# Basic Complex Integrals

1. Jun 22, 2011

### MurraySt

I'm integrating 1/(z-1/2) over the closed disk w/ radius = 3 centered at 0.

I've seen other problems where the final answer was i2pi times f(w) - here w =1/2.

Since f(z) is equal to 1. Is the final answer just i2pi?

Next up:

I have the integral of dt/(2 + sint) the problem then tells me expand sin to its complex definition and replace e^it with z. My question is what does dt become?

Lastly:

I'm given the integral zexp(z^2). I'm asked to provide a proof for why it equals zero on any closed curve gamma.

Do I simply perform a u sub (u=z^2, du=2zdz) and then say that since e^u is entire and thus holomorphic on any region (or curve) it is equal to 0?

2. Jun 22, 2011

### micromass

Hi MurraySt!

The answer to your first question is indeed $2\pi i$, but you might want to give a more rigourous reason for this. Maybe the Cauchy Integral formula.

For the second question, consider $\gamma$ the circle of radius 1, thus $\gamma(t)=e^{it}$ You must find a complex function such that

$$\int_\gamma{f(z)dz}=\int_0^{2\pi}{\frac{dt}{2+\sin(t)}}$$

(I just guessed the limits, as you did not provide any). So using the definition of the complex integral, you must find f such that

$$\int_0^{2\pi}{f(e^{it})ie^{it}dt} = \int_0^{2\pi}{\frac{dt}{2+\sin(t)}}$$

Regarding the third question. Can't you just say that $ze^{z^2}$ is holomorphic on $\mathbb{C}$ and thus any integral is 0.

3. Jun 23, 2011

### MurraySt

As a quick follow-up: If my curve is oriented negatively (clockwise) how does this affect my final answer? Do I simple negate it?

4. Jun 23, 2011

### micromass

Yes!

5. Jun 23, 2011

### MurraySt

So sorry to bother you one final question:

The denominator of my problem is z^2 +4iz -1. I used the quadratic equation to get the answers i(-2+rad3) and i(-2-rad3).

Does my denominator simply become the product of the roots? It scares me that I don't get back any z.

Thanks as always for your help!

6. Jun 23, 2011

### micromass

Well, you're scared for the right reason!! In general, if we have

$$az^2+bz+c$$

and if we have roots $\xi_1,\xi_2$, then we can factor it as

$$az^2+bz+c=a(z-\xi_1)(z-\xi_2)$$