seen is B0.
\(\arg(z_1) = \tan^{-1}\left(\dfrac{3}{-3}\right)\) is B0
Note: They used 3 instead of \(-3\) is B0, there are two 3’s in line 1 do they mean both should \(-3\)
It should be negative is B0

(ii) B1: See scheme

Notes

B1ft: States a correct value for \(\arg\left(\dfrac{z_1}{z_2}\right)\) Follow through on their answer to part (a) (i), do not ISW

Notes

(i)
M1: Finds the modulus and argument of \(z\)
A1: Correct modulus and argument of \(z\)
M1: Uses a correct method to find all the other 4 vertices of the pentagon. Must be doing the equivalent of adding/ subtracting multiples of \(\dfrac{2\pi}{5}\) to the argument.
A1ft: All 4 vertices following through on their modulus and argument. Does not need to be simplified for this mark.
A1: All 4 vertices correct in the required form

(corrected from the printed mark scheme: in the second A1ft list the last vertex \(r\mathrm{e}^{\left(\theta - \frac{4\pi}{5}\right)\mathrm{i}}\) is printed as \(r\mathrm{e}^{\left(\theta - \frac{8\pi}{5}\right)\mathrm{i}}\), which is the same point as \(r\mathrm{e}^{\left(\theta + \frac{2\pi}{5}\right)\mathrm{i}}\))

Notes

(ii)(a)
B1: Circle centre \((0, 2)\) and radius 2 or the two half-lines (a wedge) with the vertex on the origin.
B1: Fully correct region shaded.

Notes

(ii)(b)
M1: Writes the required area using polar coordinates
M1: Uses \(\sin^2\theta = \dfrac{1}{2} - \dfrac{1}{2}\cos 2\theta\) and integrates to the form \(A\theta + B\sin 2\theta\)
M1: Uses the limits of \(\dfrac{\pi}{4}\) and \(\dfrac{\pi}{3}\) and subtracts the correct way around. Must be some attempt at \(\text{area} = \dfrac{1}{2}\displaystyle\int\left(\alpha\sin\theta\right)^2\mathrm{d}\theta\) and integration.
A1: Correct exact area \(= \dfrac{\pi}{3} - \sqrt{3} + 2\)

Alternative
M1: Finds either area 1 or area 2
M1: A complete method to find the area 3
M1: A complete method to find the required area \(=\) Area of semi circle \(-\) area 1 \(-\) area 2 \(-\) area 3 or \(=\) Area of sector \(-\) area 1 \(-\) area 3
A1: Correct exact area \(= \dfrac{\pi}{3} - \sqrt{3} + 2\)

Notes

(a)(i)
B1: Deduces \(|z_1 z_2| = 3\sqrt{2}\)
(ii)
B1: Deduces \(\arg(z_1 z_2) = \dfrac{\pi}{4}\) o.e.
These marks may be awarded for \(z_1 z_2 = 3\sqrt{2}\left(\cos\dfrac{\pi}{4} + \mathrm{i}\sin\dfrac{\pi}{4}\right)\)

Notes

(b)(i)
B1ft: \(2\pi\) divided by their \(\arg(z_1 z_2)\) found in part (a)(ii) to give an integer. Alternatively smallest positive integer multiple required to make their argument a multiple of \(2\pi\)
(ii)
M1: Their answer to (a)(i) to the power of their \(n\)
A1: 104 976

Notes

(Corrected from the printed mark scheme: the second method prints \(x^2 = 9 - 2r^2 \Rightarrow 9 - 2r^2 \gt 0\); from \(2x^2 + 18 - 4r^2 = 0\), \(x^2 = 2r^2 - 9\), so the condition is \(2r^2 - 9 \gt 0\).)

B1: Correct equations for each loci of points

M1: A complete method to find a 3TQ involving one variable using equations of the form \((x \pm 3)^2 + (y \pm 5)^2 = (2r)^2\) or \(2r^2\) or \(r^2\) and \(y = \pm x \pm 2\)

A1: Correct quadratic equation

dM1: Dependent on previous method mark. A complete method uses \(b^2 - 4ac \gt 0\) or rearranges to find \(x^2 = \mathrm{f}(r)\) and uses \(\mathrm{f}(r) \gt 0\) to the minimum value of \(r\).

