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I have solved the following exercise, in which I had to compute $$\int _\gamma \! \overline z ^2 \, \mathrm{d}z,$$ where $\gamma$ is

  1. The circumference $\left| z\right| =1$
  2. The circumference $\left|z-1 \right|=1$

Let's call each curve $\gamma_1$ and $\gamma_2$ respectively. Then I have parametrized them as $\gamma_1 \left( t \right) = e^{it}$ and $\gamma_2 \left( t \right) = 1+ e^{it}$ for $t\in\left[0,2\pi \right)$.

After using the following formula for complex integration around a curve $$\int _\gamma \! f\left( z \right) \, \mathrm{d} z = \int _0 ^{2\pi} f \left( \gamma \left( t \right) \right) \gamma ' \left( t \right) \, \mathrm{d}t$$ I have found that $$\int_{\gamma_1} \! \overline z ^2 \,\mathrm d z = 0$$ and $$\int_{\gamma_2} \! \overline z ^2 \,\mathrm d z = 2\pi i.$$

Then here go the actual questions:

  1. Can we say by Cauchy theorem that $\overline z ^2$ is holomorphic in $\mathrm{supp}\left(\gamma_1\right)$ but not in $\mathrm{supp}\left(\gamma_2\right)$?
  2. If it is holomorphic in $\mathrm{supp}\left(\gamma_1\right)$, then there exists $F\left( z \right)$ such that $F'\left(z \right)=\overline z ^2$ defined in $\mathrm{supp}\left(\gamma_1\right)$? Can we compute it explicitly?

I have trouble seeing how this function has different behaviour in two overlapping sets.

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Cauchy's integral theorem says that if $f$ is holomorphic (on a simply connected domain $D$) then $\int_\gamma f(z)\,dz = 0$ for every closed curve in $D$. Not the other way around.

Just because the integral of a particular function along a particular curve happens to be $0$, it doesn't tell us that $f$ is holomorphic. On the other hand, if we find a closed curve $\gamma$ such that $\int_\gamma f(z)\,dz \neq 0$, we can be sure that $f$ is not holomorphic on (a neighbourhood of) the interior of $\gamma$.

If you want some kind of converse, Morera's theorem says that if $f$ is continuous and $\int_\gamma f(z)\,dz = 0$ for every closed curve $\gamma$ in $D$, then $f$ is holomorphic on $D$.

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