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

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Summary:

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

Main Question or Discussion Point

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 [itex] e^{iz} [/itex] over the semicircle [itex]= Re^{i\theta}, \; 0 \le \theta \le \pi[/itex]. We note that:
[tex]
\left\vert e^{iz} \right\vert = e^{\text{Re}(iz)} = e^{-R \sin(\theta)}
[/tex]
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?
 
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Answers and Replies

  • #2
BvU
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Yes (IMHO)
 
  • #3
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Thanks!
 
  • #4
BvU
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How does one determine a modulus of a complex number ?
 
  • #5
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Use that to prove ##|\bf ab | = |a|\,|b|##
 
  • #6
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I offer the following observation:$$\left| e^{iz} \right|= \sqrt{e^{iz}e^{-iz^*}}=\sqrt {e^{i(z-z^*)}}\\
=e^{\frac{iR}{2}((\cos(\theta)+i\sin(\theta))-(\cos(\theta)-i\sin(\theta)))}=e^{-R\sin(\theta)}$$
 
  • #7
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Got it !
 

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