Question

If z = $i^i$ . Express ${log}_e$z in A+iB form.Find the value of A and B.

• A. A = $\frac{1}{e^{\frac{\pi }{2}}}$, B = $\frac{1}{e^{\frac{\pi }{2}}}$
• B. A = 0 , B = $\frac{1}{e^{\frac{\pi }{2}}}$
• C. A = $\frac{1}{e^{\frac{\pi }{2}}}$ , B = 0
• D. A = $e^{\frac{\pi }{2}}$ , B = $e^{\frac{\pi }{2}}$

Answer 1

The correct option is C

A = $\frac{1}{e^{\frac{\pi }{2}}}$ , B = 0

Given z = $i^i$

${log}_e$z = i${log}_e$i

= i${log}_e[cos\frac{\pi }{2}+isin\frac{\pi }{2}]$

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

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

= $\frac{-\pi }{2}$

z = $e^{\frac{-\pi }{2}}$ = $\frac{1}{e^{\frac{\pi }{2}}}$

So, A = $\frac{1}{e^{\frac{\pi }{2}}}$ and B = 0