I Estimate the magnitude of a line integral exp(iz) over a semicircle

[hotvette]

TL;DR Summary

Clarify statement in Complex Variables (S. Fisher) about magnitude of a line integral exp(iz) over a semicircle

Not homework, just trying to understand a statement in the book. On page 158 in Fisher, the following statement is made:
In these applications of the Residue Theorem, we often need to estimate the magnitude of the line integral of e^{iz} over the semicircle = Re^{i\theta}, \; 0 \le \theta \le \pi. We note that:
<br /> \left\vert e^{iz} \right\vert = e^{\text{Re}(iz)} = e^{-R \sin(\theta)}<br />
To understand the above, I did the following.
\begin{align}
&z = Re^{i\theta} = R \cos(\theta) + i R \sin(\theta)
\\
&iz = i R \cos(\theta) - R \sin(\theta)
\\
&e^{iz} = e^{i R \cos(\theta) - R \sin(\theta)} = e^{i R \cos(\theta)} e^{-R \sin(\theta)}
\\
&\left\vert e^{iz} \right\vert = \left\vert e^{i R \cos(\theta)} e^{-R \sin(\theta)} \right\vert
= \left\vert e^{i R \cos(\theta)} \right\vert \; \left\vert e^{-R \sin(\theta)} \right\vert
\\
&e^{i R \cos(\theta)} = e^{ix} && \text{where $x = R \cos(\theta)$, real}
\\
\therefore \, &\left\vert e^{i R \cos(\theta)} \right\vert = 1
&& \cos^2 x + \sin^2 x= 1
\\
&\left\vert e^{-R \sin(\theta)} \right\vert = e^{-R \sin(\theta)}
&& \text{since } e^{-R \sin(\theta)} \ge 0
\\
\therefore \, &\left\vert e^{iz} \right\vert = e^{-R \sin(\theta)} =e^{\text{Re}(iz)}
\end{align}
Is it correct?

 

[BvU]

Yes (IMHO)

 

[hotvette]

Thanks!

 

[BvU]

How does one determine a modulus of a complex number ?

 

[BvU]

Use that to prove $|\bf ab | = |a|\,|b|$

 

[Fred Wright]

I offer the following observation:$$\left| e^{iz} \right|= \sqrt{e^{iz}e^{-iz^*}}=\sqrt {e^{i(z-z^*)}}\\
=e^{\frac{iR}{2}((\cos(\theta)+i\sin(\theta))-(\cos(\theta)-i\sin(\theta)))}=e^{-R\sin(\theta)}$$

 

[BvU]

Got it !