Writing complex number $c$ in form $e^{iz}$

Question

Let $c\in \mathbb C$. Then we can write $c = r\cdot e^{i\varphi},$ where $\varphi\in [0,2\pi)$ and $r\in [0, \infty)$.

My professor stated, however, that if $z \in \mathbb C$, then $c = e^{i\cdot z}$ holds as well, i.e., there exists a representation for $c$ in the "pure" form $c = e^{iz}$ with $z\in \mathbb C$, even though the modulus of $c$, i.e. $|c|$, might be different from $1$. Could sb please prove this?

Thanks!

Answer 1

Only if $r \ne 0$.

Then $r*e^{i\phi} = e^{\ln r}e^{i\phi} = e^{\ln r + i\phi} = e^{i\frac {\ln r+ i\phi}i}$ for $z = \frac {\ln r+ i\phi}i$ if you want.

This assumes we are allowing $e^z= e^{Re(z) + iIm(z)} = e^{Re(z)}e^{iIm(z)}=e^{Re(z)}(\cos (Im(z)) + i\sin (Im(z)))$ to be true by fiat definition.

Which we are.