Let $z=x+yi$ be a complex number, $x,y\in\mathbb{R}$. I have the following expression $$\sin(x)\sinh(y) + i\cos(x)\cosh(y)$$ and I would like to express it in terms of $z$ (not $\overline{z}$, for instance).
I think I got to show that $$\sin(x)\sinh(y) + i\cos(x)\cosh(y)=e^{\overline{z}}+e^{-\overline{z}}$$but I don’t know how to follow.
Note,
$$\cos(x+iy)=\cos x \cos(iy) - \sin x \sin (iy)$$ $$=\cos x \cosh y +\frac{1}{i}\sin x \sinh y$$ $$=\frac1i( i\cos x \cosh y + \sin x \sinh y)$$
where $\cos{iy}=\cosh y$ and $\sin(iy) = -\frac1i \sinh y$ are used. Thus,
$$ \sin x \sinh y + i\cos x \cosh y = i\cos z $$