M1: Realises there will be an upper limit for \(r\) and uses Pythagoras theorem
\((2r)^2 = (y\text{ coord of centre})^2 + (x\text{ coord of centre} - 2)^2\)
condone \((r)^2 = (y\text{ coord of centre})^2 + (x\text{ coord of centre} - 2)^2\)

A1: One correct limit, either \(\dfrac{3\sqrt{2}}{2} \lt r\) or \(r \lt \dfrac{\sqrt{26}}{2}\) o.e.

A1: Fully correct inequality

Alternative

B1: Using a circle with centre (3, 5) and radius \(2r\) and \(y = -x + 2\)

M1: A complete method to find the point of intersection of the line \(y = \pm x \pm 2\) and circle where the line is a tangent to the circle.

A1: Correct point of intersection

dM1: Finds the distance between the point of intersection and the centre and uses this to find the minimum value of \(r\). Condone radius of r.

M1: Realises there will be an upper limit for \(r\) and uses Pythagoras theorem
\((2r)^2 = (y\text{ coord of centre})^2 + (x\text{ coord of centre} - 2)^2\)

A1: One correct limit, either \(\dfrac{3\sqrt{2}}{2} \lt r\) or \(r \lt \dfrac{\sqrt{26}}{2}\) o.e.

A1: Fully correct inequality

Notes

B1: Seen \(2 - 3\mathrm{i}\)

M1: Finds the third and fourth roots of the form \(p \pm 3\mathrm{i}\). May be seen in an Argand diagram

A1: Third and fourth roots are \(-4 \pm 3\mathrm{i}\). May be seen in an Argand diagram

dM1: Uses an appropriate method to find \(\mathrm{f}(z)\). If using roots of a polynomial at least 3 coefficients must be attempted.

A1: At least two of \(a\), \(b\), \(c\), \(d\) correct

A1: All \(a\), \(b\), \(c\) and \(d\) correct

Note: Using roots \(2 \pm 3\mathrm{i}\) and \(-2 \pm 3\mathrm{i}\) leads to \(z^4 + 10z^2 + 169\) Maximum score B1 M1 A0 M1 A0 A0

Notes

B1: Correct exact value for \(|z_1| = \sqrt{13}\) and \(\arg z_1 = \tan^{-1}\left(\dfrac{3}{2}\right)\). The value for \(\arg z_1\) can be implied by sight of awrt 0.98 or awrt 56.3°

B1ft: Follow through on \(r = |z_1|\) and \(\theta = \arg z_1\) and writes \(z_1 = r(\cos\theta + \mathrm{i}\sin\theta)\) where \(r\) is exact and \(\theta\) is correct to 4 s.f. do not follow through on rounding errors.

Notes

(Corrected from the printed mark scheme: the printed scheme has \(\tan^{-1}(-0.1974) = \dfrac{b}{a}\); it should be \(\tan(-0.1974) = \dfrac{b}{a}\), which gives \(\dfrac{b}{a} = -0.2\).)

M1: A complete method to find the modulus of \(z_2\)

A1: \(|z_2| = 3\sqrt{26}\)

M1: A complete method to find the argument of \(z_2\)

A1: \(\arg(z_2) = \dfrac{\pi}{4} - \tan^{-1}\left(\dfrac{3}{2}\right)\) or \(\dfrac{\pi}{4} - 0.9828\) or \(-0.1974\ldots\)

ddM1: Writes \(z_2\) in the form \(r(\cos\theta + \mathrm{i}\sin\theta)\), dependent on both previous M marks.
Alternative forms two equations involving \(a\) and \(b\) using the modulus and argument of \(z_2\) and solve to find values for \(a\) and \(b\)

A1: Deduces that \(z_2 = 15 - 3\mathrm{i}\) only

Alternative

(b) Alternative: \(z_1z_2 = (a + b\mathrm{i})(2 + 3\mathrm{i}) = (2a - 3b) + (3a + 2b)\mathrm{i}\)

M1: A complete method to find an equation involving \(a\) and \(b\) using the modulus

A1: Correct simplified equation \(a^2 + b^2 = 234\) o.e.

