Question

Statement-$1$: $i^i$ is a real number.
Statement-$2$: If $z=(\sqrt{i}{)}^{\sqrt{i}}$, then the real part of ${\log }_{e}z$ is $\frac{-\pi }{4\sqrt{2}}$

(consider principal value)

• A. Only Statement $1$ is true
• B. Only Statement $2$ is true
• C. Both Statements are true
• D. Both Statements are false

Answer 1

The correct option is C Both Statements are true

$e^{\frac{i\pi }{2}}=\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}=i$
$e^{i(2n+\frac{1}{2})\pi }=\cos (2n\pi +\frac{\pi }{2})+i\sin (2n\pi +\frac{\pi }{2})=i$
$\therefore i^i={\left(e^{i(2n+\frac{1}{2})\pi }\right)}^i=e^{-(2n+\frac{1}{2})\pi }$
which is a real number.

$z=(\sqrt{i}{)}^{\sqrt{i}}$
${\log }_{e}z=\sqrt{i}{\log }_e\sqrt{i}$
$=\frac{1}{2}{(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2})}^{\frac{1}{2}}{\log }_e(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2})$
$=\frac{1}{2}(\cos \frac{\pi }{4}+i\sin \frac{\pi }{4}){\log }_e\left(e^{\frac{i\pi }{2}}\right)$
$=\frac{1}{2}(\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}})\left(\frac{i\pi }{2}\right)$
$=\frac{-\pi }{4\sqrt{2}}+i\frac{\pi }{4\sqrt{2}}$

Real Part $=\frac{-\pi }{4\sqrt{2}}$