Find the value of
$a r g ({(1 + i)}^{i})$

\frac{1}{4} l n (2)

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\frac{1}{2} l n (2)

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\frac{1}{2} l n (\frac{1}{2})

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l n (2)

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Solution

The correct option is C $\frac{1}{2} l n (2)$

Write $(1 + i)$ as $\sqrt{2} e^{[\frac{i π}{4}]}$

Therefore

$(1 + i)^{i} = {\sqrt{2}}^{i} e^{- π / 4} = [e^{- (8 n + 1) \frac{π}{4}}] \times e^{i \frac{l n 2}{2}}$

So,
$a r g (1 + i)^{i} = a r g [e^{- (8 n + 1) \frac{π}{4}}] \times e^{i \frac{l n 2}{2}} = a r g [k e^{i \frac{l n 2}{2}}] = \frac{l n 2}{2}$ .........[where $K = e^{- (8 n + 1) \frac{π}{4}}$ ]

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