Notes

B1: See scheme. This can appear anywhere in the proof.
Accept \(2^4\sin^4\theta\) for \(16\sin^4\theta\), but not \((2\sin\theta)^4\)
Alternatively, they may instead substitute \(\sin\theta = \dfrac{1}{2\mathrm{i}}\left(z - \dfrac{1}{z}\right)\) into the expression \(8\sin^4\theta\). This can be implied but must come from correct work. This can appear anywhere in their proof.
e.g. \(8\sin^4\theta = 8\left[\dfrac{1}{2\mathrm{i}}\left(z - \dfrac{1}{z}\right)\right]^4\) or \(8\sin^4\theta = 8\left[\left(z - \dfrac{1}{z}\right)\right]^4\dfrac{1}{2^4}\)
and allow \(8\sin^4\theta = \dfrac{1}{2}\left[\left(z - \dfrac{1}{z}\right)\right]^4\)

M1: Finds the expansion of \(\left(z - \dfrac{1}{z}\right)^4\) which may be unsimplified. All five terms must be present.
Condone sign slips only

A1: Correct expansion, with terms grouped.

M1: Uses \(z^n + \dfrac{1}{z^n} = 2\cos n\theta\) to write in terms of \(\cos 4\theta\) and \(\cos 2\theta\)

A1*: Achieves the printed answer with no errors or omissions. Cso Follows A0.

Notes

B1: Correct formula \(\pi\displaystyle\int \left(\sin^2\left(\dfrac{1}{2}y\right)\right)^2\,\mathrm{d}y\) (but not \(\text{vol} = \pi\displaystyle\int x^2\,\mathrm{d}y\)) used to find a volume, stated or implied, ignore limits. If there is a missing \(\pi\) or d\(y\) in their integral, then withhold this mark only. However, this mark may be awarded if \(\pi\) and d\(y\) are seen together later in their integral.

Do not award the following marks if algebraic integration is not used. For example finding an answer of 7.3 without algebraic integration will obtain M0A0dM0A0

M1: Uses the result in part (a) to express the volume in an integrable form and attempts to integrate.
Award for an integral of the form \(\displaystyle\int \dfrac{1}{8}\big(A\cos(2y) + B\cos(y) + C\big)\ (\mathrm{d}y)\) with at least one term integrated correctly. Do not be concerned if they use a different variable such as \(\theta\) for \(y\).

Special Case:
If they have not used part (a) and instead use the double angle formulae:
\(\sin^2\alpha = \dfrac{1}{2} - \dfrac{1}{2}\cos 2\alpha\) and \(\sin^4\alpha = \left(\dfrac{1}{2} - \dfrac{1}{2}\cos 2\alpha\right)^2\) with \(\alpha = \dfrac{1}{2}y\)
In this case they must obtain an exact integral equivalent to \(\displaystyle\int \dfrac{1}{8}\big(\cos(2y) - 4\cos(y) + 3\big)\ (\mathrm{d}y)\) and proceed to integrate at least one term correctly.

A1: Correct integration. May be in terms of another variable such as \(\theta\). Ignore \(\pi\)

dM1: Dependent on previous method mark. Finds the required volume using \(\pi\displaystyle\int_0^{\frac{8\pi}{5}} x^2\,\mathrm{d}y\) and applies their limits to their integral and subtracts the correct way round.
If there are no limits seen substituted, then a correct final answer implies the correct use of limits and the inclusion of \(\pi\). This is provided they have already achieved an integrated expression of \(\dfrac{1}{8}\left(\dfrac{1}{2}\sin(2y) - 4\sin(y) + 3y\right)\) oe with correct limits seen, possibly on their integral.
If their integration is incorrect, then there must be evidence of substituting both limits in each of their terms and subtracting. Allow the omission of subtracting zero provided their integration would produce zero for the lower limit.

A1: awrt 7.3

Notes

M1: Finds the mass of the ornament by multiplying their volume by 0.85

A1ft: Draws an appropriate conclusion about the suitability of the model, comparing their two masses.
If the masses differ by 10% then they must conclude it is a good model.
If the masses differ between 10% and 20% then they may conclude it is either a good model or poor model.
If the masses differ by greater than 20% then they must conclude it is a poor model.

Alternative 1

M1: Finds the volume of the ornament by dividing the mass of 6 grams by 0.85 g/cm\(^3\)

A1ft: Draws an appropriate conclusion about the suitability of the model, comparing their two volumes.
If the volumes differ by 10% then they must conclude it is a good model.
If the volumes differ between 10% and 20% then they may conclude it is either a good model or poor model.
If the volumes differ by greater than 20% then they must conclude it is a poor model.

Alternative 2

M1: Finds the density of the ornament by dividing the mass of 6 grams by 7.3 cm\(^3\) (corrected from the printed mark scheme: printed as 7.3 g/cm\(^3\))

A1ft: Draws an appropriate conclusion about the suitability of the model, comparing their two densities.
If the densities differ by 10% then they must conclude it is a good model.
If the densities differ between 10% and 20% then they may conclude it is either a good model or poor model.
If the densities differ by greater than 20% then they must conclude it is a poor model.

