# Confusion with finding the range on complex number functions

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!

• For all ? That's quite a lot...The way or adguing above is nice, and every singl;e case must be dealt with separatedly. Mar 21, 2021 at 14:35
• 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.
– lulu
Mar 21, 2021 at 14:36

$$e^x > 0$$, 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 $$1=0$$.