Question

$i^{i}$ is equal to

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

Answer 1

Answer: (B)

$e^{- \pi/2}$

Answer 2

Sol. Let $A = i^{i}$ then $\log A = \log i^{i} = i\log i$ ⇒ $\log A = i\log(0 + i)$

⇒ $\log A = i\lbrack\log 1 + i\pi/2\rbrack$ $(\because(i) = 1$and arg$(i) = \pi/2)$

log A $= i\lbrack 0 + i\pi/2\rbrack = - \pi/2$ ⇒ $A = e^{- \pi/2}$.