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

M1: Substitutes \(z\) into the LHS and simplifies the powers as shown. Allow if they go direct to trigonometric expressions without exponentials. The mark is for sorting out the negative index.

A1*: Converts the exponential form to trigonometric form correctly and correctly completes the proof with no errors seen. The trigonometric expansion must be clearly seen. Condone missing brackets in e.g \(\cos -n\theta\) terms if intent is clear. Note the LHS of the identity may be implied.

Alt (a)

(corrected from the printed mark scheme: the denominator in the first line is printed as \(\cos n\theta + \mathrm{i}\sin\theta\))

Alt: by De Moivre

M1: Applies De Moivre on both terms and puts over a common denominator.

A1*: Complete correctly, using \(1 - \sin^2 n\theta = \cos^2 n\theta\) and cancelling \(\cos n\theta + \mathrm{i}\sin n\theta\). No errors seen.

Notes

B1: Deduces that \(\left(z + z^{-1}\right)^5 = 32\cos^5\theta\) Do not accept \(2^5\) for 32. May be implied.

M1: Attempts to expand \(\left(z + z^{-1}\right)^5\). Correct binomial coefficients must be used, terms need not be simplified. Condone at most one slip in powers.

A1: Correct expansion, terms need not be gathered but powers must have been simplified.

M1: Sets their expressions equal and applies the result from (a) – grouping must be shown.

A1*: Reaches the printed answer with no errors and relevant steps all shown.

Alt (b)

Alt: by De Moivre

B1: Correctly stated or clearly implied De Moivre statement for \(\cos 5\theta\)

M1: Attempts to expand \((\cos\theta + \mathrm{i}\sin\theta)^5\) Correct coefficients but allow one slip per main scheme. The powers of i need not be simplified for the attempt at expansion, accept if only the real terms are shown. Allow \(c\) and \(s\) notation.

A1: Correct real terms extracted with the i's removed.

M1: Applies \(\sin^2\theta = 1 - \cos^2\theta\) to reduce to an equation in \(\cos\theta\) and applies \(\cos^3\theta = \dfrac{1}{4}(\cos 3\theta + 3\cos\theta)\) (quoted or derived – allow a slip if derived) to get to an equation without powers of cos terms.

A1*: Reaches the printed answer with no errors and relevant steps all shown.

Notes

B1: Uses the result from (b) to deduce the correct equation.

M1: Must have attempted to use part (b) to obtain \(\alpha\cos^5\theta = \beta\cos\theta\) or equivalent. Collects to one side and attempts to factorise and solve. Note dividing through by \(\cos\theta\) is M0.

A1: Rejects the inappropriate solution and selects \(\cos\theta = 0\) and obtains the correct values only. The equation must have been correct. There must have been some consideration of the \(8\cos^4\theta + 1\) e.g. stating \(8\cos^4\theta \gt 0\) so no solutions, or attempting to find complex roots and deducing no answers. May be minimal, but some consideration that no roots arise from this part must have been given.

Note: The correct answer will appear from incorrect attempts – the M must be gained in order to award the A. E.g. assuming the equation reduces to \(\cos^5\theta = 0\) will score B0M0A0.
Likewise, answers only scores B0M0A0 (questions says hence so use of (b) must be seen).

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

(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

B1: Correct identity or equivalent rearrangement. This can appear anywhere in the proof.

M1: Attempts the expansion of \(\left(z + \dfrac{1}{z}\right)^6\) must have at least 3 correct terms. Combining the powers when expanding is fine.

A1: Correct expansion with \(z\) terms simplified, need not be rearranged. (So a correct expansion will score M1A1.)

M1: Uses \(z^n + \dfrac{1}{z^n} = 2\cos n\theta\) to write the expression in terms multiple angles of \(\cos 6\theta,\ \cos 4\theta\) and \(\cos 2\theta\). Pairing of terms must be seen.

A1*: Achieves the printed answer with no errors or omissions. Cso

For approaches using De Moivre B0M1A1M0A0 may be scored if the binomial expansions is attempted (and correct for the A).

Notes

B1: See scheme

Notes

Note: The question instructs use of algebraic integration and part (a), so answer only can score at most B1 for implied correct formula.

B1ft: Correct expression for the volume of the paperweight or the solid formed through 360° rotation, stated or implied, ignore limits. No need to expand, but must be applied, not just a formula in \(y\), though allow a correct formula followed by correct integral if the \(\pi\) disappears. Follow through their \(H\)

M1: Uses the result in part part (a) to express the volume in an integrable form and attempts to integrate. Note use of \(\theta\) instead of \(x\) is permissible for this mark. Allow if one term is missing or miscopied.

A1: Correct integration in terms of \(x\). Ignore \(\pi\), their \(H^2\) and the \(\dfrac{1}{32}\). Note if \(\theta\) has been used it is A0 unless a correct substitution method has been implied as the coefficients will be incorrect.

dM1: Dependent on previous method mark and must have reached and integral of the correct form -- in terms of \(x\) with correct arguments allowing for one slip. Finds the required volume using either \(\pi\int_0^4 y^2\,\mathrm{d}x\) or \(\dfrac{1}{2}\pi\int_{-4}^{4} y^2\,\mathrm{d}x\) and applies their limits - accept any value following a valid attempt at the integration as an attempt at applying limits.

