Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Please can somebody give me hint for this?

For $n\ne -1$ $$\frac{1}{2\pi i}\int_{C} z^ndz=0$$

Where C is a simple closed curve with the usual positive orientation and its inside.

share|cite|improve this question
Where is $f$ mentioned in the problem? – Alex Becker Apr 19 '12 at 2:42
@AlexBecker: It is just a typo, I fix it. Thanks. – Hassan Muhammad Apr 19 '12 at 3:19
Is $n$ an integer? For $n=1/2$, this is false... – J. M. Apr 19 '12 at 3:25

For $n\neq -1$, $f(z)=z^n$ has a primitive $F(z)=\frac{z^{n+1}}{n+1}$. Parameterize $\gamma$ with $t$ from 0 to 1.

$\displaystyle \int_\gamma f(z) \ dz = \displaystyle \int_\gamma F'(z)\ dz = \displaystyle \int_0^1 F'(\gamma(t)) \gamma'(t)\ dt = \displaystyle \int_0^1 (F(\gamma(t)))'\ dt =F(\gamma(1))-F(\gamma(0))=0$ because the path is closed.

This works even if the curve is not simple.

share|cite|improve this answer
Can you give the value of the parameterize $\gamma$? – Hassan Muhammad Apr 19 '12 at 7:21
Yes. You can apply linear transformations to change the parameterization at will. Alternatively, just note that because the curve is closed, the values at the starting point and end point must be the same. – Potato Apr 19 '12 at 18:28

let my $C(t)= re^{it}$, $0\le t\le 2\pi$ and n is an integer, It follows that $$I=r^{n+1} \int_{0}^{2\pi}ie^{(n+1)it}dt= r^{(n+1)}\frac{e^{it(n+1)}}{n+1}|_{0}^{2\pi} \text {when }n\ne -1$$ and $2\pi i$ when $n=-1$, that is $\int f(z)dz= 0$ when $n\ne -1$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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