2020 H2 Mathematics Paper 1 Question 4

Complex Numbers

Answers

(i)

122(cos⁡5π12+isin⁡5π12){\frac{1}{2} \sqrt{2} \left( \cos \frac{5 \pi}{12} + \mathrm{i} \sin \frac{5 \pi}{12} \right)}

(ii)

2(cos⁡π12+isin⁡π12),{\sqrt{2} \left( \cos \frac{\pi}{12} + \mathrm{i} \sin \frac{\pi}{12} \right), } 2(cos⁡−11π12+isin⁡−11π12){\sqrt{2} \left( \cos \frac{- 11 \pi}{12} + \mathrm{i} \sin \frac{- 11 \pi}{12} \right)}

Full solutions

(i)

z1=1+3i=2 eiπ3z2=1−i=2 ei−π4z3=2(cos⁡π6+isin⁡π6)=2 eiπ6\begin{align*} z_1 &= 1 + \sqrt{3} \mathrm{i} \\ &= 2\,\mathrm{e}^{ \mathrm{i} \frac{\pi}{3} } \\ z_2 &= 1 - \mathrm{i} \\ &= \sqrt{2}\,\mathrm{e}^{ \mathrm{i} \frac{- \pi}{4} } \\ z_3 &= 2 \left( \cos \frac{\pi}{6} + \mathrm{i} \sin \frac{\pi}{6} \right) \\ &= 2\,\mathrm{e}^{ \mathrm{i} \frac{\pi}{6} } \end{align*}
z1z2z3=2 eiπ3(2 ei−π4)(2 eiπ6)=2 eiπ322 ei−π12=ei(π3+π12)2=122 ei5π12=122(cos⁡5π12+isin⁡5π12)  ■\begin{align*} \frac{z_1}{z_2z_3} &= \frac{2\,\mathrm{e}^{ \mathrm{i} \frac{\pi}{3} }}{(\sqrt{2}\,\mathrm{e}^{ \mathrm{i} \frac{- \pi}{4} })(2\,\mathrm{e}^{ \mathrm{i} \frac{\pi}{6} })} \\ &= \frac{2\,\mathrm{e}^{ \mathrm{i} \frac{\pi}{3} }}{2 \sqrt{2}\,\mathrm{e}^{ \mathrm{i} \frac{- \pi}{12} }} \\ &= \frac{\mathrm{e}^{\mathrm{i} (\frac{\pi}{3}+\frac{\pi}{12})}}{\sqrt{2}} \\ &= \frac{1}{2} \sqrt{2}\,\mathrm{e}^{ \mathrm{i} \frac{5 \pi}{12} } \\ &= \frac{1}{2} \sqrt{2} \left( \cos \frac{5 \pi}{12} + \mathrm{i} \sin \frac{5 \pi}{12} \right) \; \blacksquare \end{align*}

(ii)

∣z1z4z2z3∣=1∣z1z2z3∣∣z4∣=1122r=1r=2\begin{align*} \left| \frac{z_1z_4}{z_2z_3} \right| &= 1 \\ \left| \frac{z_1}{z_2z_3} \right| \left| z_4 \right| &= 1 \\ \frac{1}{2} \sqrt{2} r &= 1 \\ r &= \sqrt{2} \end{align*}
Since z1z4z2z3{\displaystyle \frac{z_1z_4}{z_2z_3}} is purely imaginary, for k∈Z,{k \in \mathbb{Z},}
arg⁡(z1z4z2z3)=π2+kπarg⁡z4+arg⁡(z1z2z3)=π2+kπθ+5π12=π2+kπθ=π12+kπ\begin{gather*} \arg \left( \frac{z_1z_4}{z_2z_3} \right) = \frac{\pi}{2} + k \pi \\ \arg z_4 + \arg \left( \frac{z_1}{z_2z_3} \right) = \frac{\pi}{2} + k \pi \\ \theta + \frac{5 \pi}{12} = \frac{\pi}{2} + k \pi \\ \theta = \frac{\pi}{12} + k \pi \end{gather*}
Since −π<θ≤π,{-\pi < \theta \leq \pi, } k=0,−1{k = 0, -1}
θ=π12   or   θ=−11π12\theta = \frac{\pi}{12} \; \textrm{ or } \; \theta = \frac{- 11 \pi}{12}
z4=2(cos⁡π12+isin⁡π12)  ■or z4=2(cos⁡−11π12+isin⁡−11π12)  ■\begin{align*} z_4 &= \sqrt{2} \left( \cos \frac{\pi}{12} + \mathrm{i} \sin \frac{\pi}{12} \right) \; \blacksquare \\ \textrm{or } \quad z_4 &= \sqrt{2} \left( \cos \frac{- 11 \pi}{12} + \mathrm{i} \sin \frac{- 11 \pi}{12} \right) \; \blacksquare \\ \end{align*}