Euler's formula for complex $z$

Given Euler's formula $e^{ix} = \cos x + i \sin x$ for $x \in \mathbb{R}$, how does one extend this definition for complex $z$? I.e for $z \in \mathbb{C}, $ show that $e^{iz} = \cos(z) + i\sin(z)$?

I can reexpress $z = x+$i$y\,$ for $x,y \in \mathbb{R}$ and then substitute into the above to get $e^{ix}e^{-y} = (\cos x + i\sin x)e^{-y}$, but I am unsure of how to progress. I am hoping this can be done without appealing to the complex $\sin$ and $\cos$.

Thanks.

Answer

Let $z=x+iy$, and you want to show that $e^{iz}=\cos(z)+i\sin(z)$ i.e. Euler's formula applies to complex $z$.

I will prove the formula starting from the right hand side.

I don't think you can get away from complex trigonometry, where the cosine/sine of a purely imaginary term is defined as a hyperbolic function as follows:-

$\cos(ix)=\cosh(x)$

$\sin(ix)=i\sinh(x)$

Using the above definitions in conjunction with the sum of angles identities, we have

$\cos(x+iy)=\cos(x)\cos(iy)-\sin(x)\sin(iy)=\cos(x)\cosh(y)-i\sin(x)\sinh(y)$

$\sin(x+iy)=\sin(x)\cos(iy)+\cos(x)\sin(iy)=\sin(x)\cosh(y)+i\cos(x)\sinh(y)$

So we have

$\cos(x+iy)+i\sin(x+iy)\\=\cos(x)\cosh(y)-\cos(x)\sinh(y)+i(\sin(x)\cosh(y)-\sin(x)\sinh(y))\\=\cos(x)e^{-y}+i\sin(x)e^{-y}=(\cos(x)+i\sin(x))e^{-y}=e^{ix}e^{+i^2y}=e^{i(x+iy)}=e^{iz}$

where we have used the fact that $\cosh(y)-\sinh(y)=e^{-y}$