1. ## Complex Trig Function

Show that $\overline{sin(iz)}=sin(i\bar{z})$ if and only if $z=n \pi i, (n=0, \pm 1, \pm 2, ...)$.

Anyways, I got from the LHS to the RHS but didn't find the restriction. Where should that pop up?

2. Originally Posted by davesface
Show that $\bar{sin(iz)}=sin(i\bar{z})$ if and only if $z=n \pi i, (n=0, \pm 1, \pm 2, ...)$. (There should be a complex conjugate bar over the whole first expression, but I'm not sure how to do that in Latex.)

Anyways, I got from the LHS to the RHS but didn't find the restriction. Where should that pop up?
$\sin(z)=\cos(x)\sinh(y)+i\sin(y)\cosh(x)$

3. Originally Posted by davesface
Show that $\bar{sin(iz)}=sin(i\bar{z})$ if and only if $z=n \pi i, (n=0, \pm 1, \pm 2, ...)$. (There should be a complex conjugate bar over the whole first expression, but I'm not sure how to do that in Latex.)

Anyways, I got from the LHS to the RHS but didn't find the restriction. Where should that pop up?

You must have made the same mistake I did the first part:

$\overline{\frac{e^{i\cdot iz}-e^{-i\cdot iz}}{2i}}=\overline{\sin iz}$ $=\sin i\overline{z}=\frac{e^{i\cdot i\overline{z}}-e^{-i\cdot i\overline{z}}}{2i}\Longleftrightarrow$ $\overline{e^{-z}-e^{z}}=e^{-\overline{z}}-e^{\overline{z}}$ ...and the mistake's already here! It must be

On the left hand: $\overline{\frac{e^{-z}-e^z}{2i}}=\overline{-i\,\frac{e^{-z}-e^z}{2}}=i\,\frac{e^{-z}-e^z}{2}$

On the right hand: $\frac{e^{-\overline{z}}-e^{\overline{z}}}{2i}=-i\,\frac{e^{-\overline{z}}-e^{\overline{z}}}{2}$ ,and mulitiplying by $\frac{2}{i}$ and assuming the equality (as in the beginning) we continue:

$e^{-z}-e^z=e^{\overline{z}}-e^{-\overline{z}}\Longleftrightarrow$ putting $z=x+iy\,,\,\,x,y\in\mathbb{R}$ and comparing real and imaginary parts I get $x=0$ as unique condition...

Tonio