M1: A complete method to find an equation involving \(a\) and \(b\) using the argument.
Note \(\tan^{-1}\left(\dfrac{2a - 3b}{3a + 2b}\right) = \dfrac{\pi}{4}\) this would score M0 A0 ddM0 A0

A1: Correct simplified equation \(a = -5b\) o.e.

ddM1: Dependent on both the previous method marks. Solves their equations to find values for \(a\) and \(b\)

A1: Deduces that \(z_2 = 15 - 3\mathrm{i}\) only

Notes

(a)
B1: Correct coordinates of centre
M1: Fully correct strategy for identifying the radius. If the diameter is calculated this must be halved to achieve this mark.
A1: Correct equation using the required notation

Notes

(b)
M1: Begins the process of finding \(z_1\) and \(z_2\) by using the Cartesian equations to obtain the equation of the line of intersection
M1: Substitutes back into the equation of one of the circles to obtain an equation in one variable
A1: Correct 3 term quadratic
M1: Solves their 3TQ
M1: Substitutes to find values of the other variable to complete the process of finding \(z_1\) and \(z_2\)
A1: Correct complex numbers

Notes

(a)
B1: Identifies the correct complex conjugate as another root
B1: Correct values for the sum and product for the conjugate pair
M1: Correct application of the pair sum
M1: Identifies a complete and correct strategy for identifying the third root
A1: Deduces the correct third root
B1: \(3 \pm 2\sqrt{2}\,\mathrm{i}\) plotted correctly, in quadrants 1 and 4 which are reflections in the real axis. Do not be concerned about labelling or scaling.
B1ft: Their real root plotted correctly, in correct relative position to the two complex roots. Scales are not needed but if correct, the real root must be close to the origin compared to the complex roots.

(a) Alternative:

Alternative:
B1: Identifies the correct complex conjugate as another root
B1: Correct quadratic factor obtained
M1: Expands their quadratic\(\times\)(\(3z\) + “\(a\)”) or attempts to factor out the quadratic, or use long division, leading to a factor (\(3z\) + “\(a\)”). Implied by seeing \(\left(z^2 - 6z + 17\right)(3z + a)\) with any value of \(a\) (or with their quadratic).
M1: Proceeds to extract the root from their third factor of from (\(3z\) + “\(a\)”).
A1: Deduces the correct third root. If not explicitly stated, look for it on their diagram.
B1: \(3 \pm 2\sqrt{2}\,\mathrm{i}\) plotted correctly, as above
B1ft: Their real root plotted correctly as above.

Note: some may attempt to use the factor theorem with the complex root.
\(\mathrm{f}\left(3 - 2\mathrm{i}\sqrt{2}\right) = 36 + p + q + \mathrm{i}\left(-228\sqrt{2} - 12\sqrt{2}p\right) = 0\)
2^nd B1: equate real and imaginary components to 0 to get correct equations
\(36 + p + q = 0,\ -228\sqrt{2} - 12\sqrt{2}p = 0\)
1^st M1: solves their equations \(\Rightarrow p = -19,\ q = -17\)
2^nd M1: Solves the cubic (may be from calculator). The 1^st B1 may then be implied for the second complex root, and the rest as main scheme.

Notes

(b)
M1: Correct strategy used for identifying at least one of \(p\) or \(q\)
A1: At least one value correct
A1: Both values correct

(b) Alternative:

Alternative:
M1: Correct strategy by expanding their quadratic and linear factors to identifying at least one of \(p\) or \(q\)
A1: At least one value correct
A1: Both values correct

Notes

M1: Identifies a suitable method to rotate the given point by 120° (or equivalent) about the origin. May see equivalent work with modulus/argument or exponential form e.g. an attempt to multiply by \(\cos\dfrac{2\pi}{3} + \mathrm{i}\sin\dfrac{2\pi}{3}\) or \(\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}}\)

A1: Correct real part or correct imaginary part

A1: Completely correct complex number

M1: Identifies a suitable method to rotate the given point by 240° (or equivalent e.g. rotate their \(B\) by 120°) about the origin

May see equivalent work with modulus/argument or exponential form e.g. an attempt to multiply \(6 + 2\mathrm{i}\) by \(\cos\dfrac{4\pi}{3} + \mathrm{i}\sin\dfrac{4\pi}{3}\) or \(\mathrm{e}^{\frac{4\pi}{3}\mathrm{i}}\) or their \(B\) by \(\cos\dfrac{2\pi}{3} + \mathrm{i}\sin\dfrac{2\pi}{3}\) or \(\mathrm{e}^{\frac{2\pi}{3}\mathrm{i}}\)

