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

B1: Recalls the fact that if \(\mathrm{f}(-5) = 0\) then \((z + 5)\) is a factor

M1: Finds the quadratic factor with a \(4z^2\) term, any method. Condone using a factor of \((z - 5)\)

A1: Correct factorised expression for \(\mathrm{f}(z)\)

Notes

M1: Uses a correct method to show where the complex roots have come from.

Note just uses the calculator is M0

A1: Completes the method to show the complex roots. There needs to be an intermediate line of work with either 4i or \(4\sqrt{-1}\) term

Note: \(z = \dfrac{32 \pm \sqrt{-16}}{8} = \dfrac{8 \pm \mathrm{i}}{2}\) is M1 A0

Alternative

(Corrected from the printed mark scheme: the alternative prints \(z^2 - 8z + \dfrac{64}{4} = 0\); the product of the roots is \(\dfrac{65}{4}\).)

M1: uses the roots and expands

A1: Achieves the correct equation with = 0 no errors and a conclusion

Notes

B1: Deduces one correct root

B1: Deduces all 3 correct roots

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: 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

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

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

(a)
M1: Attempts to expand \((\cos\theta + \mathrm{i}\sin\theta)^7\) including a recognisable attempt at binomial coefficients
Some simplification may be done at this stage. (May only see imaginary terms)
M1: Identifies imaginary terms with \(\sin 7\theta\)
A1: Correct expression with coefficients evaluated and i’s dealt with correctly
M1: Replaces \(\cos^2\theta\) with \(1 - \sin^2\theta\) and applies the expansions of \(\left(1 - \sin^2\theta\right)^2\) and \(\left(1 - \sin^2\theta\right)^3\) to their expression
A1*: Reaches the printed answer with no errors and expansion of brackets seen.

Notes

(b)
M1: Makes the connection with part (a) and realises the need to solve \(\sin 7\theta = -1\)
A1: At least one correct value for \(7\theta\)
M1: Divides by 7 and deduces that \(x\) values are found by finding at least one value for \(\sin\theta\)
A1: Awrt 2 correct values for \(x\)
A1: Awrt all 4 \(x\) values correct and no extras

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: Equates sum of roots to 8 and obtains an equation in just \(\alpha\).

A1: Obtains a correct equation in \(\alpha\).

M1: Forms a three term quadratic equation in \(\alpha\) and attempts to solve this equation by either completing the square or using the quadratic formula to give \(\alpha = \ldots\)

A1: \(\alpha = 2 \pm \mathrm{i}\sqrt{2}\)

M1: Any correct method for finding the remaining root. There are various routes possible. See scheme for common ones.
Allow this mark if \(-24\) is used as the product.
See note below for a less common approach.

A1: Third root found with all three roots correct. Note \(\alpha\) and \(\beta\) need not be identified.

Note for (a) final M – it is possible to find the second and third roots using only one initial root (e.g. if second root forgotten or error leads to only one initial root being found).
Product of roots \(= \alpha\beta\left(\alpha + \dfrac{12}{\alpha} - \beta\right) = 24 \Rightarrow \alpha\beta^2 - (\alpha^2 + 12)\beta + 24 = 0\), substitutes in \(\alpha\) and attempts to solve the quadratic in \(\beta\) to achieve remaining roots. The final M can be gained once three roots in total have been obtained. (This is unlikely to be seen as part of a correct answer.) Allow if \(-24\) has been used for the product.

Notes

M1: Any correct method of finding \(p\). For example, applies the factor theorem, process of finding the pair sum of roots, or uses the roots to form \(\mathrm{f}(z)\).

A1: \(p = 22\) by correct solution only. Note: this can be found using only their complex roots from (a) (e.g. by factor theorem)

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 \(q\)

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

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

\((x^2 + 25)\)

Typical solution

Typical solution

(i)

\(\therefore\) \((1 + \mathrm{i})\) is a root of \(\mathrm{f}(z) = 0\)

(ii)

(iii)

2nd linear factor is \(\big(z - (1 - \mathrm{i})\big)\)

Quadratic factor is

(iv)

2nd quadratic factor is \(z^2 + 2z + 5\)

(v)

\(\mathrm{f}(z) = 0\) has no real roots.

Hence \(y = \mathrm{f}(x)\) does not intersect the \(x\)-axis.

