2013 H2 Mathematics Paper 1 Question 8

Complex Numbers

Answers

(i)

∣w∣=2r,  arg⁡w=θ−π3{\left|w\right| = 2r, \; \arg w = \theta - \frac{\pi}{3}}

(ii)

Out of syllabus

(iii)

θ=π24{\theta = \frac{\pi}{24}}

Full solutions

(i)

w=(1−3i)z=(2 ei−π3)(reiθ)=2rei(θ−π3)\begin{align*} w &= (1 - \sqrt{3} \mathrm{i})z \\ &= \left(2\,\mathrm{e}^{ \mathrm{i} \frac{- \pi}{3} }\right)\left( r \mathrm{e}^{\mathrm{i}\theta}\right) \\ &= 2r \mathrm{e}^{\mathrm{i}(\theta - \frac{\pi}{3})} \end{align*}
∣w∣=2r  ■arg⁡w=θ−π3  ■\begin{gather*} \left|w\right| = 2r \; \blacksquare \\ \arg w = \theta - \frac{\pi}{3} \; \blacksquare \end{gather*}

(ii)

Out of syllabus

(iii)

arg⁡(z10w2)=π10arg⁡z−2arg⁡w=π10θ−2(θ−π3)=π\begin{gather*} \arg \left( \frac{z^{10}}{w^2} \right) = \pi \\ 10\arg z - 2 \arg w = \pi \\ 10 \theta - 2 \left( \theta - \frac{\pi}{3} \right) = \pi \\ \end{gather*}
8θ=π3θ=π24  ■\begin{align*} 8 \theta &= \frac{\pi}{3} \\ \theta &= \frac{\pi}{24} \; \blacksquare \end{align*}