A1: cao 24.56

Notes

B1: States an appropriate limitation. See scheme for some examples. The limitation should refer to the paperweight, not to paper. Do not accept “it does not take into account thickness of material” as it is a solid, not a shell, being modelled. Award the mark for a correct reason if two reasons are given and one is incorrect.

Notes

(a)
B1: See scheme

Notes

(b)(i)
M1: Substitutes \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into at least 3 terms of the series and applies de Moivre’s theorem.
M1: Substitutes \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into their answer to part (a) and rationalises the denominator.
M1: Equates the imaginary terms.
M1: Multiplies out the denominator and simplifies by using the identity \(\cos^2\theta + \sin^2\theta = 1\)
A1*: cso. Achieves the printed answer having substituted \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into 4 terms of the series.

Alternative
M1: Substitutes \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into at least 3 terms of the series and applies de Moivre’s theorem.
M1: Substitutes \(z = \dfrac{1}{2}\mathrm{e}^{\mathrm{i}\theta}\) into their answer to part (a) and rationalises the denominator.
M1: Uses \(\mathrm{e}^{-\mathrm{i}\theta} = \cos\theta - \mathrm{i}\sin\theta\) and \(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta} = 2\cos\theta\) to express in terms of \(\sin\theta\) and \(\cos\theta\)
M1: Select the imaginary terms.
A1*: cso Achieves the printed answer having substituted \(z = \dfrac{1}{2}\left(\cos\theta + \mathrm{i}\sin\theta\right)\) into 4 terms of the series.

(corrected from the printed mark scheme: in the alternative, the denominator \(1 - \frac{1}{2}\mathrm{e}^{\mathrm{i}\theta} - \frac{1}{2}\mathrm{e}^{-\mathrm{i}\theta} + \frac{1}{4}\) is printed with \(\frac{1}{4}\mathrm{e}^{\mathrm{i}\theta}\) and \(\frac{1}{4}\mathrm{e}^{-\mathrm{i}\theta}\))

Notes

(b)(ii)
M1: Setting the real part of the series \(= 0\) and rearranges to find \(\cos\theta = \ldots\)
A1: See scheme

Alternative 1
M1: Rearranges the real part so that \(\cos\theta\) only appears once
A1: Uses \(-1 \leqslant \cos\theta \leqslant 1\) to show that the sum must always be positive so must contain a real part

Alternative 2
M1: Sets sum as purely imaginary and rearranges to make \(z\) the subject
A1: Shows a contradiction and draws an appropriate conclusion

(corrected from the printed mark scheme: in Alternative 1 the expression \(\dfrac{4 - 2\cos\theta}{5 - 4\cos\theta}\) is the real part but is printed as “Imaginary part”, and the note says “Rearranges imaginary part”; the bounds \(\dfrac{1}{6}\) and \(\dfrac{3}{2}\) are reached at \(\cos\theta = -1\) and \(\cos\theta = 1\), so they are printed as strict inequalities but should be \(\leqslant\))

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

(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

Way 1

M1: Combines the two series by pairing the multiples of \(\theta\) (At least up to \(5\theta\))

A1: Converts to Euler form correctly (At least up to \(5\theta\))

M1: Recognises that C + iS is a convergent geometric series and uses the sum to infinity of a GP

A1*: Reaches the printed answer with no errors

Alternative: Way 2

M1: Combines the two series by pairing the multiples of \(\theta\) (At least up to \(5\theta\))

A1: Converts to power form correctly (At least up to \(5\theta\))

M1: Recognises that C + iS is a convergent geometric series and uses the sum to infinity of a GP

A1*: Reaches the printed answer with no errors

Notes

Way 1

M1: Multiplies numerator and denominator by \(2 - \mathrm{e}^{-4\mathrm{i}\theta}\)

A1: Correct fraction in terms of exponentials

dM1: Converts back to trigonometric form

A1*: Reaches the printed answer with no errors

Alternative: Way 2

M1: Converts back to trigonometric form and realises the need to make the denominator real and multiplies numerator and denominator by the complex conjugate of the denominator which is correct for their fraction

A1: Correct fraction in terms of trigonometric functions

dM1: Uses the correct addition formula to obtain \(\sin 3\theta\) in the numerator

A1*: Reaches the printed answer with no errors

Typical solution

Typical solution

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

Equating real parts

Typical solution

Equating imaginary parts

Typical solution

Typical solution

Also

Typical solution

\(z^6\)

Typical solution

By de Moivre’s theorem,

Typical solution

Typical solution

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

Typical solution

(in addition to given solutions)

Typical solution

By de Moivre’s theorem

Equating real parts

Typical solution

from part (b)

\(\cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}\) \(\quad\therefore \left(\cos\theta - \frac{1}{\sqrt{2}}\right)\) is a factor of \(32\cos^6\theta - 48\cos^4\theta + 18\cos^2\theta - 1\)

Similarly \(\cos\frac{3\pi}{4} = -\frac{1}{\sqrt{2}}\) and \(\left(\cos\theta + \frac{1}{\sqrt{2}}\right)\) is also a factor of the expression.

