# Complex Logarithm Calculation

1. Nov 4, 2014

### Bashyboy

Hello everyone,

I am asked to calculate $\log (e^{1+2i})$, and I would appreciate it if someone could verify my calculation..

My textbook defines $\log z$ as $\log z = \ln |z| + i \arg z$.

$\log (e^{1+2i}) = \ln |e^{1+2i}| + i \arg(e^{1+2i}) \iff$

$\log (e^{1 + 2i}) = \ln|e e^{2i}| + i \arg (e e^{2i}) \iff$

$\\log (e^{1+2i}) = \ln e + i \arg (e e^{2i}) \iff$

$\log (e^{1+2i}) = 1 + i \arg (e e^{2i})$

Here is the step that I am not entirely certain about. I know that $e^{2i}$ is the exponential representation of the complex point whose angle is $2$. But I am wondering, would the coefficient $e$ just represent the distance of the complex point from the origin? More generally, the complex exponential function $e^z$ is defined as $e^z = e^x e^{iy}$, and so $e^x$ would represent the distance of the complex point $e^{iy}$?

If this is so, then the answer would seem to be

$\log (e^{1+2i}) = 1 + i(2 + 2 n \pi)$

Is this correct?

Last edited: Nov 4, 2014
2. Nov 4, 2014

### haruspex

Yes, but it is rather a long way about. You can straight away write down that one value of ln(ez) is z. The only question is what other values are there. If w is another value then w = ln(ez). Exponentiate both sides and see what develops.

3. Nov 5, 2014

### Bashyboy

I am not certain I see where the quantity $\ln e^z$ arises. If recall correctly, my professor said that we do not use the natural logarithm when dealing with complex numbers.

4. Nov 5, 2014

### haruspex

OK, I see - you have to start with the definition you quoted, and your method is fine.