# Integral of z^n on a closed contour

1. Oct 14, 2007

### futurebird

I'm trying to show that
$$\int_{c}z^{n}dz= \left\{\frac{0, n\neq-1}{2\pi i, n=-1}\right$$

I did a change of variables with $$z=e^{i\theta}$$ and $$dz=ire^{i\theta}d\theta$$:

$$=i\int^{2\pi}_{0}r^{n+1}e^{i(n+1)\theta}d\theta$$

$$=ir^{n+1}\int^{2\pi}_{0}e^{i(n+1)\theta}d\theta$$ Moving the constant out.

$$=-(n+1)r^{n+1}\int^{2\pi}_{0}\frac{e^{i(n+1)\theta}d\theta}{i(n+1)}$$ Getting ready to integrate.

$$=-r^{n+1}(n+1)\left[e^{i(n+1)\theta}\right]^{2\pi}_{0}$$

$$=-r^{n+1}(n+1)$$???

This is nothing like the answer... where am I going wrong?

2. Oct 14, 2007

### Hurkyl

Staff Emeritus
Firstly, your work is only valid when n + 1 is nonzero.

Thirdly, you computed the antiderivative wrong when going from your antepenultimate step to your penultimate step.

3. Oct 14, 2007

### futurebird

Okay thanks.

4. Oct 14, 2007

### HallsofIvy

If $n+ 1\ne 0$, what IS $$e^{i(n+1)(2\pi)}$$ and $$e^{i(n+1)(0)}$$?

(It's NOT 1!)

5. Oct 14, 2007

### futurebird

Wait why isn't $$e^{i(n+1)(0)}=1?$$ or are you talking about the whole integral?

$$e^{i(n+1)(2\pi)}$$

$$=e^{2\pi i}e^{2\pi in}$$

$$=e^{2\pi i}(e^{2\pi i})^n$$

$$=(1)(1)^n$$

$$=1$$

I thought that I fixed my errors since I had the correct solution after integrating in the right way. Then I looked at what happened when n=-1 and everything seemed fine.

But, if this is wrong, then I'm still missing something.

****Nevermind****** I see what you mean now!

Last edited: Oct 15, 2007
6. Oct 14, 2007

### futurebird

$$=i\int^{2\pi}_{0}r^{n+1}e^{i(n+1)\theta}d\theta$$

$$=ir^{n+1}\int^{2\pi}_{0}e^{i(n+1)\theta}d\theta$$ Moving the constant out.

$$=\frac{r^{n+1}}{n+1}\int^{2\pi}_{0}i(n+1)e^{i(n+1)\theta}d\theta$$ Getting ready to integrate, the right way

$$=\frac{r^{n+1}}{n+1}\left[e^{i(n+1)\theta}\right]^{2\pi}_{0}$$

$$=\left[e^{i(n+1)2\pi}-e^{0}\right]$$

$$=0$$

But this whole process only makes sense if $$n\neq-1$$, so for that case we just plug in -1 for n to:

$$i\int^{2\pi}_{0}r^{0}e^{i(0)\theta}d\theta=2\pi i$$

So,

$$\int_{c}z^{n}dz= \left\{\frac{0, n\neq-1}{2\pi i, n=-1}\right$$