I have that $f(z)=\frac{e^{3iz}}{z^2+12}$ and what I actually need to show is that the integral below goes to $0$ as $R\rightarrow \infty$. The $\gamma_R$ curve is the semi circle, counterclockwise with radius $R$. Using the triangle inequality for integrals I get.

$$\left|\oint_{\gamma_R}\frac{e^{3iz}}{z^2+12} \ dz\right|\leq\max_{z\in\gamma_R}\left|\frac{e^{3iz}}{z^2+12}\right||\gamma_r|, \tag 1$$

In the book, there is an intermediate step before they proceed with $(1)$, which is

$$|e^{3iz}|\le e^{-3y} \le 1 \tag 2$$ $$|z^2+12|\ge |z|^2-12=R^2-12 \tag 3$$

This then implies that

$$\max_{z\in\gamma_R}\left|\frac{e^{3iz}}{z^2+12}\right||\gamma_r|\le \frac{\pi R}{R^2-12}\rightarrow 0, \quad \text{as} \quad R\rightarrow\infty.$$

I understand everything, except equation $(2)$. Can someone explain that? What is $y$ and why is $|e^{3iz}|\le e^{-3y}$?

  • 1
    $\begingroup$ $y$ is the imaginary part of $z$. In general,$\lvert e^z\rvert^2=e^z\overline{e^z}=e^{2x}$ for $z=x+iy$. Now use $z=3iz$. $\endgroup$ – ThePuix Oct 22 '18 at 19:10
  • $\begingroup$ I'm sorry, I still don't understand $(1)$. Why is the second part raised to a negative power? How can $z=3iz?$ This does not make sense since it implies $3i=1$. Can you please elaborate? $\endgroup$ – Parseval Oct 22 '18 at 19:18
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    $\begingroup$ $(1)$ follows from the estimation lemma and you are correct, I was being sloppy. Apply what I wrote above for $e^{3iz}$ or $z=3iz'$ or however you want to do it. $\endgroup$ – ThePuix Oct 22 '18 at 19:53

You can write $z=x+iy$. Then $e^{3iz}=e^{-3y+3ix}=e^{-3y}e^{3ix}$. Since $|e^{3ix}|=1$, one has that $|e^{3iz}|=e^{-3y}$.

  • $\begingroup$ Perfect! Thanks a lot it makes sense now. $\endgroup$ – Parseval Oct 23 '18 at 11:59

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