Say I have to solve the following equation $$ \operatorname{Log}(z) = -2+3i $$ where this is the principal value of the logarithm. I can think of two ways to solve it; first is just raising both sides from $e$ to get $$ z = e^{-2+3i} \stackrel{?}{=} e^{-2+3i+2i\pi k} $$ Where I'm not sure if the second equality is necessary. Second is noting $$ \operatorname{Log}(z) = \operatorname{Log}|z|+i\operatorname{Arg}(z) = -2+3i $$ where $\operatorname{Arg}(z)$ is the principl; branch. This means for polar coordinates $r = e^{-2}$ and $\theta = 3$ and the solution $e^{-2+3i}$ is unique. (mini-question: would this mean if my imaginary part was 5 there would be no solution because this is outside $(-\pi,\pi]$?)

Of course, if this was $\log(z)$ then I would have $\arg(z) = 3+2\pi k$ and there would be multiple solutions. I think the unique solution is correct, but at the same time is $e^{-2+3i+2i\pi}$ not also in the domain of $\operatorname{Log}(z)$? I thought $\operatorname{Log}(z)$ had a domain of everything except $(-\infty,0]$.


There is no difference between your approaches. Remember that the exponential function is periodic with period $2\pi i$. Therefore, $e^{-2+3i+2i\pi k}$ is always the same number, no matter the choice of $k\in\mathbb Z$.

  • $\begingroup$ I understand the periodic nature of $e^{z}$, but part of my issue comes from defining $\operatorname{Log}(z)$ in terms of the principal value of $\arg(z)$. My understanding is that equating imaginary parts gives $\operatorname{Arg}(z)=3$ and only 3, but this seems to imply the only solution is $e^{-2+3i}$. Maybe I'm misunderstanding how $\operatorname{Arg}(z)$ is defined. $\endgroup$ – Krish100 Jun 23 '18 at 17:50
  • $\begingroup$ A logarithm of a complex number $w\neq0$ is a complex number $z$ such that $e^z=w$. When $w\notin(-\infty,0]$, we call principal value of the logarithm of $w$ to the only logarithm whose imaginary part is in $(-\pi,\pi)$. In other words, among all logarithms of $w$, we pick the only logarithm $z$ of the form $a+bi$ with $b\in(-\pi,\pi)$. This means, of course, that $b=\operatorname{Arg}(z)$. $\endgroup$ – José Carlos Santos Jun 23 '18 at 18:08
  • $\begingroup$ Alright, so for this equation we should also have $e^{-2+3i+2i\pi}$ as another solution? But why does this solution seem hidden in the second approach? Doing it that way it seems like my solution can only have an argument of 3, because $\operatorname{Arg}(z)$ does not get the multiple $2\pi k$ factor. $\endgroup$ – Krish100 Jun 23 '18 at 18:24
  • $\begingroup$ How can $e^{-2+3i+2i\pi}$ be another solution when it turns out that $e^{-2+3i+2i\pi}=e^{-2+3i}$? $\endgroup$ – José Carlos Santos Jun 23 '18 at 18:27
  • $\begingroup$ I think see what you're saying. Ok... so if the question were to ask "find all solutions", would only $e^{-2+3i}$ be adequate, or is that an ill formatted question? What if changed from $\operatorname{Log}$ to $\log$? $\endgroup$ – Krish100 Jun 23 '18 at 18:32

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