Let $f $ be complex value defined in some bounded closed region $\gamma $ (rectangle) but we don't know whether it has poles or not in $\gamma$.
Let us consider the integral $I$ given by
$$I=\int_{\gamma} f(z)dz $$
Now we have two cases:
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If $I=c$ where $c \in \mathbb{C}$ and $c$ different from $0$ so in this case, we deduce that $f$ is not analytic in the region $\gamma$ because if it is analytic the Cauchy theorem says that $I$ should be zero but we find it different from $0$.
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Now the second case when $I=0 $, in this case, we can't say that $f$ is analytic or not in $\gamma$ (because there are some functions which are not analytic but they integral is zero).
So the Cauchy theorem works in one direction if $I$ is different from $0$ then $f$ it is not analytic in $\gamma$.
My question is what is the condition that makes the Cauchy theorem work in the other direction that means
If $I=0$ $\Longrightarrow$ $f$ is analytic and has no poles in $\gamma$
I suggest the condition for any
$$z_{0} \in \gamma \hspace{0.2cm}\& \hspace{0.2cm} \lim\limits_{z \to z_0} \frac{1}{(k-1)!} \left( \frac{d^{k-1}}{dz^{k-1}} (z-z_{0})^{k}f(z) \right) \neq 0 $$
Because the residue theorem tells us that $I=\sum \text{res} (f;z_{0})$
Is this condition is enough to say that theorem work in the other direction?
Best Answer
Nope. This won't work. Consider $\displaystyle\int_\gamma \bar z{}^2\,dz$ with $\gamma$ the unit circle.