Question

Find the value of i.$i^i$ is _______.

• A. i.$e^{\frac{\pi }{2}}$
• B. i.$e^{-\frac{\pi }{2}}$
• C. i.$e^{\frac{\pi }{4}}$
• D. i.$e^{-\frac{\pi }{4}}$

Answer 1

The correct option is B

i.$e^{-\frac{\pi }{2}}$

Let z = i.$i^i$

Taking log on both sides

${log}_e$z = ${log}_e$i + ${log}_0$$i^i$

= ${log}_e(cos\frac{\pi }{2}+isin\frac{\pi }{2})+i{log}_e(cos\frac{\pi }{2}+isin\frac{\pi }{2})$

= ${log}_e{e}^{i\frac{\pi }{2}}+i{log}_e{e}^{i\frac{\pi }{2}}$

= $\frac{i\pi }{2}$ + i.$\frac{i\pi }{2}$

${log}_e$2 = $\frac{i\pi }{2}$ - $\frac{\pi }{2}$ = $\frac{\pi }{2}$(i-1)

z = $e^{(-\frac{\pi }{2}+\frac{i\pi }{2})}$

= $e^{-\frac{\pi }{2}}.e^{i\frac{\pi }{2}}$

= $e^{-\frac{\pi }{2}}.(cos\frac{\pi }{2}+isin\frac{\pi }{2})$

= $e^{-\frac{\pi }{2}}$ (i)

= i.$e^{-\frac{\pi }{2}}$