This is a follow up question to this question: Computing $\int_{-\infty}^{\infty} \frac{\cos x}{x^{2} + a^{2}}dx$ using residue calculus. For clarity, I'll reproduce the question here: calculate $\int_{-\infty}^\infty \frac{\cos x}{x^2 + a} \ dx$ using the residue theorem. The answer given lets $\gamma$ be the semi-circle centered at $0$ with radius $R$ in the upper half of the complex plane, and we can further let $\gamma_1$ be the arc of $\gamma$, and $\gamma_2$ be the part of $\gamma$ along the real line.

I'm working through the proof, which involves using the fact that $\cos x = \Re(e^{ix})$ to solve the problem of $\cos x$ not being bounded if $x$ is allowed to be imaginary. However, it is stated simply that the integral

$$\int_{\gamma_1}\frac{e^{ix}}{x^2 + a^2}\ dx \to 0\ \ \ \text{ as } \ \ \ R \to 0,$$

which I don't fully understand. I have the following bound on the integral:

$$\left|\int_{\gamma_1}\frac{e^{ix}}{x^2 + a^2}\ dx \right| \le \sup_{s \in \gamma_1}\left|\frac{e^{ix}}{x^2 + a^2}\right| \cdot \text{length}(\gamma_1). $$

But I am having a difficult time getting the RHS to go to $0$ as $R$ goes to $0$. Clearly,

$$|e^{ix}| \le 1,$$

but then we are left with

$$\left|\int_{\gamma_1}\frac{e^{ix}}{x^2 + a^2}\ dx \right| \le \sup_{s \in \gamma_1}\left|\frac{1}{x^2 + a^2}\right| \cdot \pi R. $$

Now, we have $x = z + iy$, where $-R \le z \le R$, and $0 \le y \le R$. But how can I find a lower bound on $|x^2 + a^2|$? I'm having a hard time visualizing what it should be. I've tried expanding out $x^2$, and I have that

$$|x^2+a^2| \ge |z^2 + 2izy + a^2| - |y^2|,$$

but can't get anywhere from there. It feels as if I should eventually get $|x^2 + a^2| \ge R^2+a^2,$ but I can't really prove this.

I know this is probably a very silly question, but I seem to always be misunderstanding the details for proofs in complex analysis, and I want to make sure I understand every piece of this one.