Typical solution

(i)

(ii)

If \(\mathrm{f}(x) \lt 0\),

Typical solution

If \(\mathrm{g}(x) \leqslant 0\),

Typical solution

Typical solution

Each of the solutions has modulus 1, and their arguments increase in steps of \(2\pi/5\)

Typical solution

Let \(c = \cos\theta\), \(s = \sin\theta\)

Real parts:

as required

Typical solution

By symmetry

\(z_4^* = z_3\) so

and

\(z_5^* = z_2\) so

So \(c = a\) and \(c = b\) are both double roots of the equation \(16c^5 - 20c^3 + 5c - 1 = 0\) and, by the factor theorem, \((c - a)(c - b)\) is a repeated quadratic factor of \(16c^5 - 20c^3 + 5c - 1\)

Typical solution

\(h\) is a solution to the equation

Discard \(c = 1\)

Select the solution with the greater absolute value; so

as required

Typical solution

\(z = -\dfrac{3}{2} - \mathrm{i}\dfrac{\sqrt{11}}{2}\) is another root

Sum of complex roots = −3

Product of complex roots = 5

So \(z^2 + 3z + 5\) is a factor of \(z^4 - 3z^3 - 5z^2 + kz + 40\)

The other quadratic factor is

The other roots are

Typical solution

Typical solution

Typical solution

but \(\cosh\theta \geqslant 1 \quad \therefore \cosh\theta \neq -1\)

Typical solution

Can only assume that complex roots of a polynomial equation are in conjugate pairs if the coefficients are all real.

Typical solution

Typical solution

10

Typical solution

\(z = 1 + 3\mathrm{i}\) is another root.

\(z^2 - 2z + 10\) is a factor.

Comparing \(z^3\)-terms gives \(b = 0\)

\(\therefore\) the quartic is

Typical solution

Typical solution

\(1 + \mathrm{i}\) and \(1\)

Typical solution

\(m\) and \(p\) are real numbers.

Typical solution

Product of roots = − 91/2

One possible solution to the cubic is given by (for example)

since

This gives \(m = -1\) and \(p = -2\)

Another solution to the cubic is (for example)

since

This gives different values of \(m\) and \(p\).

Because there is more than one possible set of values of \(m\) and \(p\), there is not sufficient information to solve the problem, and Bonnie is right.

Typical solution

2^nd complex root is \(2 + 3\mathrm{i}\)

sum of roots \(= -\dfrac{b}{a}\)

Alternative

2^nd complex root is \(2 + 3\mathrm{i}\)

product of roots \(= -\dfrac{d}{a}\)

Typical solution

Alternative

Notes

M1*: De Moivre’s theorem with \(n = 5\)
Assume \(c = \cos\theta\) and \(s = \sin\theta\) (so condone not explicitly stated)

M1dep*: Expanding \((c + \mathrm{i}s)^5\) to obtain six terms - coefficients must be numerical and correct (so binomial coefficients must be evaluated) and correctly in terms of i only (so not powers of i). Allow at most one sign error and at most one term with an incorrect index

M1: Correctly taking real and imaginary parts of their expanded \((c + \mathrm{i}s)^5\) and dividing correctly (numerator and denominator must contain the correct number of terms and no i’s).
Dependent on both previous M marks

A1: AG - www
Must see mathematically each term in both the numerator and the denominator being divided by \(c^5\) oe (e.g. multiplying by \(c^{-5}\)) so \(\dfrac{c^{-5}\left(5c^4s - 10c^2s^3 + s^5\right)}{c^{-5}\left(c^5 - 10c^3s^2 + 5cs^4\right)}\) or \(\dfrac{\frac{5c^4s - 10c^2s^3 + s^5}{c^5}}{\frac{c^5 - 10c^3s^2 + 5cs^4}{c^5}}\) is fine but \(\dfrac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4} \div c^5\) or \(\dfrac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4} \div \dfrac{c^5}{c^5}\) etc. or in words (e.g. ‘divide both by \(c^5\)’) is A0
Note that \(\dfrac{5c^4s - 10c^2s^3 + s^5}{c^5 - 10c^3s^2 + 5cs^4} = \dfrac{\frac{5s}{c} - \frac{10s^3}{c^3} + \frac{s^5}{c^5}}{1 - \frac{10s^2}{c^2} + \frac{5s^4}{c^4}}\) is A0 (as AG)
Final answer must be in terms of tan not \(t\)

Notes

(b)(i)

M1*: Equates result from part (a) to 1, multiplies through by denominator and then rearranges to get all terms on one side. Condone using tan rather than \(t\) - Allow sign errors only.