A1: Correct real part or correct imaginary part

A1: Completely correct complex number

Notes

In general, the marks in (b) should be awarded as follows:
M1: Attempts to find the area of a relevant triangle
dM1: completes the problem by multiplying by an appropriate factor to find the area of \(DEF\)
Dependent on the first method mark
A1: Correct exact area
In some cases it may not be possible to distinguish the 2 method marks. In such cases they can be awarded together for a direct method that finds the area of \(DEF\)

Examples:

Way 1

M1: A correct strategy for the area of a relevant triangle such as \(ABC\) or \(AOB\)

dM1: Completes the problem by linking the area of \(DEF\) correctly with \(ABC\) or with \(AOB\)

A1: Correct value

Alternative: Way 2

M1: A correct strategy for the area of a relevant triangle such as \(DOF\)

dM1: Completes the problem by linking the area of \(DEF\) correctly with \(DOF\)

A1: Correct value

Alternative: Way 3

M1: A correct strategy for the area of a relevant triangle such as \(ABC\)

dM1: Completes the problem by linking the area of \(DEF\) correctly with \(ABC\)

A1: Correct value

Alternative: Way 4

M1dM1: A correct strategy for the area of \(DEF\). Finds 2 midpoints and attempts one side of \(DEF\) and uses a correct triangle area formula. By implication this scores both M marks.

A1: Correct value

Alternative: Way 5

M1: A correct strategy for the area of \(ABC\) using the “shoelace” method.

dM1: Completes the problem by linking the area of \(DEF\) correctly with \(ABC\)

A1: Correct value

Note the marks in (b) can be scored using inexact answers from (a) and the A1 scored if an exact area is obtained.

Notes

M1: Some evidence that complex roots occur as conjugate pairs shown, e.g. stated as in scheme, or e.g. identifying if \(-1 + 2\mathrm{i}\) is a root then so is \(-1 - 2\mathrm{i}\). Mere mention of complex conjugates is sufficient for this mark.

A1: A complete argument, referencing that a quartic has at most 4 roots, but would need at least 5 for all of \(z_1\), \(z_2\) and \(z_3\) as roots. There should be a clear statement about the number of roots of a quartic (e.g. a quartic has four roots), and that this is not enough for the two conjugate pairs and real root.

Notes

M1: Substitutes the numbers in expression and attempts multiplication of numerator and denominator by the conjugate of their denominator or uses calculator to find the quotient. (May be implied.)
NB Applying the difference of arguments and using decimals is M0 here.

A1: Obtains \(\dfrac{1}{2} + \dfrac{1}{2}\mathrm{i}\). (May be from calculator.) Accepted equivalent Cartesian forms.

A1*: Uses arctan on their quotient and makes reference to first quadrant or draws diagram to show they are in the first quadrant, to justify the argument.

Notes

M1: Applies the formula for the argument of a difference of complex numbers and substitutes values (may go directly to arctans if the arguments have already been established). If used in (b) it must be seen or referred to in (c) for this mark to be awarded. Allow for \(\arg(z_2 - z_1) - \arg(z_3 - z_1)\) if \(z_2 - z_1\) and \(z_3 - z_1\) have been clearly identified in earlier work.

A1*: Completes the proof clearly by identifying the required arguments and using the result of (b). Use of decimal approximations is A0.

Notes

B1: Draws a line through \(z_2\) and passing through negative imaginary axis.

B1: Correct side of bisector shaded. Allow this mark if the line does not pass through \(z_2\). But it should be an attempt at the perpendicular bisector of the other two points – so have negative gradient and pass through the negative real axis.

Ignore any other lines drawn for these two marks.

Notes

M1: Identifies at least one correct complex conjugate as another root (can be seen/implied by Argand diagram)

A1: Both complex conjugate roots identified correctly (can be seen/implied by Argand diagram)

For the next two marks allow either a cross, dot or line drawn where the end point is labelled with the correct coordinate, corresponding complex number or clearly plotted with correct numbers labelled on the axis or indication of the correct coordinates by use of scale markers. Condone (3, i) etc. The axes do not need to be labelled with Re and Im.

B1: One complex conjugate pair correctly plotted.

B1: Both complex conjugate pair correctly plotted. The \(3 \pm \mathrm{i}\) must be closer to the real axes than the \(-1 \pm 2\mathrm{i}\)

If there is no indication of the coordinates, scale or complex numbers on the Argand diagram this is B0 B0.

