Show the existence of a complex differentiable function defined outside $|z|=4$ with derivative $\frac{z}{(z-1)(z-2)(z-3)}$

Question

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): http://math.rice.edu/~idu/Sp05/cx_ucb.pdf

Thanks for your time.

Answer 1

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 $\{C\in\mathbb C\mid |z|>4\}$.

Answer 2

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 $\{z \in \Bbb{C} : |z| > 4\}$