The question is about finding the range and domain of $f(z)=e^z$, and I have read the solution and the reason why many times but I just don't understand it. My professor simply taught domain and range with real numbers and gave us complex number question which I don't exactly know how to do, so here's the answer:

The complex exponential function has natural domain $\mathbb{C}$; hence $f$ has natural domain $\mathbb{C}$. If $z=x+iy$ with $x, y∈\mathbb{R}$ then $e^z=e^x e^{iy} = e^x(\cos y+i\sin y)$. Since $e^x$ can take any positive value, and $y$ can take any value in $\mathbb{R}$ including all those in $(−π, π]$ then $f(z)$ can take any non zero modulus and all arguments. Hence $f(z)$ has the range $\mathbb{C}\setminus\{0\}$.

I think I have a clear understanding on most of it other than the reason why the range would be $\mathbb{C}\setminus\{0\}$. Could you also please tell me a way to find the range for all complex functions too? That would make my life so much easier!

  • $\begingroup$ For all ? That's quite a lot...The way or adguing above is nice, and every singl;e case must be dealt with separatedly. $\endgroup$
    – DonAntonio
    Mar 21, 2021 at 14:35
  • $\begingroup$ Did you mean $f(z)=e^z$? To see that this has a range of $\mathbb C-\{0\}$, just use polar coordinates (as described). And you can't be serious about "a way to find the range for all complex functions"...there are an awful lot of functions. $\endgroup$
    – lulu
    Mar 21, 2021 at 14:36

1 Answer 1


$e^x > 0$, view it as the non-zero radius.

$e^{iy}$ is a point on the unit circle, it decides the direction.

Given any non-zero point, find the radius $r$ and the angle $\theta$, we can solve $e^x=r$ and let $y$ be $\theta$.

There is no solution for $e^xe^{iy}=0$, since $e^x$ and $e^{iy}$ are non-zero, we can divide them by their multiplicative inverses and get a contradiction of $1=0$.


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