2021 H2 Mathematics Paper 1 Question 4

Complex Numbers

Answers

(a)

∣z∣=1{\left|z\right|=1}
arg⁡z=π4{\arg z=\frac{\pi}{4}}
z2=i{z^2=\mathrm{i}}
(bii)
0{0}

Full solutions

(a)

z=(eiπ16)2ei−π8=eiπ8ei−π8=eiπ4\begin{align*} z &= \frac{\left(\mathrm{e}^{ \mathrm{i} \frac{\pi}{16} }\right)^2}{\mathrm{e}^{ \mathrm{i} \frac{- \pi}{8} }} \\ &= \frac{\mathrm{e}^{ \mathrm{i} \frac{\pi}{8} }}{\mathrm{e}^{ \mathrm{i} \frac{- \pi}{8} }} \\ &= \mathrm{e}^{ \mathrm{i} \frac{\pi}{4} } \end{align*}
Hence ∣z∣=1  ■{\left|z\right|=1 \; \blacksquare} and arg⁡z=π4  ■{\arg z = \frac{\pi}{4} \; \blacksquare}
z2=(eiπ4)2=eiπ2=i  ■\begin{align*} z^2 &= \left( \mathrm{e}^{ \mathrm{i} \frac{\pi}{4} } \right)^2 \\ &= \mathrm{e}^{ \mathrm{i} \frac{\pi}{2} } \\ &= \mathrm{i} \; \blacksquare \end{align*}
(bi)
(cos⁡θ+isin⁡θ)(1+cos⁡θ−isin⁡θ)=eiθ(1+e−iθ)=eiθ+eiθe−iθ=eiθ+1=1+cos⁡θ+isin⁡θ  ■\begin{align*} & (\cos \theta + \mathrm{i}\sin\theta)(1+\cos \theta - \mathrm{i}\sin\theta) \\ &= \mathrm{e}^{\mathrm{i}\theta}(1+\mathrm{e}^{-\mathrm{i}\theta}) \\ &= \mathrm{e}^{\mathrm{i}\theta} + \mathrm{e}^{\mathrm{i}\theta}\mathrm{e}^{-\mathrm{i}\theta} \\ &= \mathrm{e}^{\mathrm{i}\theta} + 1 \\ &= 1 + \cos \theta + \mathrm{i}\sin\theta \; \blacksquare \end{align*}
(bii)
From (a) and (bi), we have
z(1+z∗)=1+zz2=i\begin{align} &&\quad z(1+z^*) &= 1 + z \\ &&\quad z^2 &= \mathrm{i} \end{align}
(1+z)4+(1+z∗)4=(z(1+z∗))4+(1+z∗)4=(1+z∗)4(z4+1)=(1+z∗)4((z2)2+1)=(1+z∗)4(i2+1)=0  ■\begin{align*} & (1+z)^4 + (1+z^*)^4 \\ & = \Big( z(1+z^*) \Big)^4 + (1+z^*)^4 \\ &= (1+z^*)^4 (z^4+1) \\ &= (1+z^*)^4 \Big((z^2)^2+1\Big) \\ &= (1+z^*)^4 (\mathrm{i}^2+1) \\ &= 0 \; \blacksquare \end{align*}