Using Cauchy Integral Formula for Laurent Series Coefficients

In summary, the Laurent series for $f(z)$ at the point $z=i$ is $\frac{1}{2i\pi}\left[\frac{1}{e^z-i}\right]$, where $e^z$ is the complex exponential of $z$.
  • #1
cbarker1
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Dear Everyone,

I am wondering how to use the integral formula for a holomorphic function at all points except a point that does not exist in function's analyticity. For instance, Let $f$ be defined as $$f(z)=\frac{z}{e^z-i}$$. $f$ is holomorphic everywhere except for $z_n=i\pi/2+2ni\pi$ for all $n$ in the integers. Let curve $C$ be closed positively oriented simple curve. \[ f(z_0)=\frac{1}{2i\pi}\int_C\frac{f(z)}{z-z_0}dz \], I want to find $z_0=i$, if it is possible.Thanks,

Cbarker1
 
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  • #2
I'm not sure what you mean by "find $z_0= i$". Do you mean find the value of the integral $\frac{1}{2\pi i}\int_C \frac{dz}{z- i}$?
Since that is equal to $f(z_0)= f(i)$, and $f(z)= \frac{z}{e^z- i}$ that is $\frac{i}{e^i- i}$. In general, $e^{ix}= cos(x)+ i sin(x)$ so $e^i= cos(1)+ i sin(1)$. $f(i)= \frac{i}{cos(1)+ i(sin(1)- 1}$. If you like you can "rationalize" the denominator by multiplying both numerator and denominator by $cos(1)- i(sin(1)- 1)$. That gives $\frac{1- sin(1)- i}{cos^2(1)+ sin^2(1)- 2sin(1)+ 1}= \frac{1- sin(1)- i}{2- 2sin(1)}$.
 
  • #3
Country Boy said:
I'm not sure what you mean by "find $z_0= i$". Do you mean find the value of the integral $\frac{1}{2\pi i}\int_C \frac{dz}{z- i}$?
Since that is equal to $f(z_0)= f(i)$, and $f(z)= \frac{z}{e^z- i}$ that is $\frac{i}{e^i- i}$. In general, $e^{ix}= cos(x)+ i sin(x)$ so $e^i= cos(1)+ i sin(1)$. $f(i)= \frac{i}{cos(1)+ i(sin(1)- 1}$. If you like you can "rationalize" the denominator by multiplying both numerator and denominator by $cos(1)- i(sin(1)- 1)$. That gives $\frac{1- sin(1)- i}{cos^2(1)+ sin^2(1)- 2sin(1)+ 1}= \frac{1- sin(1)- i}{2- 2sin(1)}$.
I want to find the coefficient of Laurent series at the singularity point. Sorry for the confusion...
 

FAQ: Using Cauchy Integral Formula for Laurent Series Coefficients

1. What is the Cauchy Integral Formula?

The Cauchy Integral Formula is a theorem in complex analysis that relates the values of a holomorphic function inside a closed contour to the values of the function on the contour itself. It is given by the formula: f(z) = \frac{1}{2 \pi i} \int_C \frac{f(\zeta)}{\zeta - z} d\zeta where f(z) is the holomorphic function, C is the closed contour, and z is a point inside the contour.

2. What is a Laurent series?

A Laurent series is a representation of a complex function as an infinite sum of powers of z, with both positive and negative powers. It is given by the formula:f(z) = \sum_{n=-\infty}^{\infty} a_n(z-z_0)^n where z_0 is the center of the series and the coefficients a_n can be calculated using the Cauchy Integral Formula.

3. How do you use the Cauchy Integral Formula to find Laurent series coefficients?

To find the Laurent series coefficients using the Cauchy Integral Formula, we first need to identify the center of the series, z_0. Then, we can use the formula:a_n = \frac{1}{2 \pi i} \int_C \frac{f(\zeta)}{(\zeta - z_0)^{n+1}} d\zeta to calculate the coefficients. This involves evaluating the integral along the contour C, which can be done using techniques from complex analysis.

4. What are the applications of using Cauchy Integral Formula for Laurent series coefficients?

The Cauchy Integral Formula for Laurent series coefficients has many applications in complex analysis, including calculating residues, finding poles and zeros of a function, and solving differential equations. It is also useful in solving problems in physics and engineering, such as calculating electric fields and fluid flow.

5. Are there any limitations to using the Cauchy Integral Formula for Laurent series coefficients?

One limitation of using the Cauchy Integral Formula for Laurent series coefficients is that it only works for functions that are holomorphic inside the contour C. This means that the function must be differentiable at every point inside the contour. Additionally, the contour C must be simple and closed, and the function must not have any singularities inside the contour.

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