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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.

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

$$ \sin x \sinh y + i\cos x \cosh y = i\cos z $$

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