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

In summary, on page 158 in Fisher, it is stated that in applications of the Residue Theorem, there is a need to estimate the line integral of e^{iz} over a semicircle. The magnitude of e^{iz} is equal to e^{-R \sin(\theta)}, which can be proven using the modulus of a complex number.
  • #1
hotvette
Homework Helper
996
5
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 [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?
 
Last edited:
  • Like
Likes Delta2
Physics news on Phys.org
  • #2
Yes (IMHO)
 
  • #3
Thanks!
 
  • #4
How does one determine a modulus of a complex number ?
 
  • #5
Use that to prove ##|\bf ab | = |a|\,|b|##
 
  • #6
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
Got it !
 

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

1. What is a line integral?

A line integral is a mathematical concept that measures the integral of a function along a curve or path. It takes into account both the values of the function and the length of the curve.

2. What is the magnitude of a line integral?

The magnitude of a line integral is the absolute value of the line integral. It represents the overall size or magnitude of the integral, regardless of its direction.

3. What is a semicircle?

A semicircle is a half of a circle, formed by cutting a circle along its diameter. It has a curved edge and a straight edge, with a total of 180 degrees.

4. How do you estimate the magnitude of a line integral over a semicircle?

To estimate the magnitude of a line integral over a semicircle, you can break the curve into smaller segments and approximate the integral using numerical methods such as the trapezoidal rule or Simpson's rule. Alternatively, you can use a computer program to calculate the integral accurately.

5. What is the significance of exp(iz) in the line integral over a semicircle?

The function exp(iz) is a complex exponential function, where z is a complex number. It is often used in physics and engineering to represent oscillatory behavior. In the context of a line integral over a semicircle, it represents the function being integrated along the curve.

Similar threads

  • Calculus
Replies
8
Views
338
Replies
4
Views
515
Replies
4
Views
2K
Replies
16
Views
1K
Replies
1
Views
1K
Replies
2
Views
2K
  • Calculus
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
674
Replies
3
Views
1K
  • Advanced Physics Homework Help
Replies
1
Views
554
Back
Top