So \(\left(\cos\theta - \frac{1}{\sqrt{2}}\right)\left(\cos\theta + \frac{1}{\sqrt{2}}\right) = \left(\cos^2\theta - \frac{1}{2}\right)\) is a factor and \((2\cos^2\theta - 1)\) is a factor

Typical solution

Let \(c = \cos\theta\)

For \(\theta = \frac{\pi}{4}\) and \(\theta = \frac{3\pi}{4}\), \(2c^2 - 1 = 0\)

So for the other four roots, which are the cosines of the other four solutions to \(\cos 6\theta = 0\),

Solving this as a quadratic in \(c^2\) gives

So

Of the angles \(\frac{\pi}{12}, \frac{5\pi}{12}, \frac{7\pi}{12}\), and \(\frac{11\pi}{12}\),

\(\frac{11\pi}{12}\) has the negative cosine of the greatest magnitude

Therefore \(\cos\left(\frac{11\pi}{12}\right) = -\sqrt{\frac{2 + \sqrt{3}}{4}}\)

Typical solution

\(\therefore w^n\) satisfies the equation \(z^7 = 1\)

Typical solution

The roots of \(z^7 - 1 = 0\) are \(1\), \(w\), \(w^2\), \(w^3\), \(w^4\), \(w^5\), and \(w^6\)

\(z^6\) term \(= 0 \therefore\) sum of roots \(= 0\)

and

as required.

Typical solution

Typical solution

and

which is the conjugate of \(w\).

Similarly \(\quad w^2 + w^5 = 2\cos\frac{4\pi}{7}\)

and \(\quad w^3 + w^4 = 2\cos\frac{6\pi}{7}\)

From part (b),

as required.

Typical solution

Typical solution

Typical solution

Typical solution

Typical solution

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

B1: Diagram should clearly show equal angular distance between the roots and an equal distance (of 1) from \(O\) to each root.
At least 3 points including \(\omega_0\) and \(\omega_1\) shown.
For this B1 allow an \(n\)-gon with a specific value of \(n\) chosen.

B1: Dealing with modulus (could be incorporated in the below).
\(|\omega_k| = 1\) since it is a root of unity
\(|\omega_1| = 1 \Rightarrow \left|\omega_1^k\right| \left(= |\omega_1|^k\right) = 1\)

B1: Dealing with argument. Accept a well-reasoned argument based on multiplication by \(\omega_1\) representing a pure rotation (by the required angle).
Could be argued by induction if rigorous.

\[\begin{aligned} &\arg(\omega_k) = \frac{2\pi k}{n} \\ &\arg(\omega_1) = \frac{2\pi}{n} \\ &\Rightarrow \arg\left(\omega_1^k\right) = k\arg(\omega_1) = \frac{2\pi k}{n} \end{aligned}\]

Alternative method for last B1B1

B1: \(\omega_1\) can be implied if appearing in the equation below

Notes

M1: Using the identity from (a) and recognising the GP (can be implied by the formula).
\(\displaystyle\sum_{k=0}^{n-1}\omega_k = \displaystyle\sum_{k=0}^{n-1}\omega_1^k\) and recognition of \(\omega_1^n - 1 = (\omega_1 - 1)\left(\omega_1^{n-1} + \omega_1^{n-2} + \ldots + \omega_1 + 1\right)\)

A1: AG so reasoning must be shown,
If GP not recognised then justification for \(\omega_1 - 1 \neq 0\) must also be given.

Notes

B1: AG

Notes

B1: Or the (non-real) roots of unity come in complex conjugate pairs (since they are roots of the real polynomial \(z^n = 1\)).
Could use symmetry of cos function in geometric context (eg \(\cos\frac{2}{5}\pi = \cos\left(2\pi - \frac{2}{5}\pi\right)\) etc)

Notes

(f)(i)

M1: Finding the first argument. Could be embedded.
Could see sum of all 5 real parts.

A1: \(a\) and \(b\) can be embedded.

(f)(ii)

M1: Using correct double angle formula to derive and solve a quadratic equation in \(\cos\frac{2\pi}{5}\) (must have real solutions). Could be BC.

A1: Clear rejection after valid argument leading to correct answer.

Notes

M1: DR. Correctly using formula or completing the square. Condone “\(-6 \pm \sqrt{\Delta}\)” for M1. Condone missing brackets under root if \(-6\) squares to 36 (ie \(\Delta = -196\) rather than \(-268\)).

B1FT: FT their negative discriminant. Writing square root of negative number as the correct multiple of i. “\(-196\)” (or “\(-49\)”) must be seen.

A1: Must be \(a + b\mathrm{i}\).

Alternative method

B1: Using the fact that the roots of a real quadratic form a complex conjugate pair. May be embedded.

M1: Finding the numerical value of the sum and product of the roots.
Could also be found by expanding \((x - (a + b\mathrm{i}))(x - (a - b\mathrm{i}))\) and comparing with equation.
Could also substitute \((a + b\mathrm{i})\) into the equation to derive \((2ab - 6b) = 0\) and \(a^2 - b^2 - 6a + 58 = 0\).

