# Limit problem!

1. Dec 2, 2011

### blahblah8724

1. The problem statement, all variables and given/known data

Using $C_R$ as,

($C_R$ doesn't include the bottom line between R and -R)

Prove that,

$\lim_{R \to \infty} \int_{C_R} \frac{1}{(z+i)^2} dz = 0$

2. The attempt at a solution

Using the estimation lemma and trying to find a bound on $\frac{1}{(z+i)^2}$

So if $|z| = R$ then,

$|(z + i)^2| = (|z+i|)^2 \geq (|z| - |i| )^2 = (|z| - 1)^2 = (R - 1)^2$

>>>(Not sure of the integrity of this step)

Thus,

$\frac{1}{(z+i)^2} \leq \frac{1}{(R - 1)^2}$

So using estimation lemma and the fact that arclength of $C_R$ is $\pi R$ then,

$\left| \int_{C_R} \frac{1}{(z+i)^2} dz \right| \leq \int_{C_R} \left|\frac{1}{(z+i)^2} \right| |dz| \leq \frac{\pi R}{(R-1)^2} \to 0$ as $R \to \infty$