# Bounding a complex integral over a square

I'm solving the following exercise:

Use the estimate lemma to prove that $$\left|\oint_\gamma \frac{z-2}{z-3}\,{\rm d}z\right| \leq 4\sqrt{10},$$where $\gamma$ is the square with vertices $\pm 1 \pm i$.

Clearly the lenght of the square is $8$. Starting from $-1-i$, counterclockwise, call the sides of the square $\gamma_1,\ldots,\gamma_4$.

• For $\gamma_1$ and $\gamma_3$, we have $|z-2| \leq \sqrt{10}$ and $|z-3| \geq \sqrt{5}$, whence: $$\left|\frac{z-2}{z-3}\right|\leq \frac{\sqrt{10}}{\sqrt{5}} = \sqrt{2}.$$

• For $\gamma_2$, we have $|z-2| \leq \sqrt{2}$ and $|z-3| \geq 2$, so: $$\left|\frac{z-2}{z-3}\right|\leq \frac{\sqrt{2}}{2}.$$

• For $\gamma_4$, we have $|z-2| \leq \sqrt{10}$ e $|z-3| \geq 4$, so: $$\left|\frac{z-2}{z-3}\right| \leq \frac{\sqrt{10}}{4}$$ The greatest of these upper bounds is $\sqrt{2}$. So that inequality is good on all of $\gamma$. We get:$$\left|\oint_\gamma \frac{z-2}{z-3}\,{\rm d}z\right| \leq 8\sqrt{2}.$$

I got these inequalities geometrically, looking at maximum and minimum distances from $2$ and $3$ to said curves $\gamma_j$. I don't see how he got $4\sqrt{10}$. Can someone help me please?

Here's a figure to make your life easier: • Since $8\sqrt{2} < 4\sqrt{10}$, you have a stronger bound. – JimmyK4542 Jun 20 '15 at 3:35
• I thought that for a second there, but I got insecure. So far the book had given the best bounds. But I guess you're right: $$8\sqrt{2} < 4\sqrt{10} \iff 64 \cdot 2 < 16 \cdot 10 \iff 128 < 160 \quad \checkmark$$ – Ivo Terek Jun 20 '15 at 3:38
• Yes, @JimmyK4542 beat me to it (+1). You have shown the desired result. – MPW Jun 20 '15 at 3:38
• Thanks guys. I thought I was crazy. I'll delete the question then. :) – Ivo Terek Jun 20 '15 at 3:38
• I'll make a CW answer to wrap up this. – Ivo Terek Jun 20 '15 at 3:40

## 2 Answers

I'll convert my comment into an answer:

Since $8\sqrt{2} < 4\sqrt{10}$, you have found a stronger upper bound for $\left|\oint_\gamma \frac{z-2}{z-3}\,{\rm d}z\right|$ than what was required by the problem.

We can get the weaker upper bound that the problem is asking for as follows:

For all $z$ on the curve, we have $|z-2| \le \sqrt{10}$ and $|z-3| \ge 2$.

Hence, $\left|\dfrac{z-2}{z-3}\right| \le \dfrac{\sqrt{10}}{2}$, and thus, $\displaystyle\left|\oint_\gamma \frac{z-2}{z-3}\,{\rm d}z\right| \le 8 \cdot \dfrac{\sqrt{10}}{2} = 4\sqrt{10}$.

This gives us a weaker upper bound since we didn't have to consider 3 seperate cases.

We can get an even stronger upper bound by using the Residue Theorem to get $\displaystyle\left|\oint_\gamma \frac{z-2}{z-3}\,{\rm d}z\right| = 0$, (because the only pole $z = 3$ is outside $\gamma$), but that's overkill.

• I knew that it was supposed to be simple! Thanks again - not only my answer was also correct, but you even indicated how to get the proposed bound. Very helpful, thanks again. – Ivo Terek Jun 20 '15 at 4:04

In fact, I got a better bound than the one proposed, so there's no problem at all: $$8\sqrt{2} < 4\sqrt{10} \iff 64 \cdot 2 < 16 \cdot 10 \iff 128 < 160 \quad \checkmark$$

Thanks Jimmy and MPW for the useful comments.