A1: Must be \(a + b\mathrm{i}\). \(a\) real \(\Rightarrow b \neq 0 \Rightarrow a = 3 \Rightarrow b = \pm 7\)

Notes

M1: DR. Using correct formula for argument of complex number with non-zero real and imaginary parts.
Condone \(\tan\alpha = \frac{5\sqrt{12}}{-10} \Rightarrow \alpha = -\frac{\pi}{3}\) or \(\tan^{-1}\frac{5\sqrt{12}}{10} \Rightarrow \alpha = \frac{2\pi}{3}\) or \(-\frac{\pi}{3}\) for M1.

M1: Using De Moivre’s Theorem for their angle. Condone error in modulus if shown.
or using a valid method for finding \(z^5\) explicitly (eg by expansion or by writing \(-10 + 5\sqrt{12}\mathrm{i} = 20\mathrm{e}^{\frac{2}{3}\pi\mathrm{i}}\))
\(z^5 = -1600000 - 1600000\sqrt{3}\,\mathrm{i}\) or \(\left(20^5\right)\mathrm{e}^{5 \times \frac{2}{3}\pi\mathrm{i}}\)

A1: cao

Notes

B1: Or \(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta} = 2\cos\theta\) May use \(z\) without definition

M1: oe, eg. \((2\mathrm{i}\sin\theta)^4 = 16\sin^4\theta = \left(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}\right)^4\). Award this mark for \(\sin\theta\) to the power of four, and for \((2\mathrm{i})^4 = 16\). Note that 16 may appear later.

M1: Expanding \(\left(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}\right)^4\) with correct coefficients.

M1: Grouping terms and using \(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta} = 2\cos\theta\).

A1: cao, from fully correct reasoning.
Allow \(A\), \(B\), \(C\) seen in the expression only.

Notes

B1: Sight of \(\dfrac{\mathrm{d}\left(\sin^{-1}u\right)}{\mathrm{d}u} = \dfrac{1}{\sqrt{1 - u^2}}\)

M1: Uses chain rule

A1: Correct derivative, f’

M1: Uses result from (a)

M1: Uses \(\sin^4\left(\sin^{-1}\left(x^{\frac{1}{5}}\right)\right) = x^{\frac{4}{5}}\)

A1: AG Clearly shown

Notes

M1: For use of part (b) and an upper limit of \(k \lt 1\)
Integral must be found in terms of \(k\) (which could be 1)

M1: For correct use of limits

Notes

B1: AG so must show use of \(|z| = \sqrt{a^2 + b^2}\)

B1: AG. Or \(\theta = \pi - \arctan\left(\frac{\sqrt{3}}{3}\right)\), may be indicated on a diagram, but clear reasoning must be shown (eg. finding complementary angle, or use of Pythagoras’ theorem and then arcsin or arccos)

Alternative method

A1: AG Clearly shown

Notes

B1: For \(r = \sqrt{3}\) oe (including 1.73....)

M1: For their \(\frac{5}{6}\pi + 2\pi n\) from (a) divided by 5 (either in terms of \(n\), or for at least two values of \(n\)).

A1: Allow \(\sqrt{3}\mathrm{e}^{\frac{1}{30}(5 + 12n)\pi\mathrm{i}}\) for \(n = 0, 1, 2, 3, 4\).
Accept only \(r = \sqrt{3}\) or \((3)^{\frac{1}{2}}\)
For last two marks, If M0 then SC B1 for all five roots

Notes

M1: Use of DMT

A1: Must equate terms to complete demonstration

Notes

*M1: Expands correct brackets using binomial theorem. Terms can be unsimplified but must have correct numerical coefficients.
eg \(\left(\mathrm{e}^{2\mathrm{i}\theta} + 1\right)^4\) or \(\left(z + z^{-1}\right)^4\) or \(\left(\dfrac{\sqrt{3}}{2} + 1 + \dfrac{1}{2}\mathrm{i}\right)^4\) etc
if expansion seen in 9(a), must be used in 9(b) to gain mark here

dep*M1: Use of Euler’s formula to convert exponential form to trigonometric form

dep*M1: Taking real parts

dep*M1: Choice of \(\theta\) soi and substituted into identity with their 16.

dep*M1: Gives correct numerical values to all \(\cos\dfrac{n\pi}{6}\) terms. Also dependent on use of Euler’s formula and choice of \(\theta\).

A1: So \(b = 7\) and \(c = 4\) (can be embedded). cao.

Notes

M1: Use of \(z\) and de Moivre  Sight of i is necessary

A1: Both sides

M1: Attempt conversion into sin  soi

A1: All three stated

Alternative method 1

M1: Sight of i in the denominator is necessary

M1: Collection to convert back

A1: All three stated

Alternative method 2

M1: De Moivre

A1: for both

M1: Eliminate \(\sin^3\theta\)

A1: All three stated

(Corrected from the printed mark scheme: Alternative method 2 is printed with \(B = -\dfrac{5}{8}\) and with \(z^3 = (\cos\theta + i\sin\theta)^5\); these are \(B = -\dfrac{5}{16}\) and \((\cos\theta + i\sin\theta)^3\), as typed above.)

