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If $f(z)$ is continuous on $\Omega$ (open) and has a simple pole at $z=a$ on $\Omega$, then $f$ has no primitive function on the punctured open neighborhood of $z=a$ and thus the integral over a circle, say centered at $a$, will not vanish.

In the case $f(z)$ has a pole of order $> 1$ it has also a primitive function except at the poles. According to a theorem, if $f(z)$ has a primitive function then the integral over a simply closed path will vanish.

From the book I am learning complex analysis from it is not clear; if the integral of $f(z)$ over the closed path will also not vanish if $f(z)$ has a pole of order $>1$? More generally, in which cases does the integral of a complex function over a closed path vanish since a holomorphic function must not have a primitive function?

Thanks for your comment.

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I assume you mean "holomorphic" when you write "continuous".

Suppose $f$ has a pole at $z_0$. Let $\sum_{n=-p}^\infty a_n(z-z_0)^n$ be the Laurent series of $f$. The integral of $f$ on a loop around $z_0$ then depends only on $a_{-1}$. In fact, it is equal to $2i\pi\cdot a_{-1}$.

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  • $\begingroup$ Thanks. But f(z) is not continuous at the pole also for the pole of order >1. Does it change anything ? $\endgroup$ – ivo Jun 6 '15 at 15:59
  • $\begingroup$ @ivo No, it doesn't. $\endgroup$ – Amitai Yuval Jun 6 '15 at 16:30

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