Proving $\DeclareMathOperator{\Re}{Re}\DeclareMathOperator{\Im}{Im}\Re{(\cosh z}) = \cosh(\Re z)\cdot \cos(\Im z)$ I want to show $\Re{(\cosh z)}) = \cosh(\Re z)\cdot \cos(\Im z)$.
I did the following: 
\begin{align*} \cosh(\Re z)\cdot \cos(\Im z) &= \frac {1}{2}(e^{\Re z}+e^{-\Re z})\cdot \frac {1}{2}(e^{i\Im z}+e^{-i\Im z}) \\& = \frac {1}{4}(e^{\Re z}+e^{-\Re z})\cdot (e^{i\Im z}+e^{-i\Im z}) \\
&= \frac {1}{4}(e^{\Re z}+e^{-\Re z}) \cdot 2\\
& = \frac {1}{2}(e^{\Re z}+e^{-\Re z}) \\
&= \frac {1}{2}\Re(e^z+e^{-z}) \\
&= \Re(\cosh z)
\end{align*}
Using: $\cosh(x) = \frac {1}{2}(e^x+e^{-x})$ and $\cos(x) = \frac {1}{2}(e^{ix}+e^{-ix})$.
I would be very glad if someone could tell me if this is correct!
 A: As an alternative, but taking the path of OP so that his /her error can be seen:
$$ \cosh\left[\operatorname{Re}(z)\right] \cdot \cos\left[\operatorname{Im}(z)\right] = \left(\frac{e^{\operatorname{Re}(z)} + e^{-\operatorname{Re}(z)}}{2}\right) \left(\frac{e^{i\operatorname{Im}(z)} + e^{-i\operatorname{Im}(z)}}{2}\right)$$
$$ = \frac{e^{\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} +  e^{-\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) - i\operatorname{Im}(z)}}{4} $$
$$ = \frac{e^{\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} +  e^{-\operatorname{Re}(z) + i\operatorname{Im}(z)} + }{4} $$
$$ = \frac{e^{Re(z) + i\operatorname{Im}(z)} + e^{-\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} +  e^{-[\operatorname{Re}(z) - i\operatorname{Im}(z)]}}{4} $$
$$ = \frac{e^{\operatorname{Re}(z) + i\operatorname{Im}(z)} + e^{-[\operatorname{Re}(z) + i\operatorname{Im}(z)]}}{4} + \frac{e^{\operatorname{Re}(z) - i\operatorname{Im}(z)} + e^{-[\operatorname{Re}(z) - i\operatorname{Im}(z)]}}{4} = \frac{\cosh {z}}{2} + \frac{\cosh {\overline {z}}}{2}$$
$$  = \frac{\cosh {z} + \cosh {\overline {z}}}{2} = \frac{2\operatorname{Re}{\left[\cosh{z}\right]}}{2} = \operatorname{Re}{\left[\cosh{z}\right]}$$
A: Let $z=x+iy$, then we have that
$$\cosh z=\frac{e^z+e^{-z}}2=\frac12e^xe^{iy}+\frac12e^{-x}e^{-iy}=\frac12e^x(\cos y+i\sin y)+\frac12e^{-x}(\cos y-i\sin y) $$
$$ \implies \operatorname{Re}(\cosh z)=\frac12e^x\cos y+\frac12e^{-x}\cos y=\frac{e^x+e^{-x}}2\cos y=\cosh x \cdot\cos y$$
A: Using the sum of angles formulae,
$\cosh(x+iy)=\cosh(x)\cosh(iy)+\sinh(x)\sinh(iy)$
but $
\cosh(iy)=\cos(y)$ and $\sinh(iy)=i\sin(y)$
so:   $\cosh(x+iy)=\cosh(x)\cos(y) + i \sinh(x)\sin(y)$
QED
