Question

If $sinlog\left(i^i\right)=a+ib$, find a and b. Hence find $cos\left(log{i}^i\right)$

• A. $a=-1,b=0,0$
• B. $a=1,b=0,0$
• C. $a=0,b=1,0$
• D. $a=0,b=-1,0$

Answer 1

The correct option is A $a=-1,b=0,0$

$i=e^{i\frac{\pi }{2}}$
$i^i=(e^{i\frac{\pi }{2}}{)}^i$
$=e^{-\frac{\pi }{2}}$
$log\left(i^i\right)=-\frac{\pi }{2}$
Hence $sin\left(log\right(i^i\left)\right)$
$=sin(-\frac{\pi }{2})$
$=-1$
$a+ib=-1$
Thus,
$a=-1$ and $b=0$
Hence $cos\left(log{i}^i\right)$
$=cos\left(\frac{-\pi }{2}\right)$
$=0$