real analysis - Verifying Euler's Formula from trigonometry

I know the proof for the Euler's formula by writing $e^{iz}$ as a Taylor series and arrange the brackets so that I get: $e^{iz}=cos(z) + isin(z)$. But I wonder if there is another way from going from $cos(z) + isin(z)$ to $e^{iz}$ without using the Taylor series that is understandable for an undergraduate student?

Answer

Here's a thought:

Let $f(x)=\cos(x)+i\sin(x)$.

Then $f'(x)=-\sin(x)+i\cos(x)=if(x)$.

This would suggest that $f(x)=c\cdot e^{ix}$.

The initial condition $f(0)=\cos(0)+i\sin(0)=1$ gives $c=1$.

So $f(x)=e^{ix}$.