# Contour integral property

1. Feb 6, 2005

### cateater2000

Hi, I'm having a bit of trouble with this question.

Use the property |integral over c of f(z)dz|<=ML
to show |integral over c of 1/(z^2-i) dz|<=3pi/4

where c is the circle |z|=3 traversed once counterclockwise

thanks in advance for any tips.

2. Feb 6, 2005

### Janitor

I'm not sure what "ML" means, though presumably it is a constant. Is the "L" a subscript by chance?

At any rate, a good start would be to factor the denominator of the integrand, and determine the poles. Once you have done that, ask yourself if the poles lie inside the circle |z|=3. That should get you well on the way to proving what you are asked to prove.

3. Feb 6, 2005

### cateater2000

I think the ML
is M*L two different constants

I geuss I could do the pole thing, thanks for the reply.

4. Feb 7, 2005

### Galileo

I`m sure M means the upper bound of |f(z)| on C and L is the arc length of the contour C.

$$\left|\int_C f(z)dz\right|=\left|\int_a^bf(z(t))\frac{dz(t)}{dt}dt\right|$$.

Now use the fact that:
$$\left|\int_a^b f(t)dt\right|\leq \int_a^b \left|f(t)\right|dt$$