$y = log (cos 60 + i sin 60)$ , then $(cosh y)$ is

\frac{1}{2}

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\frac{1}{4}

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\frac{1}{8}

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\frac{3}{2}

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Solution

The correct option is B $\frac{1}{2}$
$y = log (cos 60^{0} + i sin 60^{0})$
$= log (e^{i 60})$
$= 60 i$ Or, $\frac{π}{3} i$
$cosh y = cosh \frac{π}{3} i = \frac{e^{\frac{π}{3} i} + e^{- \frac{π}{3} i}}{2}$
$= \frac{cos \frac{π}{3} + i sin \frac{π}{3} + cos (- \frac{π}{3}) + i (sin (- \frac{π}{3}))}{2}$
$= \frac{\frac{1}{2} + \frac{1}{2}}{2} = \frac{1}{2}$

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