7
$\begingroup$

There's this problem in my homework, I did it, but somehow it just doesn't seem right. I wonder where the problem is... Please help me :)

Show that $\int_\gamma z^n dz=0$ for any closed smooth $\gamma$ and any integer $n\neq -1$. [If $n$ is negative, assume that $\gamma$ does not pass through the origin, since otherwise the integral is not defined.]

$\rightarrow$ Sol. Let $z=\gamma(t)$, $dz=\gamma'(t)dt$, $a\leq t\leq b$, $\gamma(a)=\gamma(b)$. Then: \begin{align*} &\int_\gamma z^n dz\\ =&\int_a^b\gamma(t)^n\gamma'(t)dt\\ =&\int_{\gamma(a)}^{\gamma(b)} z^n dz\\ =&\frac{z^{n+1}}{n+1}\bigg|_{\gamma(a)}^{\gamma(b)}\\ =&0\\ \end{align*}

What could the problem be? Thank you!

$\endgroup$
1
$\begingroup$

Very good question!

The problem could be with $n=-1$ only, because $\int 1/z=\ln z$ (instead of $z^0$), and that the logarithm is not well defined as function on $\Bbb C$, has countably infinite 'branches' because $$e^{2\pi i+z}=e^z$$ for all $z\in\Bbb C$. And, indeed $$\int_{|z|=1}\frac1z=2\pi i$$ and is not $0$. This is the base of the residue theorem and Cauchy's integral form, and many more..

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.