Question

If sin(${log}_e$ $i^i)$ = A+iB where i = $\sqrt{-1}$.Find the value of cos(${log}_e$ $i^i)$.

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Answer 1

Given sin(${log}_e$ $i^i$) = A+iB

A+iB = sin(i${log}_{e}i)$

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

Express I in Euler's form

= sin $\left[i{log}_e{e}^{i\frac{\pi }{2}}\right]$

= sin $[i.i^{\frac{\pi }{2}}]$

= sin $(-\frac{\pi }{2})$ = -1

We know $si{n}^2\theta$ + $co{s}^2\theta$ = 1

$si{n}^2$ $\left({log}_e{i}^i\right)$ + $co{s}^2$ $\left({log}_e{i}^i\right)$ = 1

$(-1{)}^2$ + $co{s}^2$ $\left({log}_e{i}^i\right)$ = 1

$co{s}^2$ $\left({log}_e{i}^i\right)$ = 0

cos$\left({log}_e{i}^i\right)$ = 0