B1: Finds all five solutions to \(\tan 5\theta = 1\) in the interval \(0 \lt \theta \lt \pi\) - ignore any solutions outside of this interval but if any incorrect in this interval, then B0

B1dep*: Correct justification that \((t - 1)\) is a factor of the correct quintic in \(t\) using the result that \(\tan\left(\frac{5}{20}\pi\right) = 1\) and re-writing quintic as \((t - 1)\left(t^4 - 4t^3 - 14t^2 - 4t + 1\right)\) oe (e.g. by long division) or re-writes quintic as \((t - 1)\left(t^4 - 4t^3 - 14t^2 - 4t + 1\right)\) oe (e.g. by long division) and relates the factor \((t - 1)\) to \(\tan\left(\frac{5}{20}\pi\right)\) (in essence this mark is for justifying that the root \(\tan\left(\frac{5}{20}\pi\right)\) of the correct quintic equation is not a root of the given quartic equation therefore this mark is dependent on having derived the correct expression \(t^5 - 5t^4 - 10t^3 + 10t^2 + 5t - 1 \; (= 0)\))

A1: These exact four roots only (so must not include \(\tan\left(\frac{5}{20}\pi\right)\)) – this mark is not dependent on the previous B1 mark – a correct quintic equation/expression followed by these four roots only scores M1 A1 only. These four roots only without the correct quintic equation/expression seen is no marks

(b)(ii)

B1: www for correctly finding the value of \(k\) – allow for either \((t - 1)^4 = 20t^2\) or \(k = 20\) stated with no working

M1: For correct three term quadratic expressions(s)/equation(s) (in \(t\)), follow through their positive value of \(k\). Condone missing \(\pm\) - so for either \(t^2 - \left(2 \pm \sqrt{k}\right)t + 1\) or \(t^2 - \left(2 + \sqrt{k}\right)t + 1\) or \(t^2 - \left(2 - \sqrt{k}\right)t + 1\) with their \(k\)

A1: AG
Obtains all four roots www and explains that \(\tan\left(\frac{9}{20}\pi\right)\) is positive and the largest root and so \(\tan\left(\frac{9}{20}\pi\right) = 1 + \sqrt{5} + \sqrt{5 + 2\sqrt{5}}\)
(corrected from the printed mark scheme: the second pair of roots is printed as \(t = 1 - \sqrt{5} \pm \sqrt{5 + 2\sqrt{5}}\); since \(\left(1 - \sqrt{5}\right)^2 - 1 = 5 - 2\sqrt{5}\), it should be \(t = 1 - \sqrt{5} \pm \sqrt{5 - 2\sqrt{5}}\))

Notes

M1: Finding sum or product of known roots (could be embedded)
Or dividing cubic by one of \((x - (-7 + 5\mathrm{i}))\), \((x - (-7 - 5\mathrm{i}))\) and the quotient by the other
Or by multiplying \((x - (-7 + 5\mathrm{i}))(x - (-7 - 5\mathrm{i})) = \ldots\)
(corrected from the printed mark scheme: the sum is printed as \((-7 + 5\mathrm{i}) + (-7 + 5\mathrm{i}) = -14\); the two roots are \(-7 + 5\mathrm{i}\) and \(-7 - 5\mathrm{i}\))

A1: No need to reassemble if both factors correct and unambiguous
SCB1 for correct answer without evidence of use of (a) (max 1/3).

Alternative method

M1: Finding product of known roots and \(\alpha\beta\gamma = -\frac{d}{a}\) to find the real root.
(corrected from the printed mark scheme, which has \(\alpha\beta\gamma = -\frac{c}{a}\); for \(ax^3 + bx^2 + cx + d = 0\) the product of the roots is \(-\frac{d}{a}\))

A1: No need to reassemble if both factors correct and unambiguous

Notes

B1: (2nd) or eg \(\mathrm{f}(1) = 0\) if intent is clear but must be some justification.