Notes

M1: Use of \(\alpha\beta\gamma = \dfrac{3}{2}\) to find the 3rd root.
Alternatively, find \(x^2 - 2x + 3\) and divide

Notes

M1: Multiply out (can be seen in (b))

Alternative method

M1: Use of symmetry forms for roots

(corrected from the printed mark scheme: the first line is printed as \(\alpha + \beta + \gamma = -\dfrac{a}{2}\); the coefficient is \(p\))

Notes

M1: Either \(\alpha\) or \(\beta\) seen in mod/arg form

A1: For both of them – accept exponentials

M1: Derivation of \(\alpha^n\) or \(\beta^n\)

Notes

M1: Correct use of relevant formula(e). Some working must be seen.
Correct answer with no working: M0A0

A1: Not \(\pm 8\) unless later corrected
or eg \(\theta = 8\pi/3\)

B1ft: Modulus of cube root(s) is the cube root of their modulus

B1ft: Argument of (principal) cube root is one third of their argument

M1: Considering further arguments at angular distance \(2\pi\)

A1: or eg \(2\mathrm{e}^{\frac{2}{9}\pi\mathrm{i}},\ 2\mathrm{e}^{\frac{8}{9}\pi\mathrm{i}}\) and \(2\mathrm{e}^{\frac{14}{9}\pi\mathrm{i}}\)
Must be in exponential form, not just \(r =\) and \(\theta =\). Do not condone any missing i’s.

Notes

B1 B1: for one; for all three without extras
ft their angles if \(2\pi/3\) apart.
If valid alternatives, must come from clear explanation/diagram

Notes

M1: Use of binomial

M1: use of De Moivre for any one term

Notes

B1: Conversion soi

M1: Attempt to find at least one root by rooting the modulus and halving the argument soi

A1: Both

Alternative method

B1: Conversion and attempt to square

M1: \(a\) and \(b\). Ignore \(\pm\)

A1: Both

(corrected from the printed mark scheme: the last line of this method is printed as \(5\mathrm{e}^{\frac{1}{4}\pi}\) and \(5\mathrm{e}^{\frac{5}{4}\pi}\), without the \(\mathrm{i}\) in the exponents)

Notes

B1: All three but no extras. Scales etc are not required but if no scale then the lines representing the roots should be at \(45^\circ\) to axis.
Accept points.
No extras
Line representing \(25\mathrm{i}\) must be at least two times as long

Notes

M1: Finding \(\omega^5\)

A1: AG

Use of exponentials is satisfactory. Could be argued backwards

Notes

Alternative: Roots are \(\cos\dfrac{2k\pi}{5} + \mathrm{i}\sin\dfrac{2k\pi}{5}\) for \(k = 2, 3, 4\) and 1 (or \(k = 5\))

Exponentials satisfactory

Notes

M1: Use equation and \(\omega\)

A1: AG

Alternative method

Alternative

sum of roots \(= -\dfrac{b}{a}\) where \(b = 0\) M1 – needs explanation – i.e. coefficient of \(z^4\) term \(= 0\)

Notes

M1: Multiply out

Alternative method

M1: For extraction of \(\omega^2\)

A1: For dealing with the 2

Notes

B1: \(\omega + \dfrac{1}{\omega}\) may be seen in (d)

B1: BC

M1: Equating

A1: For taking the valid value and presenting in correct form oe. No other forms acceptable

(corrected from the printed mark scheme: the printed scheme says “From (iii)”; the quadratic is the one in part (d).)

Notes

*B1: Condone \(2\mathrm{i}\sin\theta = \mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}\)

M1: Raising expression for \(\sin\theta\) to the power 6 and \((2\mathrm{i})^6 = -64\).
Allow use of \(\sin\theta = \frac{\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta}}{2\mathrm{i}}\) for 1st two M marks only

M1: Genuine attempt to use binomial expansion with correct evaluated binomial coefficients. Condone sign errors.
If i omitted from denominator their expression for \(\sin\theta\) then only this M mark can still be awarded

M1: Collecting terms and using \(\mathrm{e}^{\mathrm{i}\phi} + \mathrm{e}^{-\mathrm{i}\phi} = 2\cos\phi\) at least once.

dep*A1: AG. Fully correct argument

Notes

*M1: Choice of \(\theta\) soi and calculation of at least one cos term.

dep*M1: Substitution and calculation of all cos terms. Terms must be shown distinct either in this line or in the form of \(\cos n\frac{\pi}{8}\)

A1: AG Some intermediate working must be seen

Notes

M1: expanding correctly and fully to give at least three terms, allow \(10 - 3\left(\mathrm{e}^{4\mathrm{i}\theta} + \mathrm{e}^{-4\mathrm{i}\theta}\right)\). Must be seen. Condone \(\mathrm{e}^0 = 1\).

A1: www. Condone only incorrect values quoted for \(a\) and \(b\). No intermediate step required.

Notes

M1: at least two terms of series in exponential form soi by correct GP formula

A1: writing series as powers of \(\mathrm{e}^{\mathrm{i}\theta}\) or identifying geometric series with correct first term and common ratio soi by correct GP formula

M1: correct use of sum to infinity formula. Must be seen.

A1: AG www

Notes

M1: multiplying numerator and denominator by a multiple of \(3 - \mathrm{e}^{-4\mathrm{i}\theta}\); multiplication of numerator must be shown but denominator could be given immediately as \(10 - 6\cos 4\theta\).

