Quick question on Jordan's Lemma

Question

The key equation in Jordan's Lemma is:

$$I_\Gamma = \int_\Gamma e^{imz}f(z) dz \rightarrow 0$$

as $R \rightarrow \infty$.

Why is $|\exp(imz)| = |\exp(-mR\sin\theta)|$?

Answer 1

Assuming $\Gamma$ is a circular arc centred at zero, we use the parameterisation $z = Re^{i\theta} = R\cos\theta + iR\sin\theta$ which gives

$$|\exp(imz)| = \exp(\operatorname{Re}[imz]) = \exp(\operatorname{Re}[-mR\sin\theta + imR\cos\theta]) = \exp(-mR\sin\theta).$$