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

[blahblah8724]

Homework Statement

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 $R \to \infty$

 

[mighty2000]

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



Thank you for your post. Your attempt at a solution is on the right track, but there are a few things that could be improved. First, your use of the estimation lemma is correct, but it would be helpful to include the specific inequality you are using. The estimation lemma states that if a function f is continuous on a contour C and |f(z)| \leq M for all z \in C, then \left| \int_C f(z) dz \right| \leq ML, where L is the length of the contour. In this case, we have f(z) = \frac{1}{(z+i)^2} and the inequality is \left| \int_{C_R} \frac{1}{(z+i)^2} dz \right| \leq \frac{\pi R}{(R-1)^2}.

Second, your use of the arclength of C_R is not quite correct. The arclength of a circle with radius R is 2\pi R, not \pi R. This does not affect your final result, but it is important to use the correct value for L in the estimation lemma.

Finally, your use of the inequality \frac{1}{(z+i)^2} \leq \frac{1}{(R-1)^2} is not justified. While it is true that |z+i| \geq |z| - |i| for all z, this does not necessarily mean that \frac{1}{|z+i|^2} \leq \frac{1}{|z|^2} for all z. In fact, this inequality is only true for z on the real line. For complex numbers, we need to use the triangle inequality to get a better bound. Specifically, we can write |z+i| = |z + i - i + i| \leq |z| + |i| = |z| + 1. Combining this with your previous estimation, we have \frac{1}{|z+i|^2} \leq \frac{1}{(|z| + 1)^2}. This gives us a tighter bound and allows us to use the estimation lemma more effectively.