How is $\left| \exp(iaRe^{i\theta}) \right|\le e^{-aR\sin\theta}$?

Question

In one book on complex variables, in the proof of Jordan's Lemma, For any constant $a>0$, and any radius $R>0$, it is stated that $\left| \exp(iaRe^{i\theta}) \right|\le e^{-aR\sin\theta}$. I think this is plain false because $\left| \exp(iaRe^{i\theta}) \right|=1$, while $e^{-12 \sin(\pi/4)}<1$.

Can someone please clarify what I'm not seeing? Or is it such a mistake printed in an expensive book?

Answer 1

It is not necessarily true that $\left| \exp(iaRe^{i\theta}) \right|=1$, since $e^{i \theta}$ might not be real. Note that:

$\exp(iaRe^{i\theta})$ $=\exp(iaR(\cos\theta+i\sin\theta))$ $= \exp(iaR\cos\theta-aR\sin\theta)$

$= \exp(iaR\cos\theta) \cdot \exp(-aR\sin\theta)$ $= \exp(iaR\cos\theta) \cdot \exp(-aR\sin\theta)$

Since $iaR\cos\theta$ is purely imaginary, we have $|\exp(iaR\cos\theta)| = 1$. Since $-aR\sin\theta$ is purely real, we have $|\exp(-aR\sin\theta)| = \exp(-aR\sin\theta)$.

Therefore, $|\exp(iaRe^{i\theta})| = |\exp(iaR\cos\theta)| \cdot |\exp(-aR\sin\theta)| = \exp(-aR\sin\theta)$.

Answer 2

Remember that for any complex number $z$, $\;\lvert\exp z\rvert=\mathrm e^{\mkern1mu\operatorname{Re}z}$. Now $$\mathrm iaR\mathrm e^{i\theta}=aR\mathrm e^{\mathrm i(\theta+\tfrac\pi2)},$$ whence $$\operatorname{Re}\bigl(\mathrm iaR\mathrm e^{i\theta}\bigr)=aR\cos(\theta+\tfrac\pi2)=-aR\sin\theta.$$