# Thread: Integration in a complex plane

1. ## Integration in a complex plane

Hello,

I have a problem:

Let gamma: z=e^(it) (0<= t <= pi). Show that

|integral (e^z / z) dz | <= pi e^2

I thought to use the Estimation theorem:

But at the end of |f(gamma(t))| i found e^(2cost)/2

And at the end of |(gamma'(t)| i founf 2.

But i don't understand how can i integrate e^(2cost)

Thank you

Richard

2. Use the ML inequality right? If $|f(z)|\leq M$ on the contour $\gamma$ and the length of the contour is L then:

$\left|\int_{\gamma} f(z)dz\right| \leq ML$

On the unit circle $z=e^{it}=x+iy$, $|e^z|=|e^x| |e^{iy}|\leq e^1$. Therefore $\left|\frac{e^z}{z}\right|\leq e$.

3. Yes thank you very much.

Here z=2e^(it) so is true that |e^z|<=e^2 ??

So M=e^2 but why L should be pi??

4. I need help plz..

5. Originally Posted by rickgoz
Yes thank you very much.

Here z=2e^(it) so is true that |e^z|<=e^2 ??

So M=e^2 but why L should be pi??
If $z=2e^{it}$ then $|e^z|\leq e^2$ but can you show that?

L is still the length of the contour which is $2\pi r$ but I believe in the original problem, since 2<e, then it was true for $\pi e^2$ right?

6. Okay thank you very much. It's fine!

Last question: do you know how can I evaluate

|14 + z + z^2| because I don't know if I can use the triangle inequality here..

$|z_1+z_2+\cdots+z_n|\leq |z_1|+|z_2|+\cdots+|z_n|$
so $|14+z+z^2|\leq 14+|z|+|z^2|$