Lecture Notes

User Tools

Site Tools

You are here: Course Notes » Math for physics Sciences » Contour Integral
math121a-f23:september_13_wednesday

Differences

This shows you the differences between two versions of the page.

Link to this comparison view

Next revision 		Previous revision
math121a-f23:september_13_wednesday [2023/09/10 23:38]
pzhou created 		math121a-f23:september_13_wednesday [2026/02/21 14:41] (current)
Line 1: Line 1:
 ====== Contour Integral ====== ====== Contour Integral ======
 +Reading: Boas, Ch14, section 1-5
 +
  
 So, you have learned what holomorphic function looks like, and you know there are functions which are 'bad' only at a few points. What do you want to do with these functions?  So, you have learned what holomorphic function looks like, and you know there are functions which are 'bad' only at a few points. What do you want to do with these functions? 
Line 12: Line 14:
 where we set the initial condition that $F(x_0) = C$, and $F'(x)= f(x)$.  where we set the initial condition that $F(x_0) = C$, and $F'(x)= f(x)$. 
  
-Can we do the same here? Say $f(z)$ is any complex valued function, $f: \C \to \C$, can we try to get a function $F(z)$, such that $F(z_0) = C$, and $F'(z) = f(z)$? We can try  +Can we do the same here? Say $f(z)$ is a holomorphic function, we can define 
-$$ F(z) = \int_{z_0}^z f(u) du $$ +$$ F(z) = C + \int_{z_0}^z f(u) du $$ 
-Now, we immediately run into trouble: how do we go from $z_0$ to $z$? Does the integration depends on how we choose the path from $z_0$ to $z$?  +Now, we immediately run into trouble: how do we go from $z_0$ to $z$? Does the integration depends on how we choose the path from $z_0$ to $z$? Thanks to the fact that $f$ is holomorphic, the integration is independent of the choice of path. 
- +
-You immediately would get, if $F$ is a holomorphic function (since we say its complex derivative $F'(z)$ exists), then $f(z)$ is also holomorphic. That is a necessary condition. So you cannot just use any $f(z)$ to integrate.  +
- +
-OK, if $f: \C \to \C$ is holomorphic, then you can find its primitive $F(z)$ up to a shift of constant. +
  
 ===== primitive of $1/z$ ===== ===== primitive of $1/z$ =====
Line 34: Line 32:
  
 Example: $f(z) = 1/ [(z-1)(z-2)(z-3)]$, $\gamma$ is a contour  around the two poles $1$ and $2$. Example: $f(z) = 1/ [(z-1)(z-2)(z-3)]$, $\gamma$ is a contour  around the two poles $1$ and $2$.
 +
 +===== Exercise =====
 +1. For $t \in [0, 2\pi]$, let $z(t) = e^{it}$. Compute $\int_{0}^{2\pi} 1 / (z(t)) d z(t). $
 +
 +2. For $t \in [0, 2\pi]$, let $z(t) = e^{i2t}$. Compute $\int_{0}^{2\pi} 1 / (z(t)) d z(t). $
 +
 +3. For $t \in [0, 2\pi]$, let $z(t) = e^{-it}$. Compute $\int_{0}^{2\pi} 1 / (z(t)) d z(t). $
  
      
math121a-f23/september_13_wednesday.1694389085.txt.gz · Last modified: 2026/02/21 14:44 (external edit)

Page Tools

Except where otherwise noted, content on this wiki is licensed under the following license: CC Attribution-Share Alike 4.0 International
CC Attribution-Share Alike 4.0 International Donate Powered by PHP Valid HTML5 Valid CSS Driven by DokuWiki