For what values does the property $a \space \ln(i) = \ln(i^a) $ hold? I found to my dismay that $ 4 \ln(i) $ was not returning the same results as $ \ln(i^4) = 0$. In addition, does there exist a log base where if $a$ is a multiple of $4$, and a base $b$ is chosen, then $ a \space \log_b(i) = \log_b(i^4) = 0$? Alternatively, is there another function $f$ of the form $a \space f(i) = f(i^a) $ that exists? I would be eager to know about either option.

Thank you in advance.

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    $\begingroup$ There's nothing to be gained by changing bases, since that just amounts to scaling the natural logarithm: $\log_b c = \frac{\log c}{\log b}$. $\endgroup$ Jul 30, 2022 at 6:23

2 Answers 2


When we write a function $z \mapsto b^z$ in the context of complex analysis, we mean $$z \mapsto \exp(z \log b) ,$$ but as you know $\log$ has infinitely many branches (owing to the fact that $\exp$ is periodic), and so our function depends on our choice of branch for $\log$. Correspondingly, so do the values for which, e.g., $\log(i^a) = a \log i$ holds.

Example Consider the branch of $\log$ with imaginary part in $\left[0, 2 \pi \right)$. Then $\log i = \frac{\pi i}{2}$, so $4 \log i = 2 \pi i$, which is not in the codomain of our branch, hence we cannot have $4 \log i = \log(i^4)$. On the other hand, for real $a$ such that $\Im(a \log i) \in [0, 2 \pi)$, that is, $a \in [0, 4)$, the identity $\log(i^a) = a \log i$ does hold.

  • $\begingroup$ Interesting. Thanks for specifying. Two noob questions. 1. I'm very new to complex analysis, I've just begun reading Visual Complex Analysis by Needham, so I'm not entirely familiar with these branches yet. Any quick reading resources/websites would be highly valuable, should you know any places. 2. Do you happen to know (offhand) if there is any way to extend or change $\Im(a \space \log{i}) $ in a programming environment to allow any arbitrary integer value of $a$ to satisfy the desired property? Is there any mathematical consequence or "penalty" of doing this? All the best. $\endgroup$
    – Nate
    Jul 30, 2022 at 7:19
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    $\begingroup$ @Nate Try this out! complex-analysis.com $\endgroup$ Jul 30, 2022 at 7:42
  • $\begingroup$ @Nate There's no function that lets you recover $a$ from $i^a$: If $a f(i) = f(i^a)$ for all real numbers $a$ (for some fixed choice of $\log$) then $0 f(i) = f(i^0) = f(1) = f(i^4) = 4 f(i)$, so $f(i) = 0$ and hence $f(i^a) = a f(i) = 0$ for all $a$. What is it that you're trying to achieve? (Your question may be an instance of the XY problem: meta.stackexchange.com/questions/66377/what-is-the-xy-problem ) $\endgroup$ Jul 30, 2022 at 11:00

The property $a\ln(i) = \ln(i^a)$ or more generally, $a\ln(z) = \ln(z^a)$ for any $z\in\mathbb C$, is true only for the Principal Branch of the Complex Logarithm.

This is due to the fact that the complex Logarithm is a multi-valued function. In $\mathbb C$, a branch of the (complex) logarithm is generally defined as $$\log z=\{\ln|z|+i\arg(z)+2\pi ik\ :\ k\in\mathbb Z \}.$$ Clearly, this is neither analytic nor even continuous at 0. To get rid of this problem and obtain an analytic branch of the logarithm, we often use $$\operatorname{Log} z:= \ln|z|+i\operatorname{Arg}(z),$$ where $\operatorname{Arg}(\cdot)$ denotes the Principal Argument. This branch is commonly called the Principal Branch.

It is this branch of the logarithm that satisfies all nice properties of the usual (real) logarithm such as:

  • $\operatorname{Log}(zw) =\operatorname{Log}z + \operatorname{Log}w $.
  • $\operatorname{Log}(z^a)=a\operatorname{Log}(z)$. Etc.
  • $\begingroup$ Thank you so much! I'm using i as my default value, so since the length of i is simply 1, that means that ln|(1|) = 0. Then the only thing that I would worry about it i * Arg(z), right? In that instance, would $ln(i^2)$ for instance be written as i * Arg(z^2), and, without loss of generality, i * Arg(i^2)? I'm running into issues getting a numerical match within the Julia programming language, so I will probably go to a coding forum to straighten the rest of this out. $\endgroup$
    – Nate
    Jul 30, 2022 at 6:55
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    $\begingroup$ The claim at the end of the answer isn't true. The principal branch of the argument function takes values in $(-\pi, \pi]$, so, e.g., $\operatorname{Log} [(-1)^2] = \operatorname{Log} 1 = 0$ but $2 \operatorname{Log} (-1) = 2 (\pi i) = 2 \pi i$. $\endgroup$ Jul 30, 2022 at 7:05
  • $\begingroup$ I generally take the principal branch with the cut of the negative real axis and 0 and the principal Argument to be in the open interval. With that convention the above issue may be resolved. $\endgroup$ Jul 30, 2022 at 7:16
  • $\begingroup$ @Nate I think you meant to say $\ln{|i|} = \ln{1} = 0$. And if we're treating $\ln{}$ as the complex definition on any general branch, then $\ln{\left(i^2\right)} = \ln{\left|i^2\right|} + i\text{arg}\left(i^2\right) = 0 + i(\pi + 2n\pi) = i(\pi + 2n\pi)$. The value of $\log{z}$ depends on what branch you choose. $\endgroup$ Jul 30, 2022 at 7:17
  • $\begingroup$ @sadman-ncc Even if one omits the negative real axis from the branch, OP's example still shows that the claim in general is false: $\operatorname{Log}(i^4) = \operatorname{Log} 1 = 0$ but $4 \operatorname{Log} i = 4 \cdot \frac{\pi i}{2} = 2 \pi i$. $\endgroup$ Jul 30, 2022 at 10:37

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