Integral of z^n on a closed contour

[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?

 

[Hurkyl]

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

Secondly, your ultimate step doesn't follow from your penultimate step.

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

 

[futurebird]

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

Secondly, your ultimate step doesn't follow from your penultimate step.

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

Okay thanks.

 

[HallsofIvy]

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

(It's NOT 1!)

 

[futurebird]

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

(It's NOT 1!)

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!

 

[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