B1: (3rd) Some justification must be given that there are no more real roots.
Allow for finding the two complex roots \((-1 \pm 2\mathrm{i})\) - must have seen the correct \(x^2 + 2x + 5\)
If B1B0B0 or B0B0B0 then SC1 for “\(a = 1\) and no others” or “\(a = 1\), \((-1 \pm 2\mathrm{i})\)” without justification.

M1: Condone small errors in calculation of coefficients in equation
Need both the pair of complex roots SOI and the quartic shown (allow sign slips). Only need one case for the M1.

M1: oe eg expanding \((z - (2 + 3\mathrm{i}))(z - (2 - 3\mathrm{i}))\) or \((z - (3 + 3\mathrm{i}))(z - (3 - 3\mathrm{i}))\)
Attempt to find quadratic factor from the complex roots. Only one case needed for M1
Allow with no working

M1: Attempt to factorise their quartic with their quadratic factor (at least \(z^3\) and constant terms consistent).
Only one case needed.
MUST see some evidence of factorisation here

A1: All four roots SOI for the \(a = 0\) case
DR so need to see evidence of where the roots came from i.e. factorisation into two quadratics

A1: All four roots SOI for the \(a = 1\) case
If extra values of \(z\) found (from complex \(a\) or incorrect \(a\) values) then A0

Alternate for 1st and 2nd M mark

M1: Quadratic factor found in general case. Can also be found by expanding \([z - (a + 2 + 3\mathrm{i})][z - (a + 2 - 3\mathrm{i})]\)
Constant term can be either \((a + 2)^2 + 9\) or \([a^2 + 4a + 13]\)

Alternate 2 if only the quartic in \(z\) with \(a = 0\) considered via linear factors

B1: (2nd) or eg \(\mathrm{f}(1) = 0\) if intent is clear but must be some justification.

B1: (3rd) Some justification must be given that there are no more real roots. Allow for finding the two complex roots \((-1 \pm 2\mathrm{i})\) (must have seen correct quadratic)
If B1B0B0 or B0B0B0 then SC1 for “\(a = 1\) and no others” or “\(a = 1\), \((-1 \pm 2\mathrm{i})\)” without justification.

M1: (1st) Identifying linear factor and factorising

A1: Identifying 4 roots
Final 2 marks unavailable

Notes

M1: Attempt to expand the determinant. If using standard method must see at least two terms, at least one of which comprises \(\pm\) a number multiplied by the residual determinant. eg
\(a(a(a - 1) - 4 \times (-13)) - (a + 9)((-6)(a - 1) - (a - 3)(-13))\)
This mark can be awarded with \(a = 2\) substituted (ie attempt to find \(\begin{vmatrix} 2 & -6 & -1 \\ 11 & 2 & 4 \\ 0 & -13 & 1 \end{vmatrix}\)) but working must be shown. eg \(2(2 + 52) - 11(-6 - 13)\) \((= 317)\)
Allow other correct methods.

M1: Setting up equation equating their determinant to their specific determinant value for \(a = 2\)

M1: Rearranging to = 0 and use of factor theorem to derive a quadratic equation in \(a\) by dividing by \((a - 2)\)
Need some evidence of how to solve cubic (Not just cubic written and then roots BC). Need to see =0 on one side of cubic (but it might disappear after this, e.g. when dividing by \((a - 2)\))

M1: Attempt to solve quadratic involving \(\sqrt{-1} = \mathrm{i}\) oe
\(\dfrac{6 \pm \sqrt{6^2 - 4 \times 1 \times 10}}{2 \times 1} = \dfrac{6 \pm \sqrt{-4}}{2} = \dfrac{6 \pm 2\mathrm{i}}{2}\)

A1: Both. No need to mention \(a = 2\)
Don’t need to see “\(a =\)” explicitly
SC – if third and/or fourth method mark not awarded then allow SC B1 for sight of \(3 \pm \mathrm{i}\).

Notes

M1: Substitute \(z = w - 1\)

M1: Expanding with at least \((w - 1)^4\) seen

A1: Convincingly shown AG. Must include \(= 0\) on last line. Brackets must be fully expanded in working or evidence of collection of like terms
A0 if variable used is not \(w\).

Alternative method

M1: Substitute \(w = z + 1\) into end result

Alternative method using symmetry of roots

B1: For sums from original equation and finding the sum of the new roots

B1: For showing convincingly the sum of new roots in pairs

B1: For the last two

Notes

M1: Solving quadratic equation in \(w^2\) (or using their variable)

M1: Square rooting their \(w^2\), including \(\pm\), as long as their \(w^2\) not both non-negative and real.

