A problem with an estimation in complex analysis

Question

Consider the following estimations:

Where the "arc" is the semicircle of radius $a$ centered at the origin, and we consider $t>0$.

I don't understand the last estimation: I get that $z^2 = a^2$ on the arc, but what then?

Answer 1

The gist of the proof is shown for a similar example on the wikipedia page for the Estimation lemma.

The first step is just a general rule for Riemann Integrable functions (see here). That is $$\left|\int f(x)dx\right|\leq \int \left|f(x)\right|dx$$

The second step uses the Cauchy-Schwarz inequality . In this case $$\left|\frac{e^{-itz}}{z^2+1}\right|\leq \left|e^{-itz}\right|\left|\frac{1}{z^2+1}\right|=\left|\frac{1}{z^2+1}\right|$$ since $\left|e^{-itz}\right|=1$ for any value of $z$.

Step 3 uses the triangle inequality and is essentially shown in my wikipedia link above, but I will reproduce their result here (the triangle inequality comes in at the third step): $$\left|z\right|^2=\left|z^2\right|=\left|z^2+1-1\right|\leq \left|z^2+1\right|-1 \to \left|z^2+1\right|\geq\left|z\right|^2+1 = a^2+1$$ since we know $\left|z\right|=a$ on the arc.

The last step simply uses that $\frac{1}{\left|a^2+1\right|}$ is an upper bound to $\left|f(x)\right|$ for the whole and so applies the estimation lemma with $M=\frac{1}{\left|a^2+1\right|}$ and the arc length $\Gamma=\pi a$.

Answer 2

I think it could be explained "geometrically", as follows (assuming you refer to the last use of the $\leq$ sign, not the last equality).

$z^2$ maps the semicircle of radius $a$ to the circumference of radius $a^2$.

Adding $1$ stands for a translation to the right.

As we want to provide an estimate from above, we look at where the denominator is smallest, and that indeed occurs if we consider the point $P$, the left-most point of the translated circumference, located at $(-a^2 +1;0)$ (smallest distance to the origin).

This explains the appearance of the term $a^2 - 1$ (as $a>1$, $\vert 1 - a^2 \vert = a^2 -1$).