# Thread: Trig. hyp. complex no.s

1. ## Trig. hyp. complex no.s

Can someone help explain why Modulus(sinZ)^2=sin^2(x)+sinh^2(x)

When Z=x+iy

many thanks

2. It is well known that $\displaystyle \sin (z) = \sin (x)\cosh (y) + i\cos (x)\sinh (y)$.

So
$\displaystyle \begin{gathered} \left| {\sin (z)} \right|^2 = \sin ^2 (x)\cosh ^2 (y) + \cos ^2 (x)\sinh ^2 (y) \hfill \\ \left| {\sin (z)} \right|^2 = \sin ^2 (x)\left( {1 + \sinh ^2 (y)} \right) + \left( {1 - \sin ^2 (x)} \right)\sinh ^2 (y) \hfill \\ \left| {\sin (z)} \right|^2 = \sin ^2 (x) + \sinh ^2 (y) \hfill \\ \end{gathered}$

3. Originally Posted by Bryn
Can someone help explain why Modulus(sinZ)^2=sin^2(x)+sinh^2(x)

When Z=x+iy

many thanks

$\displaystyle \sin z=\frac{e^{iz}-e^{-iz}}{2i}=\frac{e^{ xi-y}-e^{-xi+y}}{2i}$ $\displaystyle =\frac{e^{-y}(\cos x+i\sin x)-e^{y}(\cos x-i\sin x)}{2i}=\frac{-\cos x(e^{y}-e^{-y})}{2i}+\frac{\sin x(e^y+e^{-y})}{2i}i$ $\displaystyle = \sin x\frac{e^y+e^{-y}}{2}+\cos x\frac{e^{y}-e^{-y}}{2}i$ , so:

$\displaystyle |\sin z|^2=Re(z)^2+Im(z)^2=\sin^2\!x\cosh^2\!y+\cos^2\!x \sinh^2\!y$ $\displaystyle =\sin^2\!x(1+\sinh^2\!y)+\cos^2\!x\sinh^2\!y=$ $\displaystyle (\sin^2\!x+\cos^2\!x)\sinh^2\!y+\sin^2\!x=\sinh^2\ !y+\sin^2\!x$

Note that there's a mistake in your post: the hyperbolic sine is of y, not of x.

Tonio