# Math Help - Complex Analysis Trigonometric Identity

1. ## Complex Analysis Trigonometric Identity

Hello!

I was asked to show that the following identity holds:

|cosh(z)|2= sinh2(x) + cos2(y)

This is what I've got,

 LS= |cosh(z)|2 = |1/2(ez+e-z)|2 =1/4(ez+e-z)(ez+e-z) =1/4(e2z+2+e-2z) RS= sinh2(x) + cos2(y) = (1/2)(ex-e-x)(1/2)(ex-e-x) + (1/2)(eiy+e-iy)(1/2)(eiy+e-iy) = 1/4(e2x-2+e-2x+e2iy+2+e-2iy) = 1/4(e2z+e-2z)

So LS does not equal RS....
Where is my mistake? Is it in the Absolute value set on the LS? that's the only part that is not intuitive for me...
Any hints would be killler!!!

2. ## Re: Complex Analysis Trigonometric Identity

$|\cosh (z)| = \sqrt{\text{Cos}[y]^2 \text{Cosh}[x]^2+\text{Sin}[y]^2 \text{Sinh}[x]^2}$

then both sides ls & rs expand to:

$\frac{1}{4} e^{-2 z} \left(1+e^{2 z}\right) \left(e^{2 x}+e^{2 i y}\right)$

3. ## Re: Complex Analysis Trigonometric Identity

Thanks MaxJasper,
I've never seen this expansion of |cosh(z)| before... and not quite sure if I understand it.
Could you maybe explain the step from left side to right?
I've only seen |cosh(z)|= (cosh(z)cosh(z))1/2
I would really appreciate that!

Edit:
Are you using trig identities?
Such as
cosh(z)=cos(iz)
cos(iz)= cos(ix)cosh(iy)- (i)sin(ix)sinh(iy)

Isn't |cosh(z)|= (cosh(z)cosh(z))1/2 a legitimate expansion of |cosh(z)|. So shouldn't it work?

4. ## Re: Complex Analysis Trigonometric Identity

In expansion of |cosh(z)| you must obtain a real value f(x,y) while your method results in another complex number. You say:

$|\cosh (z)|=\frac{1}{4}\left(e^z+e^{-z}\right)^2=....$

while indeed:

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

which is another complex number! nice trick

$|\cosh (z)|$=real valued f(x,y)= $\sqrt{\sinh ^2(x) \sin ^2(y)+\cosh ^2(x) \cos ^2(y)}$

5. ## Re: Complex Analysis Trigonometric Identity

LHS: Note that $\overline{e^{z}} = e^{\bar{z}}$. (Missing that led to your incorrect expansion of $\lVert \cosh(z) \rVert^2$).

Proof: $e^{\bar{z}} = e^{x - iy} = e^x(\cos(-y) + i \sin(-y)) = e^x(\cos(y) - i \sin(y))$

$= \overline{e^x(\cos(y) + i \sin(y))} = \overline{e^{z}}$

------------------------------------------

Derivation:

$4 \left\lVert \cosh(z) \right\rVert ^2 = 4 \left\lVert \frac{e^z + e^{-z}}{2} \right\rVert ^2 = \frac{4 \left\lVert e^z + e^{-z}\right\rVert ^2}{4}$

$= \left\lVert e^z + e^{-z} \right\rVert ^2 = (e^z + e^{-z})(\overline{e^z + e^{-z}}) = (e^z + e^{-z})(\overline{e^z} + \overline{e^{-z}})$

$= (e^z + e^{-z})(e^{\bar{z}} + e^{-\bar{z}}) = e^ze^{\bar{z}} + e^ze^{-\bar{z}} + e^{-z}e^{\bar{z}} + e^{-z}e^{-\bar{z}}$

$= e^{z+\bar{z}} + e^{z-\bar{z}} + e^{-z+\bar{z}} + e^{-z-\bar{z}}= e^{2x} + e^{2yi} + e^{-2yi} + e^{-2x}$

$= [e^{2x} + e^{-2x} -2] + 2 + [e^{2yi} + e^{-2yi}] = [e^{2x} + e^{-2x} -2] + 2 + 2\cos(2y)$

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

$= 4\sinh^2(x) + 4\cos^2(y)$

6. ## Re: Complex Analysis Trigonometric Identity

$|\cosh (z)| = \frac{1}{2} [\cosh (2 x)+\cos (2 y)]$

7. ## Re: Complex Analysis Trigonometric Identity

AMAZING! Thank you!
Makes perfect sense!