(a)$|z|=1, \arg z=\frac{\pi}{4}$, $z^2=i$

(b)(ii) 0

(a) $\cos \left(\frac{\pi}{16}\right)+i \sin \left(\frac{\pi}{16}\right)=e^{i \frac{\pi}{16}}$
$\cos \left(\frac{\pi}{8}\right)-\mathrm{i} \sin \left(\frac{\pi}{8}\right)=\cos \left(-\frac{\pi}{8}\right)+\mathrm{i} \sin \left(-\frac{\pi}{8}\right)=\mathrm{e}^{-\frac{\pi}{8}}$
$\therefore z=\frac{\left(\mathrm{e}^{\left.\mathrm{i} \frac{\pi}{16}\right)^2}\right.}{\mathrm{e}^{-\frac{\pi}{8} \mathrm{i}}}=\frac{\mathrm{e}^{\mathrm{i} \frac{\pi}{8}}}{\mathrm{e}^{-\frac{\pi}{8} \mathrm{i}}}=\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{8}+\frac{\pi}{8}\right)}=\mathrm{e}^{\frac{\pi}{\mathrm{i}}}$
Hence, $|z|=1, \arg z=\frac{\pi}{4}$
$z^2=\left(e^{i \frac{\pi}{4}}\right)^2=e^{i\left(\frac{\pi}{2}\right)}=\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}=0+1 i=i$

(b) (i)$(\cos \theta+\mathrm{i} \sin \theta)(1+\cos \theta-\mathrm{i} \sin \theta)$

$= \mathrm{e}^{\mathrm{i} \theta}(1+\cos (-\theta)+\mathrm{i} \sin (-\theta))$

$= \mathrm{e}^{\mathrm{i} \theta}\left(1+\mathrm{e}^{\mathrm{-i} \theta}\right)$

$= \mathrm{e}^{\mathrm{i} \theta}+1$

$= 1+\cos \theta+\mathrm{i} \sin \theta$

(ii) Let $\theta=\frac{\pi}{4}$ in b(i),

$\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)\left(1+\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)=1+\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}$

$\left(\mathrm{e}^{\mathrm{i} \frac{\pi}{4}}\right)\left(1+\mathrm{e}^{-\frac{\pi}{4}i}\right)=1+\mathrm{e}^{\mathrm{i} \frac{\pi}{4}}$ ——(*)

Since $z=\mathrm{e}^{\mathrm{i} \frac{\pi}{4}}$, we have $z^*=\mathrm{e}^{-\frac{\pi}{4} \mathrm{i}}$

$\therefore(*) z\left(1+z^*\right)=1+z$

$\Rightarrow(1+z)^4=z^4\left(1+z^*\right)^4$

$(1+z)^4+\left(1+z^*\right)^4$ = $z^4\left(1+z^*\right)^4$ + $\left(1+z^*\right)^4$

$=\left(1+z^*\right)^4\left(z^4+1\right)$

$=\left(1+z^*\right)^4(-1+1)$

$\left(\because z^4=\left(\mathrm{e}^{i \pi}\right)^4=\mathrm{e}^{\mathrm{i} \pi}=-1\right)$

$=\left(1+z^*\right)^4(0)$

$=0$

This is a pretty tough question. To use the “hence” method, one has to observe that with $\theta$ as $\frac{\pi}{4}$, b(i) reads $z\left(1+z^*\right)=1+z$ with $z$ as $\mathrm{e}^{\mathrm{i}^{\frac{\pi}{4}}}$.