Complex Variable

**Jason Bourne** · January 3rd 2008, 04:20 PM

Show that

$\mid cosz \mid^2 = cosh^2 y - sin^2 x , ~~~~ z=x + iy$

**ThePerfectHacker** · January 3rd 2008, 04:48 PM

Originally Posted by Jason Bourne

Show that

$\mid cosz \mid^2 = cosh^2 y - sin^2 x , ~~~~ z=x + iy$

$\cos (x+iy) = \cos x \cos (iy) - \sin x \sin (iy) = \cos x \cosh y - i \sin x \sinh y$

That means, $|\cos (x+iy)|^2 = \cos^2 x \cosh^2 y + \sin^2 x \sinh^2 y$

But, $\cosh^2 y \cos^2 x + \sin^2 x\sinh^2 y = \cosh^2 y (1-\sin^2 x)+\sin^2 x \sinh^2 y =$ $\cosh^2 y - \sin^2 x (\cosh^2 y -\sinh ^2 y) = \cosh^2 y - \sin^2 x$

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