2019 H2 Mathematics Paper 1 Question 9

Complex Numbers

Answers

(ib)
itan⁡12θ{\mathrm{i} \tan \frac{1}{2} \theta}

(ii)

∣z−3i1+3iz∣=1{\left|\frac{z-3\mathrm{i}}{1+3\mathrm{i}z} \right|=1}

Full solutions

(ia)
w+1w=eiθ+1eiθ=eiθ+e−iθ=(cos⁡θ+isin⁡θ)+(cos⁡θ−isin⁡θ)=2cos⁡θ∈R\begin{align*} & w + \frac{1}{w} \\ & = \mathrm{e}^{\mathrm{i}\theta} + \frac{1}{\mathrm{e}^{\mathrm{i}\theta}} \\ & = \mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{-\mathrm{i}\theta} \\ &= (\cos \theta + \mathrm{i}\sin \theta) + (\cos \theta - \mathrm{i}\sin \theta) \\ &= 2 \cos \theta \in \mathbb{R} \end{align*}
Hence w+1w{\displaystyle w + \frac{1}{w}} is a real number ■{\blacksquare}
(ib)
w−1w+1=eiθ−1eiθ+1=eiθ2eiθ2−eiθ2e−iθ2eiθ2eiθ2+eiθ2e−iθ2=eiθ2(eiθ2−e−iθ2)eiθ2(eiθ2+e−iθ2)=2i Im(eiθ2)2 Re(eiθ2)=2isin⁡θ22cos⁡θ2=itan⁡12θ  ■\begin{align*} & \frac{w-1}{w+1} \\ & = \frac{\mathrm{e}^{\mathrm{i}\theta} - 1 }{\mathrm{e}^{\mathrm{i}\theta} + 1 } \\ & = \frac{\mathrm{e}^{\mathrm{i}\frac{\theta}{2}}\mathrm{e}^{\mathrm{i}\frac{\theta}{2}} - \mathrm{e}^{\mathrm{i}\frac{\theta}{2}}\mathrm{e}^{-\mathrm{i}\frac{\theta}{2}} }{\mathrm{e}^{\mathrm{i}\frac{\theta}{2}}\mathrm{e}^{\mathrm{i}\frac{\theta}{2}} + \mathrm{e}^{\mathrm{i}\frac{\theta}{2}}\mathrm{e}^{-\mathrm{i}\frac{\theta}{2}} } \\ & = \frac{\mathrm{e}^{\mathrm{i}\frac{\theta}{2}}\left( \mathrm{e}^{\mathrm{i}\frac{\theta}{2}} - \mathrm{e}^{-\mathrm{i}\frac{\theta}{2}} \right)}{\mathrm{e}^{\mathrm{i}\frac{\theta}{2}}\left( \mathrm{e}^{\mathrm{i}\frac{\theta}{2}} + \mathrm{e}^{-\mathrm{i}\frac{\theta}{2}} \right) } \\ & = \frac{2\mathrm{i} \, \textrm{Im}\left( \mathrm{e}^{\mathrm{i}\frac{\theta}{2}} \right)}{2 \, \textrm{Re}\left( \mathrm{e}^{\mathrm{i}\frac{\theta}{2}} \right)} \\ & = \frac{2 \mathrm{i} \sin \frac{\theta}{2}}{2 \cos \frac{\theta}{2}} \\ &= \mathrm{i} \tan {\textstyle \frac{1}{2} \theta} \; \blacksquare \end{align*}

(ii)

Let z=x+yi{z=x+y\mathrm{i}}
∣z∣=1⇒x2+y2=1\begin{align*} & \left| z \right| = 1 \\ & \Rightarrow \sqrt{x^2 + y^2 = 1} \end{align*}
x2+y2=1\begin{equation}x^2+y^2=1\end{equation}
∣z−3i1+3iz∣=∣x+yi−3i∣∣1+3i(x+yi)∣=x2+(y−3)2∣1−3y+3xi∣=x2+y2−6y+9(1−3y)2+(3x)2=1−6y+91−6y+9y2+9x2=10−6y1−6y+9=1  ■\begin{align*} & \left| \frac{z-3\mathrm{i}}{1+3\mathrm{i}z} \right| \\ & = \frac{\left| x + y\mathrm{i}-3\mathrm{i} \right|}{\left| 1 + 3\mathrm{i}(x+y\mathrm{i}) \right|} \\ & = \frac{\sqrt{x^2 + (y-3)^2}}{\left| 1 - 3y + 3x\mathrm{i} \right|} \\ & = \frac{\sqrt{x^2 + y^2 - 6y + 9}}{\sqrt{(1-3y)^2 + (3x)^2}} \\ & = \frac{\sqrt{1 - 6y + 9}}{\sqrt{1 - 6y + 9y^2 + 9x^2}} \\ & = \frac{\sqrt{10 - 6y }}{\sqrt{1 - 6y + 9}} \\ & = 1 \; \blacksquare \end{align*}