The value of $l o g (1 + i)^{12} = ?$

l o g \frac{π}{4} + \frac{i π}{2}

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6 l o g 2 + i (3 π)

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12 l o g 2 + i (9 π)

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16 l o g 2 + i (9 π)

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Solution

The correct option is B $6 l o g 2 + i (3 π)$
We have $1 + i = \sqrt{2} (e^{\frac{i π}{4}})$
Hence
$(1 + i)^{12} = 2^{6} (e^{3 i π})$
$log (1 + i)^{12} = l o g (2^{6}) + 3 i π$
$log (1 + i)^{12} = 6 l o g 2 + 3 i π$

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