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Question Number 134419 by mohammad17 last updated on 03/Mar/21

w h y t h e f u n c t i o n c o s z a n d s i n z i t i s

n o t b o u n d e d i n c o m p l e x n u m b e r ?

Answered by Olaf last updated on 03/Mar/21

\cos z = \cos (x + i y)

\cos z = \cos x \cos (i y) - \sin x \sin (i y)

\cos z = \cos x \frac{e^{i (i y)} + e^{- i (i y)}}{2} - \sin x \frac{e^{i (i y)} - e^{- i (i y)}}{2 i}

\cos z = \cos x \frac{e^{- y} + e^{y}}{2} - \sin x \frac{e^{- y} - e^{y}}{2 i}

\cos z = \cos x \cosh y + i \sin x \sinh y

\cos z is not bounded because

\cosh y and \sinh y are not bounded .

Same demonstation for \sin z .

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