Do accept correct labelling e.g.

Notes

Way 1

M1: Correct strategy for forming at least one of the quadratic factors. Follow through their roots.

A1: At least one correct simplified quadratic factor.

A1: Both simplified quadratic factors correct or a correct simplified cubic factor

M1: A complete strategy to find values for \(a\), \(b\), \(c\) and \(d\) e.g. uses their quadratic factors or cubic and linear factor to form a quartic.

A1: Correct quartic in terms of \(z\) or correct values for \(a\), \(b\), \(c\) and \(d\) stated.

Alternative: Way 2

M1: Correct strategy for finding at least three of the sum roots, pair sum, triple sum and product. Follow through their roots. This can be implied by at least three correct values for the sum roots, pair sum, triple sum and product with no working shown. If the calculations are not shown for the sums and product and they have at least two incorrect values this is M0.

A1: At least two correct values for the sum roots, pair sum, triple sum or product.

A1: All correct values for the sum, pair sum, triple sum and product.

M1: Must have real values of \(a\), \(b\), \(c\) and \(d\) and use \(a = -\)their sum roots, \(b =\) their pair sum, \(c = -\)their triple sum and \(d =\) their product.

A1: Correct quartic in terms of \(z\) or correct values for \(a\), \(b\), \(c\) and \(d\) stated.

Alternative: Way 3

M1: Substitutes two roots into \(\mathrm{f}(z) = 0\) and equates coefficients to form 4 equations

A1: At least two correct equations.

A1: All four correct equations

M1: Solve their four equation (using calculator) to find at least one value. This will need checking if incorrect equations used.

A1: Correct quartic in terms of \(z\) or correct values for \(a\), \(b\), \(c\) and \(d\) stated.

Note: Correct answer only will score 5/5

Notes

B1: Recognises that the other roots must form a conjugate pair or obtains \(z^2 + 4z + p + 12\) (or \(z^2 + 4z - \dfrac{q}{3}\)) as the quadratic factor or writes down a correct equation for \(p\) and \(q\) or writes down a correct equation involving “\(z_2\)” and “\(z_3\)”

M1: Uses the sum of the roots of the cubic or the sum of the roots of their quadratic to find a value for “\(\alpha\)”

A1: Correct value for “\(\alpha\)”

M1: Uses their value for “\(\alpha\)” and the given area to find a value for “\(\beta\)”. Must be using the area and triangle dimensions correctly e.g. \(\dfrac{1}{2} \times \beta \times 5 = 35 \Rightarrow \beta = 14\) scores M0

M1: Uses an appropriate method to find \(p\) or \(q\)

A1: A correct value for \(p\) or \(q\)

A1: Correct values for \(p\) and \(q\)

Alternative

B1: Obtains \(z^2 + 4z + p + 12\) (or \(z^2 + 4z - \dfrac{q}{3}\)) as the quadratic factor

M1: Solves their quadratic factor by completing the square or using the quadratic formula

A1: Correct value for “\(\alpha\)”

M1: Uses their imaginary part to find “\(\beta\)” in terms of \(p\)

M1: Draws together the fact that the imaginary parts of their complex conjugate pair and the real root form the sides of the required triangle and forms an equation in terms of \(p\), sets equal to 35 and solves for \(p\)

A1: A correct value for \(p\) or \(q\)

A1: Correct values for \(p\) and \(q\)

Notes

M1: Circle or arc of a circle with centre in first quadrant and with the circle in all 4 quadrants or arc of circle in quadrants 1 and 2

M1: A “V” shape i.e. with both branches above the \(x\)-axis and with the vertex on the positive real axis. Ignore any branches below the \(x\)-axis.

A1: Two half lines that meet on the positive real axis where the right branch intersects the circle or arc of a circle in the first quadrant and the left branch intersects the circle or arc of a circle in the second quadrant but not on the \(y\)-axis.

M1: Shades the region between the half-lines and within the circle

A1: Cso. A fully correct diagram including 2 marked (or implied by ticks) at the vertex on the real axis with the correct region shaded and all the previous marks scored.

