Show that $\displaystyle \overline{sin(iz)}=sin(i\bar{z})$ if and only if $\displaystyle z=n \pi i, (n=0, \pm 1, \pm 2, ...)$.
Anyways, I got from the LHS to the RHS but didn't find the restriction. Where should that pop up?
Show that $\displaystyle \overline{sin(iz)}=sin(i\bar{z})$ if and only if $\displaystyle z=n \pi i, (n=0, \pm 1, \pm 2, ...)$.
Anyways, I got from the LHS to the RHS but didn't find the restriction. Where should that pop up?
You must have made the same mistake I did the first part:
$\displaystyle \overline{\frac{e^{i\cdot iz}-e^{-i\cdot iz}}{2i}}=\overline{\sin iz}$ $\displaystyle =\sin i\overline{z}=\frac{e^{i\cdot i\overline{z}}-e^{-i\cdot i\overline{z}}}{2i}\Longleftrightarrow$ $\displaystyle \overline{e^{-z}-e^{z}}=e^{-\overline{z}}-e^{\overline{z}}$ ...and the mistake's already here! It must be
On the left hand: $\displaystyle \overline{\frac{e^{-z}-e^z}{2i}}=\overline{-i\,\frac{e^{-z}-e^z}{2}}=i\,\frac{e^{-z}-e^z}{2}$
On the right hand: $\displaystyle \frac{e^{-\overline{z}}-e^{\overline{z}}}{2i}=-i\,\frac{e^{-\overline{z}}-e^{\overline{z}}}{2}$ ,and mulitiplying by $\displaystyle \frac{2}{i}$ and assuming the equality (as in the beginning) we continue:
$\displaystyle e^{-z}-e^z=e^{\overline{z}}-e^{-\overline{z}}\Longleftrightarrow$ putting $\displaystyle z=x+iy\,,\,\,x,y\in\mathbb{R}$ and comparing real and imaginary parts I get $\displaystyle x=0$ as unique condition...
Tonio