M1: converting to sine and cosine form after denominator has been simplified to a real expression. Must be seen. No errors allowed FT their expression. Condone only missing brackets around \((-3\theta)\).

A1: AG www. Condone only missing brackets around \((-3\theta)\).

Notes

(a)(i)

B1: or \(w\mathrm{e}^{\frac{2\pi}{5}\mathrm{i}}\), \(w\mathrm{e}^{\frac{4\pi}{5}\mathrm{i}}\) and \(w\mathrm{e}^{\frac{-4\pi}{5}\mathrm{i}}\) (or \(w\mathrm{e}^{\frac{6\pi}{5}\mathrm{i}}\) or \(w^*\)) oe. Allow misattribution between C, D, E.

(a)(ii)

B1: allow any variable for \(z\)
or \(z^5 = \cos 2\pi + \mathrm{i}\sin 2\pi\)
or \(z^5 = \cos 2k\pi + \mathrm{i}\sin 2k\pi\) (or \(\mathrm{e}^{2k\pi\mathrm{i}}\)) provided \(k \in \mathbb{Z}\) seen

(a)(iii)

M1: correct use of geometric series formula in terms of \(w\) or exponentials. Cannot be implied.

A1: \(w^5 = 1\) or \(1 - \mathrm{e}^{2\pi\mathrm{i}}\) seen before completion.

Alternative method

“sum of roots \(= -\frac{b}{a} = 0\)” alone is insufficient

Notes

(b)(ii)

M1: or \(\begin{pmatrix} \cos\frac{2\pi}{5} - 1 \\ \sin\frac{2\pi}{5} \end{pmatrix}\); or equivalent for \(1 - w\). Soi by correct modulus calculation.

M1: finding the modulus or modulus² of their \(w - 1\) (or \(1 - w\))

A1: or \(\sqrt{2 - 2\cos\frac{2\pi}{5}}\) or correct expression for AB²

M1: correct use of double angle formula, must be seen. Accept \(2\left(1 - \cos\frac{2\pi}{5}\right) = 2\left(2\sin^2\frac{\pi}{5}\right)\) without intermediate step but not \(2 - 2\cos\frac{2\pi}{5} = 2\left(2\sin^2\frac{\pi}{5}\right)\).

A1: AG

Alternative method 1

M2: correct use of cosine rule; or correct expression for AB

A1: simplified expression for AB² or AB

M1: correct use of double angle formula, must be seen. Accept \(2\left(1 - \cos\frac{2\pi}{5}\right) = 2\left(2\sin^2\frac{\pi}{5}\right)\) without intermediate step but not \(2 - 2\cos\frac{2\pi}{5} = 2\left(2\sin^2\frac{\pi}{5}\right)\).

A1: AG

Alternative method 2

M1: may be stated or seen in diagram; condone missing labels if the triangle is clearly isosceles

M1: may be stated or seen in diagram; condone missing labels

A1: must be seen

A2: AG

Alternative method 3

M1*: using sine rule correctly

A1: correct expression for AB

M1dep: correct use of double angle formula

M1dep: using \(\sin\theta = \cos\left(\frac{\pi}{2} - \theta\right)\)

A1: AG

Notes

(a)(i)

B1: or \(\frac{1}{3} : 1\) only

Notes

(b)(i)

M1: Expanding correctly to give at least three terms. Condone \(e^0 = 1\).

A1: www. Condone only incorrect values quoted for \(a\) and \(b\). Intermediate step not required here.

(b)(ii)

M1: At least two terms of series in exponential form soi by correct GP formula or \(\frac{z}{1 - z}\) seen. Condone modulus-argument form.

A1: Using sum to infinity formula correctly

M1*: Multiplying their numerator and denominator by a multiple of \(3 - \mathrm{e}^{-\mathrm{i}\theta}\). Must be a clear attempt at a sum to infinity.

A1: oe

M1dep: \(\mathrm{e}^{\mathrm{i}\theta} = \cos\theta + \mathrm{i}\sin\theta\) used when denominator has been simplified to a real expression. No errors allowed.

A1: AG www

Notes

B1: seen at any stage

M1: (2nd) expansion of \(\left(z - \frac{1}{z}\right)^5\)

A1: (1st) correct expansion with \(z^n\), \(\frac{1}{z^n}\) terms paired and factorised

M1: (3rd) FT their expansion

A1: (2nd) i must appear in each term

A1: (3rd) www

Alternative method 1

M1: (2nd) expansion of \(\left(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta}\right)^5\)

A1: (1st) correct expansion with \(\mathrm{e}^{\mathrm{i}\theta}\), \(\mathrm{e}^{-\mathrm{i}\theta}\) terms paired and factorised

M1: (3rd) FT their expansion (corrected from the printed mark scheme, which has \(\mathrm{e}^{\mathrm{i}\theta} - \mathrm{e}^{-\mathrm{i}\theta} = 2\mathrm{i}\sin n\theta\); the identity is \(\mathrm{e}^{\mathrm{i}n\theta} - \mathrm{e}^{-\mathrm{i}n\theta} = 2\mathrm{i}\sin n\theta\))

