I have problems in two issues of complex variables$\ldots$
$1$) Prove that $\operatorname{Im}(\cosh z)=\sinh x\cosh y$, if $z=x+iy$.
I tried to expand $\cosh z= \cosh(x+iy)=\left[{\exp{(x+iy)}+\exp{(-x-iy)}}\right]/{2}$, then
$$\frac{e^{x+iy}+e^{-x-iy}}{2}=\frac{e^xe^{iy}+e^{-x}e^{iy}}{2}=\frac{e^x\left ( \cos y+i\sin y \right )+e^{-x}(\cos y-i\sin y)}{2}$$
and after some accounts, came to the conclusion that $$\operatorname{Im} (\cosh z)=\left ( e^{2x}-1 \right )\sin y$$
Here in Brazil we denote $\operatorname{Im} (z)$ the imaginary part of the complex $z$.
$2$) Prove that $|\sinh z|^2=\sinh^2(z)+\cosh^2(z)$
I tried from right to left and was not very useful, since from left to right, I got
$$\begin{align*}
\sinh^2(z)+\cosh^2(z)&=\left ( \frac{e^z-e^{-z}}{2} \right )^2+\left ( \frac{e^z+e^{-z}}{2} \right ) \\
&= \frac{e^{2z}-2+e^{-2z}}{4}+\frac{e^{2z}+2+e^{-2z}}{4} \\
&= \frac{2e^{2z}+2e^{-2z}}{4} \\
&= \frac{e^{2z}+e^{-2z}}{2} \\
&= \cosh(2z)
\end{align*}$$
I thank the help of you for the resolution of these two issues.
Best Answer
Use the addition formulae: $$ \cosh{(x+iy)} = \cosh{x}\cosh{iy}+\sinh{x}\sinh{iy}. $$ Now, $\cosh{iy}=\cos{y}$, and $\sinh{iy}=i \sin{y}$, and the imaginary part is hence $ \sinh{x}\sin{y} $, so in fact the first formula is false.
For the second one, it's definitely wrong (take $z=0$...). Again using the addition formulae, $$ \sinh{(x+iy)} = \sinh{x}\cosh{iy}+\cosh{x}\sinh{iy} = \sinh{x}\cos{y}+i\cosh{x}\sin{y}. $$ Now everything but the $i$ is real, so we know that $$ \lvert \sinh{(x+iy)} \rvert^2 = \sinh^2{x}\cos^2{y} + \cosh^2{x}\sin^2{y}. $$ Now, we can decide to use both of $\sin^2+\cos^2=1$, or $\cosh^2-\sinh^2=1$, to find the two equivalent forms $$ \lvert \sinh{(x+iy)} \rvert^2 = \sinh^2{x} + \sin^2{y} = \cosh^2{x} + \cos^2{y}. $$