I'm reviewing old homework problems.

Let a function $f(z)$ be some branch of $z^{1/n}$. Show that $$f'(z) = \frac{f(z)}{nz} \textbf{.}$$

I wrote:

Let $z = re^{i(\theta+2\pi k)}$ and fix $k$ thus for $f(z) = z^{1/n}$, the branch is fixed. Since $f(z)$ is analytic everywhere, we can differentiate thus $f'(z) = \frac{1}{n} z^{1/n \, -1}\, {\color{red} =} \,\frac{1}{n} \frac{z^{1/n}}{z}$, for a fixed branch.

On my homework, the professor pointed at the equal sign that I've marked in red and said: both sides are multivalued so it is unclear what this line means.

So I don't know how I would go about this problem. In general, branch cuts confuse me. I understand that the function $f(z) = z^{1/n}$ has different branches and that we need to choose one branch to allow the function to be continuous and single valued.

  • $\begingroup$ Not all branches of $z^{1 / n}$ arise by fixing $k$ and taking the branch defined by writing $z = r e^{i \theta + 2 \pi k} \mapsto e^{[i (\theta + 2 \pi k)] / n}$. Indeed, this given rise only to $n$ of the infinitely many branches. $\endgroup$ Dec 26, 2015 at 0:41

1 Answer 1


Hint By definition, any branch $f(z)$ of $z \mapsto z^{1 / n}$ satisfies $$f(z)^n = z .$$ In particular, both sides of this expression are single-valued. Now, differentiate w.r.t. $z$.

Additional hint Differentiating w.r.t. $z$ gives $$n f(z)^{n - 1} f'(z) = 1 .$$ (Strictly speaking, we only do this on the interior of the set of points where the branch $f$ is continuous.) Multiplying through by $f(z)$ gives gives $$f(z) = n f(z)^n f'(z) = n z f'(z) .$$

  • $\begingroup$ Okay, I see how this does work. Can you elaborate a little more on my solution. Specifically, is there a way I could have chose a specific branch or is your way the only way? $\endgroup$
    – astudent
    Dec 26, 2015 at 0:47
  • $\begingroup$ The problem statement suggests to me that one wants to show the claim for a general branch of $z^{1 / n}$, so one has to start with the equation characterizing a general branch, namely $f(z)^n = z$ (or something equivalent). Put another way, in this context $z^{1 / n}$ a priori denotes a "multivalued function" and not (despite the notation) an exponential function. We use this notation because its branches behave like exponential functions in a certain way, and the aim of this problem is to establish one of those properties, namely that they obey (more or less) the usual derivative rule. $\endgroup$ Dec 26, 2015 at 1:25

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