Express $f (z) = u + iv$ as a function of $ z = x + iy$? I have $u(x,y)=\sin(x)\cosh(y)$ and I got conjugate harmonic of $u(x,y)$ as $v(x,y) = \cos(x)\sinh(y)+c$.
So,
$f(z) = \sin(x)\cosh(y)+\mathrm i(\cos(x)\sinh(y)+c)$.  
How to express $f(z) = u + iv$ as a function of $z = x + iy$ ?
 A: Note that $\cos(ix)=\cosh(x)$ and $\sin(ix)=i\sinh(x)$, so that
$$f(z) = \sin(x)\cosh(y)+ i(\cos(x)\sinh(y)+c)=\sin(x)\cos(iy)+ \cos(x)\sin(iy)+ic$$
$$=\sin(x+iy)+ic=\sin(z)+ic$$
A: In this case the answer is $f(z)=\sin(z)=\frac{\mathrm e^{\mathrm iz}-\mathrm e^{-\mathrm iz}}{2\mathrm i}$. 
A: The following identities hold:
\begin{align}
\sin(x) &= \frac1{2i}(\exp(ix)-\exp(-ix)) \\
\cosh(x) &=\frac12(\exp(x)+\exp(-x)) \\
\cos(x) &= \frac12(\exp(ix)+\exp(-ix)) \\
\sinh(x) &=\frac12(\exp(x)-\exp(-x))
\end{align}
Thus we have 
\begin{align}
\sin(x)\cosh(y)& = \frac1{4i}(\exp(ix)-\exp(-ix))(\exp(y)+\exp(-y)) \\
i(\cos(x)\sinh(y))&=\frac{1}{4i}(\exp(ix)+\exp(-ix))(\exp(y)-\exp(-y))
\end{align}
And finally
\begin{align}
&\sin(x)\cosh(y)
+i(\cos(x)\sinh(y))\\
&=\frac{1}{4i}
(\exp(ix)\exp(y)+\exp(ix)\exp(-y)-\exp(-ix)\exp(y)-\exp(-ix)\exp(-y) \\
 &\quad +\exp(ix)\exp(y)+\exp(-ix)\exp(y)-\exp(ix)\exp(-y)-\exp(-ix)\exp(-y))\\
&=\frac1{4i}(2\exp(ix)\exp(y)-2\exp(-ix)\exp(-y)) \\
&= \frac{1}{2i}(\exp(y+ix)-\exp(-(y+ix))) \\
&=\sin(y+ix) = \sin(z)
\end{align}
