Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

My attempt

I wrote the given function as a sum of rational functions (via partial fraction decomposition), namely $$ \frac{z}{(z-1)(z-2)(z-3)} = \frac{1/2}{z-1} + \frac{-2}{z-2} + \frac{3/2}{z-3}. $$

This then allows me to formally integrate the function. In particular, I find that $$ F(z) = 1/2 \log(z-1) - 2 \log(z-2) + 3/2 \log(z-3) $$ is a complex differentiable function on the set $\Omega = \{z \in \mathbb{C}: |z| > 4\}$ with the derivative we want. So this seems to answer the question, as far as I can tell.

The question then asks if there is a complex differentiable function on $\Omega$ whose derivative is $$ \frac{z^2}{(z-1)(z-2)(z-3)}. $$ Again, I can write this as a sum of rational functions and formally integrate to obtain the desired function on $\Omega$ with this particular derivative. Woo hoo.

My question

Is there more to this question that I'm not seeing? I was also able to write the first given derivative as a geometric series and show that this series converged for all $|z| > 3$, but I don't believe this helps me to say anything about the complex integral of this function. In the case that it does, perhaps this is an alternative avenue to head down?

Any insight/confirmation that I'm not overlooking something significant would be much appreciated. Note that this an old question that often appears on study guides for complex analysis comps (one being my own), so that's in part why I'm thinking (hoping?) there may be something deeper here. For possible historical context, the question seems to date back to 1978 (see number 7 here):

Thanks for your time.

share|cite|improve this question
Watch out for the complex logarithms. They are not defined on a half line starting from $0$, and that halfline surely intersects the domain $|z|>4$. – Beni Bogosel Apr 9 '12 at 20:28
As Beni Bogosel says, it is the logarithm which is the culprit. You might need to do the branch cuts differently, but wikipedia… states that you need a simply connected region to define such a cut. Your domain is not simply connected. – Per Alexandersson Apr 9 '12 at 20:34
Ah, indeed I was being careless regarding the branch cuts. That's good news for me. Thanks @Paxinum, and Beni. – tentaclenorm Apr 9 '12 at 20:58
up vote 3 down vote accepted

The key seems to be that the coefficients $1/2$, $-2$ and $3/2$ sum to 0. So if you choose branch cuts for the three logarithm such that the cuts coincide through the $|z|>4$ region, then jumps at the cuts will cancel each other out (the amount each raw logarithm jumps is always $2\pi i$) and leave a single continuous function on $\{C\in\mathbb C\mid |z|>4\}$.

share|cite|improve this answer
Wonderful, thanks so much. :) – tentaclenorm Apr 9 '12 at 21:08

Using Morera's theorem you can prove that your function has an antiderivative on any disk $D \subset \{z \in \Bbb{C} : |z| > 4\}$.

Morera's theorem states that if $f: D \to \Bbb{C}$ is a complex continuous function such that $\int_T f(z)dz=0$ on any triangle contained in $D$, then $f$ has an antiderivative.

The idea is that for any such disk your function is holomorphic on $D$, and the integral on any closed(usually triangles or rectangles are chosen) path is zero (Cauchy's theorem). Therefore, by Morera's theorem you can find a antiderivative of $f$ on $D$.

I do not know how to prove or disprove that there exists a antiderivative on your whole domain $\{z \in \Bbb{C} : |z| > 4\}$

share|cite|improve this answer
Ah, Morea's theorem! I believe this will be the first time I've actually used that theorem to show something. How nice. Thanks for this! Also, @HenningMakholm has the right idea for defining a single continuous function on this domain of interest. – tentaclenorm Apr 9 '12 at 21:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.