# existence of antiderivative of a complex analytic function [duplicate]

$$(a)$$ Show that there is a complex analytic function defined on the set $$U=\{z\in\mathbb{C} \ | \ |z|>4\}$$ whose derivative is $$\displaystyle\frac{z}{(z-1)(z-2)(z-3)}$$.

$$(b)$$ Is there a complex analytic function on $$U$$ whose derivative is $$\displaystyle\frac{z^2}{(z-1)(z-2)(z-3)}$$?

The problem clearly reduces to find a analytic function $$f(z)$$ on $$U$$ such that $$f'(z)=\displaystyle\frac{z}{(z-1)(z-2)(z-3)}$$. But if I integrate both sides over the contour of $$U$$, the RHS is analytic on $$U$$. So $$f(z)=0$$. But this result is too absurd to consider. Again same goes for $$(b)$$, but which is not the case since we have to prove no such $$f$$ would exist. I considered the CR-equations but the given function is too large to handle for this. What is the best approach for this problem? Any help is appreciated.

• why don't you just use partial fractions..? Feb 24, 2020 at 6:42
• it becomes -2/(z-2) + 1/[2(z-1)) + 3/[2(z-3)] Feb 24, 2020 at 6:43
• I believe there should be an easier method for this problem, but I can't figure it out. Feb 24, 2020 at 6:43
• I used it above. But it yields no good result. Feb 24, 2020 at 6:46
• there's a shortcut lol. remove the (z-1) from the denominator, and then plug in z=1 to z/((z-2)(z-3) to get 1/2 Feb 24, 2020 at 6:46