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I am trying to prove the identity $|\cos z|^2 = \cos^2 x + \sinh^2 y$, where $z=x+iy$.

I know $|\cos z|^2=(\cos z)(\cos \bar{z})$, since cosine is analytic for all $z \in \mathbb{C}$. Thus,

$$|\cos z|^2=(\cos (x+iy))(\cos (x-iy))=(\cos x \cos iy-\sin x \sin iy)(\cos x \cos iy+\sin x \sin iy)=\cos^2 x \cosh^2 y + \sin^2 x \sinh^2 y$$

The last equality follows (with some algebra) because $\cos iy=\cosh y$ and $\sin iy=i\sinh y$.

I'd like to know if my work is correct so far, if I'm heading in the right direction, and what I could do to finish.

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    $\begingroup$ You are heading in the right direction. Use identities such as $1=\cos^2+\sin^2$ and $\cosh^2-\sinh^2=1$. You only need the second one here but I gather you are doing a problem from Ahlfors and the other will be needed later. $\endgroup$
    – dustin
    Commented Feb 20, 2015 at 20:06
  • $\begingroup$ Thank you for reminding me of the hyperbolic identity. The first one is engrained in my memory because circular trig functions are focused on in school, but I always forget the important hyperbolic trig identities; for some reason they aren't taught much anymore. $\endgroup$
    – Ducky
    Commented Feb 21, 2015 at 2:20
  • $\begingroup$ Not a problem. I go in spats of remembering them and forgetting. There will be points when I can recall them and times when I need to look it up. $\endgroup$
    – dustin
    Commented Feb 21, 2015 at 2:22

2 Answers 2

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You are almost there. Note that:

$$\cosh^2(y)=\left(\frac{e^y+e^{-y}}{2}\right)^2=\frac{e^{2y}+e^{-2y}+2}{4}$$ $$\sinh^2(y)=\left(\frac{e^y-e^{-y}}{2}\right)^2=\frac{e^{2y}+e^{-2y}-2}{4}$$ Hence $\cosh^2(y)=1+\sinh^2(y)$. Now since $\sin^2x+\cos^2x=1$ we find: $$\cos^2x\cosh^2y+\sin^2x\sinh^2y=\cos^2x(1+\sinh^2y)+\sin^2x\sinh^2y=\cos^2x+\sinh^2y$$

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  • $\begingroup$ Thank you. It seems the simple-looking steps always elude me. $\endgroup$
    – Ducky
    Commented Feb 20, 2015 at 20:25
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All fine, continue further using $ \cosh ^2 x -\sinh ^2 x = 1 $.

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