# How to calculate $\overline{\cos \phi}$

How do you calculate $\overline{\cos \phi}$? Where $\phi\in\mathbb{C}$.

I try to proof that $\cos \phi \cdot \overline{\cos \phi} +\sin \phi \cdot \overline{\sin \phi}=1$?

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$$\cos(x+iy) = \cos x \cos (iy) - i \sin x \sin(iy)$$ $$\overline {\cos(x+iy)} = \cos x \cos (iy) + i \sin x \sin(iy) = \cos x \cos (-iy) - i \sin x \sin(-iy) = \cos(x-iy)$$

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Hint:$$\cos z=\frac{e^{iz}+e^{-iz}}{2}$$ and $$\overline{e^{iz}}=\overline{e^{xi-y}}=\overline{\frac{e^{xi}}{e^y}}= \frac{\overline {\cos x+i\sin x}}{e^y}=\frac{e^{-xi}}{e^y}=e^{-xi-y}=e^{i(-x+iy)}=e^{-i\bar{z}}$$ Therefore, $$\overline{\cos z}=\frac{e^{-i\bar{z}}+e^{i\bar{z}}}{2}=\cos \bar{z}$$

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$$\cos(a+ib)=\frac{e^{i(a+ib)}+e^{-i(a+ib)}}2$$
$$=\frac{e^{-b}(\cos a+i\sin a)+e^b(\cos a-i\sin a)}2$$
$$=\cos a\frac{e^b+e^{-b}}2-i\sin a\frac{e^b-e^{-b}}2=\cos a\cosh b-i\sin a\sinh b$$
We know $f(X+iY)=A+iB\iff f(X-iY)=A-iB$
So, $$\cos(a-ib)=\cos a\cosh b+i\sin a\sinh b$$