Question

(a) $\sin \frac{6\pi }{5}+i(1+\cos \frac{6\pi }{5})$
(b) $\frac{i-1}{i(1-\cos \frac{2\pi }{5})+\sin \frac{2\pi }{5}}$

Answer 1

$2\sin \frac{3\pi }{5}\cos \frac{3\pi }{5}+i.2{\cos }^2\frac{3\pi }{5}$
$2\cos \frac{3\pi }{5}\left[\sin \frac{3\pi }{5}+i\cos \frac{3\pi }{5}\right]$
But $r=2\cos \frac{3\pi }{5}=-ive$
Hence are rewrite as under:
$z=-2\cos \frac{3\pi }{5}[-\sin \frac{3\pi }{5}-i\cos \frac{3\pi }{5}]$
$=-2\cos \frac{3\pi }{5}\left[\cos (\frac{3\pi }{2}-\frac{3\pi }{5})+i\sin (\frac{3\pi }{2}-\frac{3\pi }{5})\right]$
$=-2\cos \frac{3\pi }{5}\left[\cos \frac{9\pi }{10}+i\sin \frac{9\pi }{10}\right]$
$r=-2\cos \frac{3\pi }{5}(+ive)$
and $\theta =\frac{9\pi }{10}$ and which lives between $-\pi$ and $\pi$