A1: cao
Answers with no working is 0

Notes

B1: Need to see statement that \(3 + 5\mathrm{i}\) is a root.
May happen at end of question

M1: (1st) Attempt to use the conjugate pair to derive a real quadratic
May see \((3 + 5\mathrm{i})(3 - 5\mathrm{i}) = 9 + 25 = 34\) and \((3 + 5\mathrm{i}) + (3 - 5\mathrm{i}) = 6\) instead of expansion

M1: (2nd) Attempt to factorise or divide resulting in \(x^2\) and one other term
NB: This question required detailed reasoning

A1: (3rd) \(3 + 5\mathrm{i}\) may be mentioned as a root earlier in the solution

Notes

DR: Detailed reasoning required for this question

B1: Can be implied by \((z - (2 - \mathrm{i}))\)

M1: (1st) Correctly determining the form of the quadratic in factor form

A1: (1st) Working needs to be shown here (Show detailed reasoning). At least 1 step between factor form and final answer

M1: (2nd) With non-zero \(z\) term. Or attempt at symbolic division resulting in at least \(4z^2\). Or genuine attempt at method comparing coefficients leading to at least one useful equation (eg \(z^4\): \(A = 4\))
Must be a product of two quadratics
Multiplication table might be used (backwards)
Inspection is fine here, so candidates might just write down the correct two quadratic factors.

Notes

M1*: Correct method for finding a root of a quadratic (\(\pm\) not necessary)
or \(4\left(\left(z + \tfrac{1}{2}\right)^2 - \tfrac{1}{4}\right) + 37\) oe
Need to see some method

A1dep(*): Simplification of the square root to include i (\(\pm\) is now necessary)
Must be of the form \(z = a + b\mathrm{i}\) (or \(z = \frac{a + b\mathrm{i}}{c}\)) No square roots left in.

A1(ft) dep(*): All four - aef
Follow through on incorrect roots from previous mark as long as method mark awarded i.e. this mark is for realising that there are four roots and including the two from part (a). A0 if more than 4 roots given.
SC: If M0 awarded but all four correct roots are given with no extra then award SC B1

Notes

B1: Need to see \(|2 + \mathrm{i}|\) or \(|2 - \mathrm{i}|\)
Or \(\sqrt{2^2 + 1^2} = \sqrt{5}\) etc.

B1ft: ft the magnitude of the other of their conjugate pairs
Same comment as above.

M1: Either way round. Their \(r_1\) and/or \(r_2\).

A1ft: \(\pi(r_1^2 - 5)\) simplified
www
Positive area needed here. Must follow B1B1(ft)M1

Notes

B1: or \(\dfrac{-1 + 6\mathrm{i}}{2} + \dfrac{-1 - 6\mathrm{i}}{2} + 2 + \mathrm{i} + 2 - \mathrm{i} = 3\)
Some justification needed, not just \(\omega = 3\). Allow \(3 + 0\mathrm{i}\)

E1: Or \(5 \lt 9 \lt 37/4\)
Or \(2.23\ldots \lt 3 \lt 3.04\ldots\)
Both end comparisons needed for E1
No follow through given in this part
Conclusion needed
Can be smaller radius is 2.24, larger radius is 3.04 so \(\omega\) is in \(R\)
Diagram is ok for E1 if it implies both end comparisons

Notes

B1: Soi by correct quadratic \(x^2 - 4x + 29\)

M1: Attempt to factorise using complex conjugate.
Any valid method to find real root by reasoning, including division, or listing or using the factor theorem or sum of roots.

A1: Shown convincingly oe (i.e. \((x + 2)\) seen)

NB. A DR question so full reasoning must be shown.

Notes

M1: Binomial expansion. Must be 4 terms with 1, 3, 3, 1 soi and correct powers. Condone missing brackets
Or by \((2 + 3\mathrm{i})^2 \times (2 + 3\mathrm{i})\) but marks only to awarded once all binomial brackets expanded.

