Cauchy Integral Formula with a singularity

In summary, the integral formula for a holomorphic function may be used to determine the residues at the poles of the function, but it may not be used to determine the value of the function at a singularity.
  • #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 think the standard thing to do is calculate the Laurent series of the function. The standard way to do this if I remember correctly is to write out the Taylor series of ##e^z-i## in a small neighborhood around the singularity, factor out all the multiples of ##z##, and then use the Taylor series expansion of ##\frac{1}{1+z}## where z is actually going to be your remaining infinite series. Have you seen this kind of thing done before?

I encourage you to try it, get as far as you can, and then post what you have here.
 
  • #3
Would the center of taylor series be at z=i?
 
  • #4
You want to calculate it at the singularities you calculated. Just pick one to start.
 
  • #5
ok. I would pick the z=ipi/2.
 
  • #6
So far, the Taylor series is the following:
##f(z)=i\sum_{n=1}^\infty \frac{(z-\frac{i\pi}{2})^n}{n!}##
 
Last edited:
  • #7
I think the sum starts at n=0.

Now try putting that in the denominator. You should be able to pull out a factor of ##(z-\frac{i\pi}{2}##. Then you'll be left with a holomorphic function that you need for the theorem.
 
  • #8
when n=0, the ##a_0=0##, because ##f(\frac{i\pi}{2})=e^{\frac{i\pi}{2}}-i=0##.
 
  • #9
Oh, sure. Sorry for some reason I was thinking you were writing the Taylor series down just for ##e^z##.

Anyway, just write out the function with that in the denominator, and factor out ##z-i\pi/2## from the denominator. Then you should have your integrand in the right form.
 
  • #10
I am thinking that I am doing the long division wrong.

1|i(z-ipi/2)/1!+...
 
  • #11
Here is the function's series so far: $$\frac{1}{f(z)}=\frac{i}{(z-\frac{i\pi}{2})}-\frac{i}{2!}-\frac{i(z-\frac{i\pi}{2})}{2!^2}+\dots$$
 
  • #12
See if you can understand why this is true given what you've written. Do you see how to evaluate a line integral around the singularity from here?

$$\frac{z}{e^z-i} =\frac{1}{z-\frac{i\pi}{2}} \frac{z}{i\sum_{n=1}^\infty \frac{(z-\frac{i\pi}{2})^{n-1}}{n!}}$$
 
  • #13
I would say the summation part of the function, right?
 
  • #14
Can you say what the ##f(z)## in cauchy's integral formula is? Can you evaluate it at ##\frac{i\pi}{2}##?
 
  • #15
F(z) is the z/ the sum.
 
  • #16
Ok. So can you evaluate the integral?
 
  • #17
##f(z_0)=i\pi/4##, I think.
 
  • #18
cbarker1 said:
$$f(z_0)=\frac{1}{2i\pi}\int_C\frac{f(z)}{z-z_0}dz$$,
This integral only applies when ##z_0## is a pole of ##f(z)## and not an essential singularity. The residue theorem doesn't apply for your function. The coefficients of the negative power of z terms in the Laurent expansion give the residues for the polar order. Your function gives ##f(0)=0##, therefore the Laurent expansion can have no negative powers of z terms and thus no poles with residues of any order. Your function doesn't resonate; it's just a short circuit.
 
  • #19
Fred Wright said:
Your function doesn't resonate; it's just a short circuit.
π/(2 (z - (i π)/2)) + (-π/4 + -i) + 1/24 (π + 12 i) (z - (i π)/2) - 1/12 i (z - (i π)/2)^2 - (π (z - (i π)/2)^3)/1440 + 1/720 i (z - (i π)/2)^4 + O((z - (i π)/2)^5)
I know that the function is not define at that point. I want to sure that I know that how the pi/2 in the first negative power came to be.
 
  • #20
cbarker1 said:
π/(2 (z - (i π)/2)) + (-π/4 + -i) + 1/24 (π + 12 i) (z - (i π)/2) - 1/12 i (z - (i π)/2)^2 - (π (z - (i π)/2)^3)/1440 + 1/720 i (z - (i π)/2)^4 + O((z - (i π)/2)^5)
I know that the function is not define at that point. I want to sure that I know that how the pi/2 in the first negative power came to be.
You made a mistake in calculating the Laurent series. Let's calculate it by long division:
$$f(z)=\frac{z}{(e^z-i)}=\frac{iz}{(1-e^{(z-\frac{i\pi}{2}-2\pi i n)})}$$We make the substitution ##z'=z-\frac{i\pi}{2}-2\pi i n## and revert back after the calculation of the Laurent series.$$f(z')=\frac{-i(z'+\frac{i\pi}{2}+2\pi i n)}{z' + \frac{z'^2}{2!} + \frac{z'^3}{3!} + \frac{z'^4}{4!} +...}$$We calculate by long division,$$\frac{1}{z' + \frac{z'^2}{2!} + \frac{z'^3}{3!} + \frac{z'^4}{4!} +...}=\frac{1}{z'}-\frac{1}{2}+\frac{z'}{12}-...$$Reverting back to ##z'## we find$$f(z)= \frac{iz}{(z-\frac{i\pi}{2}-2\pi i n )} + iz(\frac{1}{2}-\frac{1}{12}(\frac{i\pi}{2}+2\pi i n)) + \frac{iz^2}{12} + ...$$ Thus the Laurent series has no ##\frac{1}{z}## terms and ##f(0)=0##.
 
  • #21
So how do we get the value for the Laurent series as given in the wolfromalpha?
 
  • #22
cbarker1 said:
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
Note that Cauchy's integral formula holds when ##f(z)## is holomorphic everywhere inside and on the contour ##C##. So in your case, it will hold for ##z_0=i## only if ##C## encircles ##z=i## but does not encircle any of the singularities of ##f##. Besides selecting an appropriate contour ##C##, it isn't clear to me what you mean when you ask how to "use" the integral formula. What are you trying to do?

jason
 

Related to Cauchy Integral Formula with a singularity

1. What is the Cauchy Integral Formula with a singularity?

The Cauchy Integral Formula with a singularity is a mathematical theorem that relates the values of a complex function inside a closed contour to the values of the function on the contour itself. It is used to evaluate complex integrals and is an important tool in complex analysis.

2. What is a singularity in the context of the Cauchy Integral Formula?

In the context of the Cauchy Integral Formula, a singularity refers to a point where the function being integrated is not defined or is undefined. This can occur at points where the function has a pole, branch point, or essential singularity.

3. How is the Cauchy Integral Formula with a singularity derived?

The Cauchy Integral Formula with a singularity is derived from the Cauchy Integral Theorem, which states that the integral of a complex function around a closed contour is equal to the sum of the function's values inside the contour. The formula is then extended to include singularities by using the Cauchy Residue Theorem.

4. What are some applications of the Cauchy Integral Formula with a singularity?

The Cauchy Integral Formula with a singularity has many applications in mathematics and physics. It is used to evaluate complex integrals, calculate residues, and solve problems in potential theory, fluid dynamics, and electromagnetism. It is also used in the study of conformal mappings and Riemann surfaces.

5. Are there any limitations to the Cauchy Integral Formula with a singularity?

Yes, there are some limitations to the Cauchy Integral Formula with a singularity. It can only be applied to functions that are analytic (meaning they have a well-defined derivative) in the region of integration. It also cannot be used to evaluate integrals with infinitely many singularities or integrals that do not converge. Additionally, the formula may not work for certain types of singularities, such as branch points or essential singularities.

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