A1: (2nd) i must appear in each term

A1: (3rd) www

Alternative method 2

B1: using binomial expansion and de Moivre’s theorem

M1*: (1st) correct expression for \(\sin 5\theta\) and substituting in \(\cos^2\theta = 1 - \sin^2\theta\)

A1: (1st) or \(16\sin^5\theta = \sin 5\theta + 20\sin^3\theta - 5\sin\theta\) or any correct rearrangement

M1*: (2nd) using binomial expansion and de Moivre’s theorem; correct expression for \(\sin 3\theta\) in terms of \(\sin\theta\) and \(\cos\theta\)

A1: (2nd) or \(\sin 3\theta = 3\sin\theta - 4\sin^3\theta\) or any correct rearrangement

M1dep: substituting expression for \(\sin^3\theta\) into \(\sin^5\theta\).

A1: (3rd) www

Notes

M1: condone \(-\pi\) for \(\pi\)

B1: soi

A1: 1 correct

A1: all correct or \(2\mathrm{e}^{7\mathrm{i}\pi/6}, 2\mathrm{e}^{3\mathrm{i}\pi/2}, 2\mathrm{e}^{11\mathrm{i}\pi/6}\)

Notes

M1: six roots lying on approximate circle centre O

A1: forming an approximate regular hexagon

A1: root at \(2\mathrm{i}\) or \(-2\mathrm{i}\) oe indicated. If part (a) contains incorrect roots do not award final A1.

Notes

M1: For expanding correctly

Notes

M1: At least 2 terms of C + iS soi by correct GP formula

M1: sum to infinity of GP formula for their series (which must be geometric)

A1: oe

M1*: multiply numerator and denominator by \(3 - \mathrm{e}^{-2i\theta}\)

M1dep*: \(\mathrm{e}^{i\theta} = \cos\theta + i\sin\theta\) used when denominator has been simplified to a real expression

A1: AG

Notes

M1: on a circle centre O

A1: form an approximate equilateral triangle
B and C must be labelled

Notes

M1: sum of GP used or other correct method

Alternative solution

Notes

B1: Exponential or modulus argument form soi

B1: Soi

B1B1: B1B0 if two out of three roots given

Notes

B1: \(z = 1\)

B1: other two roots forming correct equilateral triangle

Notes

(b)(i)

B1ft: if roots correct but in cis form, withhold one mark only

(b)(iii)

M1: sum of GP formula. May use \(\cos\theta + i\sin\theta\) but must take out factor for M1

(b)(iv)

A1: AG

Notes

M1 M1: binomial expansions oe; expanding the whole expression
or \(\left[\left(z + \dfrac{1}{z}\right)\left(z - \dfrac{1}{z}\right)\right]^3\) etc
award second M1 if changes into trig and makes some attempt at using the addition formulae

Notes

M1: modulus 2

Notes

M1: taking the \(6^{\text{th}}\) power of one of the midpoints

Notes

M1: \(\mathrm{e}^{\mathrm{i}\theta}.\mathrm{e}^{-\mathrm{i}\theta} = 1\)

M1: \(\mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta} = 2\cos\theta\) used

A1: NB AG

Notes

M1: at least 2 terms

A1: correct (\(n\)th term soi)

M1: sum of GP (condone \(S_\infty\))

A1: correct expression

M1: \(\times\) top and bottom by complex conjugate

M1: expand brackets; allow 1 slip, not on \(S_\infty\)

M1: taking real part; not on \(S_\infty\)

A1cao: NB AG; need evidence of clearing subsidiary denominators

Alternative solution

M1: substituting for cosines in terms of \(\mathrm{e}^{\mathrm{i}\theta}\)s; M1 A1: sum of GP; M1 A1: combining fractions; M1 A1: expanding; M1: expressing in cosines; A1: NB AG

(corrected from the printed mark scheme: in the printed alternative the first line has a stray factor \(\mathrm{e}^{\mathrm{i}\theta}\) after the first term and is cut off after \(\frac{1}{2^n}\); the second and third lines omit the overall factor \(\frac{1}{2}\); and the constant term in the fourth and fifth lines is printed as \(-2\) instead of \(-1\).)

Notes

M1: expanding \((-1 + \mathrm{i})^3\)

A1: correct expression; or \(-2\mathrm{i}(-1 + \mathrm{i})\)

Alternative solution

B1: \(|z| = \sqrt{2}\), \(\arg z = 3\pi/4\); or \(\sqrt{2}\,\mathrm{e}^{3\mathrm{i}\pi/4}\)

M1: \(|z^3| = |z|^3\), \(\arg(z^3) = 3\arg z\)

(corrected from the printed mark scheme: the last line is printed as \(2\sqrt{2}(1/\sqrt{2} + 1/\sqrt{2})\), missing the \(\mathrm{i}\).)

Notes

B1: 3 roots with modulus \(\sqrt{2}\) oe; oe eg \(8^{1/6}\)

B1B1B1: oe, e.g. \(8^{1/6}\mathrm{e}^{17\mathrm{i}\pi/12}\), etc; \(0 \lt \arg \lt 2\pi\) or \(-\pi \lt \arg \lt \pi\)