In my introductory complex analysis course, the professor mentioned that we can define the complex logarithm via an integral instead of the standard branch cuts method. He didn't prove it but left us with this:

On a simply connected domain $G \subset \mathbb{C}\setminus\{0\}, 1 \in G,$ define $\log z=\int_{1}^{z}\frac{1}{w} dw$, the integral is taken over an arbitrary path from $1$ to $z$. Then, we can say that $e^{\log z}≡ z$ in $G$.

Could someone help me verify this?


In THIS ANSWER, I showed in detail that if $\gamma$ is a rectifiable path in $\mathbb{C}\setminus\{0\}$ from $1$ to $z=re^{i\theta}$, then there is a $k\in\mathbb{Z}$ such that $\displaystyle\int_{\gamma}\dfrac 1w\,dw=\log r+i(\theta+2\pi k)$.

NOTE: The number $k$ is the net number of times that $\gamma$ crosses the positive real axis from the forth quadrant.

Therefore, using this result detailed in the referenced post reveals that

$$\begin{align} e^{\int_\gamma \frac1w\,dw}&=e^{\log(r)+i(\theta+2\pi k)}\\\\ &=\underbrace{re^{i\theta}}_{=z} \,\,\,\,\underbrace{e^{i2\pi k}}_{=1}\\\\ &=z \end{align}$$

as was to be shown!

  • $\begingroup$ +1 for this as well as linked answer. This is exactly how we deal with complex logarithm as an integral. $\endgroup$ – Paramanand Singh Sep 22 '17 at 16:49
  • $\begingroup$ @paramanandsingh Thank you my friend. I hope the linked answer provided the right level of detail. $\endgroup$ – Mark Viola Sep 22 '17 at 17:28

It follows from your definition of $\log$ that $\log'z=\frac1z$. Therefore$$\left(\frac{e^{\log z}}z\right)'=\frac{z\times\frac1ze^{\log z}-e^{\log z}}{z^2}=0.$$So, $e^{\log z}=kz$ for some constant $k$. But $k=k\times 1=e^{\log 1}=e^0=1$.

  • $\begingroup$ $(z e^{-\log z})' = 0$ too $\endgroup$ – reuns Sep 22 '17 at 14:38
  • $\begingroup$ @reuns Indeed... $\endgroup$ – José Carlos Santos Sep 22 '17 at 14:39

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