Would I be right in saying that: $$102^i = \exp(i \cdot \ln(102))$$

In general, what is the expansion of $$a^z$$ where $a \in \mathbb{R^+}$ and $z \in \mathbb{C}$ ?

Can I write $$a^{p+iq} = (a^p)(a^{iq})$$$$=(a^p)(e^{iq\cdot ln(a)})$$ which would be a complex number of modulus $a^p$?

But I also come to the conclusion that $$a^i=(ae^{2nπ})^i$$ where $n \in \mathbb{Z}$ .

[Since $$(ae^{2nπ})^i = (a^i)(e^{2n\pi i})$$$$=a^i\cdot 1$$]

P.S.: Please indicate if my reasoning is incorrect somewhere in the steps I have given above.

  • $\begingroup$ Over the complex $log(x)=ln|x|+iArg(x)$ $\endgroup$
    – gbox
    Commented Aug 31, 2017 at 10:30
  • $\begingroup$ The notation $$a^i$$ with $a$ real, refers to a nonexistent entity. For a simple reason, one often refers to the fact that the complex logarithm is multivalued. For example, $$102=e^{\ln(102)}=e^{\ln(102)+42i\pi}$$ "hence" $$102^i=e^{i\ln(102)}=e^{i\ln(102)-42\pi}=e^{i\ln(102)}\cdot e^{-42\pi}$$ "hence" $$1=e^{-42\pi}\ ?$$ $\endgroup$
    – Did
    Commented Aug 31, 2017 at 10:32

1 Answer 1


Yes, you are right... half of the times. The usual way of defining $a^z$, when $a\in(0,+\infty)$, is $e^{z\log a}$. It's expansion as a Taylor series centered at $0$ is$$1+\log(a)z+\frac{(\log a)^2}{2!}z^2+\frac{(\log a)^3}{3!}z^3+\cdots$$

Concrning your other remarks:

  • Yes, $|a^{p+qi}|=a^p$, and your reason for that is correct.
  • $a^i\neq(ae^{2n\pi})^i$, because $a$ and $ae^{2n\pi}$ have distinct logarithms.
  • $\begingroup$ Thanks, but can you elaborate on the other points I mentioned in my question? $\endgroup$ Commented Sep 1, 2017 at 19:13
  • $\begingroup$ @HarryWeasley Done. $\endgroup$ Commented Sep 1, 2017 at 19:31
  • $\begingroup$ Thanks a lot, @JoséCarlosSantos ! I've accepted and upvoted your answer! $\endgroup$ Commented Sep 2, 2017 at 6:24
  • $\begingroup$ BTW, why is my last point incorrect? I've added my reasoning, if it isn't too bothersome, could you look at it and explain why it's wrong? Thanks again! $\endgroup$ Commented Sep 2, 2017 at 6:28
  • $\begingroup$ @HarryWeasley You are assuming that $(a^b)^c=a^{bc}$, which is not true in general. $\endgroup$ Commented Sep 2, 2017 at 9:10

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