Question

Express $(1+i{)}^i$ in the A+iB form, Where i = $\sqrt{-1}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Let $(1+i{)}^i$ = A+iB

Taking log in both sides,we get,

${log}_e$(A+iB) = i log(1+i)

= i log $\left(\sqrt{2}(cos\frac{\pi }{4}+isin\frac{\pi }{4})\right)$

Express 1 + i in Euler's form

= i ${log}_e\left(\sqrt{2}{e}^{i\frac{\pi }{4}}\right)$

= i $[{log}_e\sqrt{2}+{log}_e{e}^{i\frac{\pi }{4}}]$

${log}_e$(A+iB) = i$[{log}_e+i^{\frac{i}{4}}]$ = i ${log}_e\sqrt{2}$ - $\frac{\pi }{4}$

= $\frac{i}{2}$ ${log}_e$2 - $\frac{\pi }{4}$

A+iB = $e^{(i{log}_e\sqrt{2}-\frac{\pi }{4})}$ = $e^{i{log}_e\sqrt{2}}.e^{\frac{-\pi }{4}}$

= $e^{\frac{-\pi }{4}}\left[cos\right({log}_e\sqrt{2})+isin({log}_e\sqrt{2}\left)\right]$

= $e^{\frac{-\pi }{4}}cos\left({log}_e\sqrt{2}\right)+i{e}^{\frac{-\pi }{4}}sin\left({log}_e\sqrt{2}\right)$ = $e^{\frac{-\pi }{4}}\left[cos\right({log}_e\sqrt{2})+isin({log}_e\sqrt{2}\left)\right]$