Why does an holomorphic function have a primitive in a simply connected space? Also, it have a primitive only in a simply connected space?
Primitive of an holomorphic function
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complex-analysis
holomorphic-functions
1 Answers
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Let $f(z)$ be your holomorphic function, defined on $U \subset \mathbb C$. Define your primitive by picking a point $z_0 \in \mathbb C$, and writing $$ F(z) = \int_{z_0}^z f(w) dw. $$ But for this to make sense, this integral needs to be independent of the choice of integration path between $z_0$ and $z$!
If $U$ is simply connected, the integral is indeed independent of the path, by Cauchy's theorem.
That's not to say that it's always impossible to define primitives on non-simply-connected spaces. For example, if $U = \mathbb C - \{0\}$, then $f(z) = 1$ has a primitive, namely $F(z) = z$, but $f(z) = 1/z$ is an example of a function which does not.
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1It may also be worth noting that some functions have primitives even when their natural domain is not simply connected (e.g., $f(z)=\frac{1}{z^2}$). This is a little less trivial than your $f(z)=1$ example... – 2017-02-26