Find the general value of ${l o g}_{e}$ (i).

$\frac{i π}{2} (2 n + 1)$ n $ϵ$ I

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$\frac{i π}{2} (2 n - 1)$ n $ϵ$ I

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$\frac{i π}{2} (4 n + 1)$ n $ϵ$ I

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$\frac{i π}{2} (n + 1)$ n $ϵ$ I

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Solution

The correct option is C
$\frac{i π}{2} (4 n + 1)$ n $ϵ$ I

Convert complex number in Euler's form

${l o g}_{e}$ (i)

= ${l o g}_{e}$ $[c o s \frac{π}{2} + i s i n \frac{π}{2}]$

= ${l o g}_{e}$ $[e^{i \frac{π}{2}}]$

= $\frac{i π}{2}$ + 2n $π$ i n $ϵ$ I

= $\frac{i π (4 n + 1)}{2}$ n $ϵ$ I

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