Question

State true or false:

Square root of $\sqrt{1+i}$ is $2^{1/4}(\cos \frac{\pi }{8}+i\sin \frac{\pi }{8})$.

Answer 1

$1\pm i=\sqrt{2}(\cos \frac{\pi }{4}\pm i\sin \frac{\pi }{4})$
$=\sqrt{2}[\cos (2n\pi +\frac{\pi }{4})\pm i\sin (2n\pi +\frac{\pi }{4})]$
$=\sqrt{2}[\cos (8n+1)\frac{\pi }{4}\pm i\sin (8n+1)\frac{\pi }{4}]$
$\therefore \sqrt{1\pm i}=2^{1/4}[\cos (8n+1)\frac{\pi }{8}\pm i\sin (8n+1)\frac{\pi }{8}]$
where n=0, 1
$=2^{1/4}[\cos \frac{\pi }{8}\pm i\sin \frac{\pi }{8}]$ for n=0
$=2^{1/4}[\cos (\pi +\frac{\pi }{8})\pm i\sin (\pi +\frac{\pi }{8})]$ for n=1
$=2^{1/4}(\cos \frac{\pi }{8}\pm i\sin \frac{\pi }{8})$