Question

Find the value of $\sin \left(\log i^i\right)$

Answer 1

Let $a+ib=\sin \left(\log i^i\right)=\sin \left(i\log i\right)$

$=\sin \left(i(\log \left|i\right|+iamp(i\left)\right)\right)\dots (i=|i|amp(i\left)\right)$

$=\sin \left(i(0+\frac{i\pi }{2})\right)$

$=\sin (0+\frac{i^2\pi }{2})$

$=\sin (0-\frac{\pi }{2})$

$a+ib=-1$

$\Rightarrow a=-1,b=0$

$\therefore \sin \left(\log i^i\right)=-1+0=-1$