Example marking for 3(a)

M1: Circle with centre in first quadrant
M0: The branches of the “V” must be above the x-axis
A0: Follows M0
M1: Shades the region between the half-lines and within the circle
A0: Depends on all previous marks

M1: Circle with centre in first quadrant
M0: The vertex of the “V” must be on the positive x-axis
A0: Follows M0
M1: Shades the region between the half-lines and within the circle (BOD)
A0: Depends on all previous marks

M1: Circle with centre in first quadrant
M0: The vertex of the “V” must be on the positive x-axis
A0: Follows M0
M1: Shades the region between the half-lines and within the circle
A0: Depends on all previous marks

M1: Circle with centre in first quadrant
M1: A “V” shape i.e. with both branches above the x-axis and with the vertex on the positive real axis. Ignore any branches below the x-axis.
A1: Two half lines that meet on the positive real axis where the right branch intersects the circle in the first quadrant and the left branch intersects the circle in the second quadrant.
M1: Shades the region between the half-lines and within the circle
A1: A fully correct diagram including 2 marked at the vertex on the real axis with the correct region shaded and all the previous marks scored.

Notes

M1: Identifies a suitable strategy for finding the \(x\) or \(y\) coordinate of the point of intersection.
Look for an attempt to solve equations of the form \((x \pm 1)^2 + (y \pm 1)^2 = 9 \text{ or } 3\) and \(y = \pm x \pm 2\)

A1: Correct coordinates for the intersection (there may be other points but allow this mark if the correct coordinates are seen). (The correct coordinates may be implied by subsequent work.)
Allow equivalent exact forms and allow as a complex number e.g. \(2 + \dfrac{\sqrt{14}}{2} + \dfrac{\sqrt{14}}{2}\mathrm{i}\)

M1: Correct use of Pythagoras on their coordinates (There must be no i’s)

A1: Correct exact value by cso

Note that solving \((x - 1)^2 + (y - 1)^2 = 9,\ \ y = x + 2\) gives \(x = \dfrac{\sqrt{14}}{2},\ y = 2 + \dfrac{\sqrt{14}}{2}\) and hence the correct answer fortuitously so scores M1A0M1A0

Typical solution

Typical solution

Only points on the real axis are equidistant from \(-2 + \mathrm{i}\sqrt{5}\) and \(-2 - \mathrm{i}\sqrt{5}\)

Typical solution

(i)

Let centre of \(C\) be \(x\) where \(x \in \mathbf{R}\)

(ii)

Locus of \(z\) is \(|z - 3| = \sqrt{30}\)

Typical solution

One root is \(-4 - 3\mathrm{i}\)

Let the other root \(= \gamma\)

Then

So the other roots are \(-4 - 3\mathrm{i}\) and \(\dfrac{1}{4}\)

Typical solution

Typical solution

Let \(z = a + b\mathrm{i}\)

Typical solution

Also

So the only solutions of the equation are

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

\(-z^*\)

Typical solution

\(3 - 5\mathrm{i}\)

Typical solution

\(c \gt 4\)

Typical solution

\(7 - \mathrm{i}\)

Typical solution

Typical solution

(i)

\(Q\) is the point representing \(z_1\)

(ii)

\(\arg(z_1)\) is acute, so

Typical solution

(i) \(a = 5\)

(ii) \(b = 3\)

Typical solution

(i)

(ii)

(iii)

Typical solution

The polynomial equation \(z^5 - 1 = 0\) is of degree 5 and has exactly one real root.

This is a counter example to Latifa’s statement.

So Sam is right.

Typical solution

Typical solution

Also

Typical solution

(i)

(ii)

As \(x\) and \(y\) are both real then \(zz^*\) is real for all \(z \in \mathbf{C}\)

Typical solution

(i)

Let \(w = x + \mathrm{i}y\) where \(x, y \in \mathbf{R}\)

Comparing real and imaginary parts:

(ii)

Typical solution

Let

where \(x, y, u, v \in \mathbb{R}\)

Re: \(x + u = 5 \quad \ldots(1)\)

Im: \(y - v = 0 \Rightarrow y = v \quad \ldots(2)\)

Re: \(3x - u = 6 \quad \ldots(3)\)

Im: \(-3y - v = 4 \quad \ldots(4)\)

(2), (4) \(\Rightarrow y = -1,\ v = -1\)

(1), (3) \(\Rightarrow x = \dfrac{11}{4},\ u = \dfrac{9}{4}\)

Typical solution

\((x^2 + 25)\)