Cauchy Integral Formula with a singularity

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  • Thread starter cbarker1
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  • #1
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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
 

Answers and Replies

  • #2
Office_Shredder
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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
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Would the center of taylor series be at z=i?
 
  • #4
Office_Shredder
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You want to calculate it at the singularities you calculated. Just pick one to start.
 
  • #5
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ok. I would pick the z=ipi/2.
 
  • #6
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So far, the Taylor series is the following:
##f(z)=i\sum_{n=1}^\infty \frac{(z-\frac{i\pi}{2})^n}{n!}##
 
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  • #7
Office_Shredder
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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
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when n=0, the ##a_0=0##, because ##f(\frac{i\pi}{2})=e^{\frac{i\pi}{2}}-i=0##.
 
  • #9
Office_Shredder
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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
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I am thinking that I am doing the long division wrong.

1|i(z-ipi/2)/1!+....
 
  • #11
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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
Office_Shredder
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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
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I would say the summation part of the function, right?
 
  • #14
Office_Shredder
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Can you say what the ##f(z)## in cauchy's integral formula is? Can you evaluate it at ##\frac{i\pi}{2}##?
 
  • #15
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F(z) is the z/ the sum.
 
  • #16
Office_Shredder
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Ok. So can you evaluate the integral?
 
  • #17
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##f(z_0)=i\pi/4##, I think.
 
  • #18
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$$f(z_0)=\frac{1}{2i\pi}\int_C\frac{f(z)}{z-z_0}dz$$,
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
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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
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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
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So how do we get the value for the Laurent series as given in the wolfromalpha?
 
  • #22
jasonRF
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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
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
 
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