A1: (1st) All correct and \(\mathrm{i}^2 = -1\) twice.
Must see evidence of \(\mathrm{i}^2\) becoming \(-1\) (could be in a table, expanding brackets etc)

A1: (2nd) SC if the only working seen is \((-5 + 12\mathrm{i})(2 + 3\mathrm{i}) = -46 + 9\mathrm{i}\) award B1

Notes

B1: May be seen in (i). May be in working below.
If only seen in (i) and not implied by working for this part award B0

M1: Attempt to substitute their \(z^2\) and \(z^3\) into \(3z^3 - 8z^2 + 23z + 52\)
\(3(2 + 3\mathrm{i})^3 - 8(2 + 3\mathrm{i})^2 + 23(2 + 3\mathrm{i}) + 52\) enough for M1

A1: Convincingly cancels to 0
Must show some collection/cancellation Could be gathering real and imaginary terms

Notes

B1: Seen or implied

M1: ...is the required quadratic factor

A1: (2nd) Can be deduced by inspection
Must be written as a product of linear factor and quadratic factor
Condone “= 0” present.

Alternative

B1: Seen or implied

M1: Using the two roots to get an equation in \(z^2\)

A1: (2nd) Can be deduced by inspection
Condone “= 0” present.

Notes

M1: sum of roots, allow a sign error. No missing terms except for those which have cancelled.

A1: any correct quadratic equation e.g. \(\alpha^2 + 2 = \alpha\); soi by correct \(\alpha\)

M1: product of roots, allow a sign error. No missing terms except for those which have cancelled.

M1: using product of roots two at a time, soi by correct numerical work. Allow only a sign error on RHS, not on \(c\). No missing terms except for those which have cancelled. Condone missing brackets. Or for substituting a root correctly into equation if \(d\) already found.

M1: using product of roots three at a time, soi by correct numerical work. Allow only a sign error on RHS, not on \(d\). No missing terms except for those which have cancelled. Condone missing brackets. Or for substituting a root correctly into equation if \(c\) already found.

Alternative method 1

M1 expanding; M1 comparing coefft of \(x^3\) and constants

(corrected from the printed mark scheme: the roots of \(z^2 - z + 2 = 0\) are printed as \(\frac{1}{2} \pm \frac{7}{2}\mathrm{i}\); they are \(\frac{1}{2} \pm \frac{\sqrt{7}}{2}\mathrm{i}\), as in the main scheme)

Alternative method 2

M1: sum of roots, allow a sign error

A1: any correct quadratic equation e.g. \(\alpha^2 + 2 = \alpha\)

M1: product of roots, allow a sign error

M1: at least one quadratic factor identified

M1: multiplying their quadratic factors

Alternative method 3

M1: sum of roots, allow a sign error

A1: Any correct quadratic equation e.g. \(\alpha^2 + 2 = \alpha\)

M1: product of roots, allow a sign error

M1: substituting any root into equation correctly

M1: equating real and imaginary parts leading to two equations

Notes

M1: factors \(x - (2 + 3\mathrm{i})\) and \(x - (2 - 3\mathrm{i})\)

A1: or \(x^2 - 2x + 3\mathrm{i}x - 2x + 4 - 6\mathrm{i} - 3\mathrm{i}x + 6\mathrm{i} - 9\mathrm{i}^2\) or better

Alternative solution 1

M1: use of factor theorem

Alternative solution 2

M1: use of quadratic formula

Alternative solution 3

M1: sum or product of roots formulae used, condone sign errors

Notes

B1: (2nd) or \(\alpha\beta + 2\alpha + 2\beta = 7\) or \(\alpha\beta + 2(\alpha + \beta) = 7\)

M1: (1st) substitution which could lead to a quadratic in \(\alpha\) or \(\beta\) or \(x\)

A1: (1st) could be in \(\alpha\) or \(\beta\)

M1: (2nd) a method to solve their quadratic

Notes

B1: (1st) must see some substitution, [so \(\mathrm{f}\left(\frac{3}{2}\right) = 0\) alone is B0]

M1: (1st) or \(z - 3/2\)

M1 A1: attempt to factorise (oe, e.g. long division)
or \((z - 3/2)(2z^2 - 4z + 10)\)

M1: (3rd) or by completing the square or using sum and prod of roots
[\(z = a \pm \mathrm{i}b\), \(2a = 2\), \(a^2 + b^2 = 5 \Rightarrow a = 1\), \(b = \pm 2\)]

B1: (2nd) [3/2 may be stated as a root earlier]
If no working shown then award no marks

Alternative solution

M1: (1st) symmetric property of roots used (condone 7, 15, 16 or sign errors) – must have at least 2 of the 3

