# | sinh (z) |^2 = sinh^2(x) + sin^2(y)

• Dec 21st 2010, 01:20 PM
dwsmith
| sinh (z) |^2 = sinh^2(x) + sin^2(y)
I am having trouble proving this identity $\displaystyle |sinh(z)|^2=sinh^2(x)+sin^2(y) \ \ z\in\mathbb{C} \ \ x,y\in\mathbb{R}$.

$\displaystyle \displaystyle \left(\sqrt{sinh^2(z)}\right)^2=\left(\frac{e^z-e^{-z}}{2}\right)^2=\frac{e^{2z}+e^{-2z}-2}{4}$

$\displaystyle \displaystyle =\frac{e^{2x}cos(2y)+e^{-2x}cos(2y)-2+\mathbf{i}(e^{2x}sin(2y)-e^{-2x}sin(2y))}{4}$

$\displaystyle \displaystyle =\frac{cos(2y)(e^{2x}+e^{-2x})-2+\mathbf{i}sin(2y)(e^{2x}-e^{-2x})}{4}$

$\displaystyle \displaystyle =\frac{cos(2y)cosh(2x)-2+\mathbf{i}sin(2y)sinh(2x)}{2}$

At this point, I am not sure how to proceed.
• Dec 21st 2010, 02:16 PM
Plato
$\displaystyle \sinh(z)=\sinh(x)\cos(y)+i~\cosh(x)\sin(y)$.
$\displaystyle |\sinh(z)|^2=\sinh^2(x)\cos^2(y)+\cosh^2(x)\sin^2( y)$
What can you do with that?
• Dec 21st 2010, 02:23 PM
dwsmith
$\displaystyle sinh^2(x)cos^2(y)+cosh^2(x)sin^2(y)$

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

$\displaystyle =sinh^2(x)(cos^2(y)+sin^2(y))+sin^2(y)=sinh^2(x)+s in^2(y)$