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Show that there is a holomorphic function defined in the set $$\Omega=\lbrace z\in \mathbb{C}:\vert z \vert >4 \rbrace $$ whose derivative is $$\frac{z}{(z-1)(z-2)(z-3)}$$ Is there a holomorphic function on $\Omega$ whose derivative is $$\frac{z^2}{(z-1)(z-2)(z-3)}\,\,?$$

I know that each holomorphic function on an open convex set has an antiderivative, but $\Omega$ isn't convex. Should I look for functions with these derivatives, such as $f(z)=\frac{1}{2}(\log(1-z)-4\log (2-z)+3\log (3-z))$ and $g(z)=\frac{1}{2}(9\log (z-3)-8\log (z-2)+\log(z-1))$, or are there any well-known propositions on finding antiderivatives in open, non-convex sets?

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This may be helpful: Suppose $f \in H(\Omega).$ Then $f$ has an antiderivative in $\Omega$ iff the integral of $f$ along any closed contour in $\Omega$ is $0.$

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  • $\begingroup$ I didn't know that. Can the antiderivative be constructed or does its existence follow from some other theorem? $\endgroup$ – The Substitute Apr 2 '15 at 13:39
  • $\begingroup$ It's constructive. Fix a point $a$ in the domain. Then define $F(z) = \int_{\gamma_z} f(w)dw,$ where $\gamma_z$ is any contour from $a$ to $z.$ Why is this well-defined? Because of the zero integral condition. Checking $F'(z)=f(z)$ is then very pleasant. $\endgroup$ – zhw. Apr 2 '15 at 17:30
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We have that: $$\frac{1}{(z-1)(z-2)}=\frac{1}{z-2}-\frac{1}{z-1},$$ hence: $$\frac{1}{(z-1)(z-2)(z-3)}=\frac{1}{2(z-1)}-\frac{1}{z-2}+\frac{1}{2(z-3)}$$ and: $$\frac{z}{(z-1)(z-2)(z-3)} = \frac{1}{2(z-1)}-\frac{2}{z-2}+\frac{3}{2(z-3)}\tag{1}$$ as well as: $$\frac{z^2}{(z-1)(z-2)(z-3)} = \frac{1}{2(z-1)}-\frac{4}{z-2}+\frac{9}{2(z-3)}\tag{2}$$ where $\frac{1}{z-1},\frac{1}{z-2},\frac{1}{z-3}$ are regular functions over the domain $D=\{z\in\mathbb{C}:|z|\geq 4\}$, so that primitives for the LHS of $(1)$ or $(2)$ over $D$ can be written in terms of $\log(z-1)$, $\log(z-2)$, $\log(z-3)$.

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  • $\begingroup$ I don't beleive you're argument for the second case is correct since when I integrate over a simple, closed $C^1$ curve in $U$, I don't get $0$. $\endgroup$ – The Substitute Apr 17 '15 at 23:42

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