Specimen 2017 H2 Mathematics Paper 1 Question 7

Complex Numbers

Answers

(a)

z=12−32i,{z=\frac{1}{2} - \frac{3}{2} \mathrm{i},}
w=35+15i.{w=\frac{3}{5} + \frac{1}{5} \mathrm{i}.}

(b)

∣1−u2∣=2sin⁡θ,{|1-u^2| = 2 \sin \theta,}
arg⁡(1−u2)=θ−π2{\arg(1-u^2) = \theta - \frac{\pi}{2}}

Full solutions

(a)

2iz+(1−2i)w=4(1+i)z+(2+i)w=3\begin{align} && \quad 2 \mathrm{i}z + (1 - 2 \mathrm{i})w &= 4 \\ && \quad (1 + \mathrm{i})z + (2 + \mathrm{i})w &= 3 \end{align}
From (1),{(1),}
z=4−(1−2i)w2i\begin{gather} z = \frac{4-(1 - 2 \mathrm{i})w}{2 \mathrm{i}} \end{gather}
Substituting (3){(3)} into (2){(2)},
(1+i)4−(1−2i)w2i+(2+i)w=3(4+4i)−(1+i)(1−2i)w+(2i)(2+i)w=6i−(3−i)w+(−2+4i)w=−4+2i(−5+5i)w=−4+2i\begin{gather*} (1 + \mathrm{i})\frac{4-(1 - 2 \mathrm{i})w}{2 \mathrm{i}} + (2 + \mathrm{i})w = 3 \\ (4 + 4 \mathrm{i}) - (1 + \mathrm{i})(1 - 2 \mathrm{i}) w + (2 \mathrm{i})(2 + \mathrm{i}) w = 6 \mathrm{i} \\ -(3 - \mathrm{i})w + (- 2 + 4 \mathrm{i})w = - 4 + 2 \mathrm{i}\\ (- 5 + 5 \mathrm{i})w = - 4 + 2 \mathrm{i} \\ \end{gather*}
w=−4+2i−5+5i×−5−5i−5−5i=30+10i50=35+15i  ■\begin{align*} w &= \frac{- 4 + 2 \mathrm{i}}{- 5 + 5 \mathrm{i}} \times \frac{- 5 - 5 \mathrm{i}}{- 5 - 5 \mathrm{i}} \\ &= \frac{30 + 10 \mathrm{i}}{50} \\ & = \frac{3}{5} + \frac{1}{5} \mathrm{i} \; \blacksquare \end{align*}
Substituting into (3),{(3),}
z=4−(1−2i)(35+15i)2i=4−(1−i)2i=3+i2i×ii=−1+3i−2=12−32i  ■\begin{align*} z &= \frac{4-(1 - 2 \mathrm{i})(\frac{3}{5} + \frac{1}{5} \mathrm{i})}{2 \mathrm{i}} \\ &= \frac{4-(1 - \mathrm{i})}{2 \mathrm{i}} \\ &= \frac{3 + \mathrm{i}}{2 \mathrm{i}} \times \frac{\mathrm{i}}{\mathrm{i}} \\ &= \frac{- 1 + 3 \mathrm{i}}{- 2}\\ & = \frac{1}{2} - \frac{3}{2} \mathrm{i} \; \blacksquare \end{align*}

(b)

1−u2=1−(cos⁡θ+isin⁡θ)2=1−(eiθ)2=1−e2θi=eiθe−θi−eiθeiθ=eiθ(e−θi−eiθ)=eiθ(−2isin⁡θ)=−2iusin⁡θ  ■\begin{align*} 1- u^2 &= 1 - (\cos \theta + \mathrm{i} \sin \theta)^2 \\ &= 1 - ( \mathrm{e}^{\mathrm{i}\theta})^2 \\ &= 1 - \mathrm{e}^{2 \theta\mathrm{i}} \\ &= \mathrm{e}^{\mathrm{i}\theta} \mathrm{e}^{- \theta\mathrm{i}} - \mathrm{e}^{\mathrm{i}\theta} \mathrm{e}^{\mathrm{i}\theta} \\ &= \mathrm{e}^{\mathrm{i}\theta} ( \mathrm{e}^{-\theta\mathrm{i}}- \mathrm{e}^{\mathrm{i}\theta}) \\ &= \mathrm{e}^{\mathrm{i}\theta} (-2\mathrm{i}\sin\theta)\\ &= -2\mathrm{i}u\sin\theta \; \blacksquare \end{align*}
∣1−u2∣=∣−2iusin⁡θ∣=∣−2i∣∣u∣∣sin⁡θ∣=2sin⁡θ  ■\begin{align*} | 1 - u^2 | &= |-2\mathrm{i}u\sin\theta | \\ &= |-2\mathrm{i}| |u| |\sin \theta| \\ &= 2 \sin \theta \; \blacksquare \end{align*}
arg⁡(1−u2)=arg⁡(−2iusin⁡θ)=arg⁡(−2i)+arg⁡(u)+arg⁡(sin⁡θ)=−π2+θ+0=θ−π2  ■\begin{align*} \arg ( 1 - u^2 ) &= \arg (-2\mathrm{i}u\sin\theta) \\ &= \arg(-2\mathrm{i}) + \arg(u) + \arg(\sin \theta) \\ &= -\frac{\pi}{2} + \theta + 0 \\ &= \theta - \frac{\pi}{2} \; \blacksquare \end{align*}