# Question 2:

## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\left(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\right)^4 = \cos\frac{20\pi}{12}+i\sin\frac{20\pi}{12}$ and $\left(\cos\frac{\pi}{3}-i\sin\frac{\pi}{3}\right)^3 = \cos(-\pi)+i\sin(-\pi)$ | M1/A1 | One correct use of de Moivre; allow $e^{i\theta}$ for $\cos\theta+i\sin\theta$ throughout |
| $(z_1=)\frac{\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3}}{\cos(-\pi)+i\sin(-\pi)} = \cos\left(\frac{5\pi}{3}-(-\pi)\right)+i\sin\left(\frac{5\pi}{3}-(-\pi)\right)$ | M1 | Correct method for division of two complex numbers; subtraction of arguments |
| $= \cos\frac{8\pi}{3}+i\sin\frac{8\pi}{3} = \cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ | A1* | Must have suitable intermediate step; do not allow going straight to $\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}$ |

## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $|z-z_1| \leqslant 1$ and $0 \leqslant \arg(z-z_1) \leqslant \frac{3\pi}{4}$ | | |
| Circle in any position (may just see the minor arc) | M1 | Accept arc if clearly arc of circle |
| Pair of half-lines in correct directions from centre, one with negative gradient, one parallel to $x$-axis (but not the $x$-axis itself) | M1 | If full lines used, M marks implied by shading |
| Area shaded inside circle between the two half-lines, anticlockwise from horizontal to negative gradient line | M1 | Half-line need not stem from circle centre for this mark |
| Fully correct shaded sector in quadrants 1 and 2; correct gradients ($-1$ and $0$); circle centre in quadrant 2 | A1 | Circle passing roughly through origin if whole circle shown |

## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\arctan\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\right) = \frac{\pi}{3}$ (or $60°$) | M1 A1 | Identifies correct point at "3 o'clock" position; attempts relevant angle; accept $\arctan\pm\frac{c}{d}$ or $\arctan\pm\frac{d}{c}$ |

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