M1: (2nd) or by completing the square or using sum and prod of roots
[\(z = a \pm \mathrm{i}b\), \(2a = 2\), \(a^2 + b^2 = 5 \Rightarrow a = 1\), \(b = \pm 2\)]

Notes

M1: Solving their quadratic to find \(\alpha\)

Notes

M1: Allow in terms of \(\alpha\) and \(\beta\), no missing terms

M1: Allow in terms of \(\alpha\) and \(\beta\), no missing terms

Alternative solution

Alternative solution

Notes

A1: (2nd) or \(5\beta^2 - 10\beta + 25 = 0\)

M1: (3rd) Or equivalent use of formula

A1: (3rd) SCB1 if ow

Alternative solution

M1: (3rd) Factorising

M1: (4th) Solving the resulting quadratic

Notes

Or \(c + d\mathrm{i}\) and \(c - d\mathrm{i}\). M1
\(4 + 2c = 10\) M1
\(\Rightarrow c = 3\) A1
\((2 + \mathrm{i})(2 - \mathrm{i})(3 + d\mathrm{i})(3 - d\mathrm{i}) = 125\)
\(\Rightarrow 5(9 + d^2) = 125\) M1A1
\(\Rightarrow d = 4\) A1

M1: (2nd) product of roots used

M1: (3rd) eliminating and solving quad

Alternative solution

M1: (1st) Attempt to multiply factors
\(x^2 - 4x + 5\)
condone sign errors

M1: (2nd) Attempt to find other quad factor
\(2x^2 - 12x + 50\)
or long division

Notes

M1: product of roots used

A1: \(\beta = 2\)

M1: sum of roots used; or \((x - 2)(x^2 + x + 1) = 0\) M1A1

A1: or equivalent quadratic (with \(\beta = 2\))

M1: solving their quadratic; \(x = \dfrac{-1 \pm \sqrt{-3}}{2} = -\dfrac{1}{2} \pm \dfrac{\sqrt{3}}{2}\mathrm{i}\)

Notes

M1: \(k\) = product of root pairs; or \((x - 2)(x^2 + x + 1) \Rightarrow k = -1\)

A1: or by direct substitution; or by factor theorem

Notes

M1: (1st) substituting \(z = \mathrm{i}\) into eqn

M1: (2nd) equating real and im parts
condone verification

E1: AG

Alternative solution

or \(c + \mathrm{i}d\), \(c - \mathrm{i}d\)

B1: (1st) \(2c = 2\)

B1: (2nd) \(\Rightarrow 1 + c^2 + d^2 = 3 \Rightarrow c = d\)

B1: (3rd) or using \(1 + \mathrm{i}\), \(1 - \mathrm{i}\) from \(\alpha^2 - 2\alpha + 2 = 0\) AG (corrected from the printed mark scheme, which has \(\alpha^2 - 2a + 2 = 0\))
if \(1 + \mathrm{i}\), \(1 - \mathrm{i}\) roots used but not established allow B0 B0 B1 B1

B1: (4th) or using \(1 + \mathrm{i}\), \(1 - \mathrm{i}\) AG

Alternative solution

B1: used (see below)

M1: comparing coeffs
may be inferred from multiplication or long division

A1: (1st) finding \(c\), \(d\)

A1: (2nd) verifying \(a\), \(b\)

Notes

B1: (2nd) working backwards:
\((z - 1 - \mathrm{i})(z - 1 + \mathrm{i}) = z^2 - 2z + 2\) B1
\((z^2 + 1)(z^2 - 2z + 2) = \ldots\)
\(= z^4 - 2z^3 + 3z^2 - 2z + 2\) B1
max 4 marks

M1: (1st) equating coeffs or long divⁿ

M1: (2nd) solving their quad factor = 0 or \((z - 1 - \mathrm{i})(z - 1 + \mathrm{i})\)

Alternative solution

M1: (1st) or \(c + \mathrm{i}d\), \(c - \mathrm{i}d\)

M1: (2nd) or \(c + \mathrm{i}d + c - \mathrm{i}d = 2\)

M1: (3rd) or \((c + \mathrm{i}d)(c - \mathrm{i}d) = 2\)

A1: or \(c^2 + d^2 = 2\)

M1: (4th) \(c = 1\), \(d = 1\)