Question

If $\cos (\log \, i^{4i}) = a + ib$, then

• A. $a = 1 , b = 2 $
• B. $a = 1 , b = - 1 $
• C. $a =- 1 , b = 1$
• D. $a = 1 , b = 0$

Answer 1

Answer: (D)

Given, $ a+i b=\cos \left(\log i^{4 i}\right)$
$=\cos [4 i \log i] $
$=\cos \left[4 i \log \left(e^{i \frac{\pi}{2}}\right)\right]=\cos \left[(4 i)\left(i \frac{\pi}{2}\right)\right]$
$=\cos (-2 \pi)=1$
$ a =1$ and $b=0$