Question

If $z=(i{)}^{(i{)}^{\left(i\right)}}$ where $i=\sqrt{-1}$, then |z| is equal to

• A. $e^{-\pi }$
• B. 1
• C. $e^{-\pi /2}$
• D. None of these

Answer 1

Answer: (B)

$log\left(z\right)=i^{i}log\left(i\right)$
$=e^{\frac{i\pi }{2}.i}log\left(e^{\frac{i\pi }{2}}\right)$
$=e^{\frac{-\pi }{2}}i\frac{\pi }{2}$
Now $e^{\frac{-\pi }{2}}\left(\frac{\pi }{2}\right)$ is constant.
Therefore
$e^{\frac{-\pi }{2}}\left(\frac{\pi }{2}\right)=\lambda$
$log\left(z\right)=\lambda \left(i\right)$
$z=e^{\lambda i}$
$\Rightarrow |z|=1$
Hence, option 'A' is correct.