Express in the form A + iB
$l o g_{e} (1 + i t a n α)$

l o g c o s α + i α

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l o g s e c α + i α

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l o g c o s e c α + i α

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l o g s i n α + i α

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Solution

The correct option is B $l o g s e c α + i α$
$l o g_{e} (1 + i t a n α) = l o g_{e} (\frac{cos α + i sin α}{cos α}) = {log}_{e} e^{i α} - {log}_{e} cos α = i α + {log}_{e} sec α$
Ans: B

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