# I don't understand this estimation lemma example

1. Dec 2, 2011

### blahblah8724

I don't understand this estimation lemma example :(

We are given the 'curve',

And part of the example is showing that the contour integral over the top semicircle $C_R$ tends to zero.

Apparently we use the estimation lemma and the fact that, $|z^2 +1| \geq |z|^2 - 1$ to show,

$\left| \int_{C_R} \frac{e^{iz}}{z^2 + 1} dz \right| \leq \int_0^\pi \frac{e^{-Rsin(t)}}{R^2 - 1} dt \leq \frac{2\pi R}{R^2 - 1} \to 0$ as $R \to \infty$

However I don't understand the part where $e^{iz}$ 'goes to' $e^{-Rsin(t)}$, is this some sort of parameterisation on the curve $C_R$?

Help would be much appreciated!

Thanks

2. Dec 2, 2011

### micromass

Re: I don't understand this estimation lemma example :(

Note that

$|e^{a+bi}|=|e^a||e^{bi}|=e^a$

Now write out

$$e^{iz}=e^{iR(\cos(t)+i\sin(t))}$$

3. Dec 2, 2011

### blahblah8724

Re: I don't understand this estimation lemma example :(

Ah thank you that makes sense, but then for the next '$\leq$' step surely as the arclength of $C_R$ is $\pi R$ and as $e^{-Rsin(t)}$ is always less than 1 for $t \in [0,\pi]$, then by the estimation lemma it should be $\frac{\pi R}{R^2 - 1}$ instead of $\frac{2\pi R}{R^2 - 1}$?