Trig proof -- Complex analysis

**JJMC89** · October 27th 2010, 04:05 PM

Prove that $|cos^2 (z)|+|sin^2 (z)| = 1$ is false if $z=x+iy$ with $y\ne 0$ .

**tonio** · October 28th 2010, 12:16 AM

Originally Posted by JJMC89

Prove that $|cos^2 (z)|+|sin^2 (z)| = 1$ is false if $z=x+iy$ with $y\ne 0$ .

Write $\cos z=\frac{e^{iz}+e^{-iz}}{2}\,,\,\,\sin z=\frac{e^{iz}-e^{-iz}}{2i}$ , so

$=\cos^2z=\frac{e^{2iz}+e^{-2iz}+2}{4}=\frac{1}{4}\left[e^{-2y}\left(\cos 2x+i\sin 2x\right)+e^{2y}\left(\cos 2x-i\sin 2x\right)\right]+\frac{1}{2}=$

$= \frac{1}{2}\left[\cos 2x\cosh 2y+1+i\sin 2x\sinh 2y\right]$

$\sin^2z=\frac{e^{2iz}+e^{-2iz}-2}{-4}=-\frac{1}{4}\left[e^{-2y}\left(\cos 2x+i\sin 2x\right)+e^{2y}\left(\cos 2x-i\sin 2x\right)\right]+\frac{1}{2}=$

$=-\frac{1}{2}\left[\cos 2x\cosh 2y-1+i\sin 2x\sinh 2y\right]$ , so

$|\cos^2z|+|\sin^2z|=\frac{1}{4}\left[(\cos 2x\cosh 2y+1)^2+(\cos 2x\cosh 2y-1)^2+2\sin^22x\sinh^22y\right]$

take it from here (and check the above carefully for mistakes)

Tonio

**JJMC89** · October 29th 2010, 09:06 AM

Thanks

I got it now.

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