# Express $\sin(x)\sinh(y) + i\cos(x)\cosh(y)$ in terms of $z=x+iy.$

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.

• I get $$\frac{i}{2}\left(e^{iz}+e^{-iz}\right)=i\cos(iz).$$ – Thomas Andrews Nov 18 '19 at 21:14

$$\cos(x+iy)=\cos x \cos(iy) - \sin x \sin (iy)$$ $$=\cos x \cosh y +\frac{1}{i}\sin x \sinh y$$ $$=\frac1i( i\cos x \cosh y + \sin x \sinh y)$$
where $$\cos{iy}=\cosh y$$ and $$\sin(iy) = -\frac1i \sinh y$$ are used. Thus,
$$\sin x \sinh y + i\cos x \cosh y = i\cos z$$