Taking logarithm of complex exponential

Question

Is taking logarithms of complex exponentials allowed/defined? If I take the natural logarithm from both sides of the equation $$ e^{i \varphi_1}=e^{i \varphi_2}; \ \varphi_1, \varphi_2 \in \mathbb{R}$$ I obtain $$i \varphi_1 = i \varphi_2$$ and therefore $\varphi_1=\varphi_2$. However, by Euler's formula, we have $$ i\sin{\varphi_1}+\cos{\varphi_1}=i\sin \varphi_2+\cos{\varphi}_2 $$ and therefore $\varphi_2 =\varphi_1+n\cdot 2\pi$ satisfies the equation $\forall n \in \mathbb{Z}$.

So is there some general rule about these things with complex numbers?

You can read about the Wikipedia article on complex logaritm — Asher2211, Oct 20 '21 at 14:01
Yes there are rules. No, these rules are not unique. Yes, there are problems with them (several branches etc) and the easiness of the real logarithm does not extend. — G. Fougeron, Oct 20 '21 at 14:02
Does this answer your question: in general, $\log(z)$ is multi-valued — ryang, Oct 20 '21 at 14:18

score 3 · Accepted Answer · answered Oct 20 '21 at 14:34

Non-zero complex number can be written as: $$z = e^{x + iy} \text{ , where } x,y\in\mathbb R^2$$ From this it follows that $$\log z = x + \log e^{iy}$$

Since $e^{iy}$ is periodic in $y$, you don't have an inverse function. However you can gice a set of possible inverse values much like when you solve equations like $\sin x = 0 \iff x =k\pi$. If you consider the equation:

$$e^{i\varphi} = e^{iy} \iff \varphi = y + k \cdot 2\pi$$

Sometimes it can be useful to use an inverse function, then you can do the same thing as with $sin$ and $cos$. You can restrict the codomain for one period only. The most useful cases would usually be $[-\pi, \pi]$ or $[0, 2\pi]$. In the latter case: $$\log z = x + \log e^{iy} = x + \left\{\frac{y}{2\pi}\right\}\cdot2\pi$$ , where $\{.\}$ is the fractional part function.

Taking logarithm of complex exponential

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