Undergrad Cauchy Integral Formula with a singularity

[cbarker1]

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

 

[Office_Shredder]

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.

 

[cbarker1]

Would the center of taylor series be at z=i?

 

[Office_Shredder]

You want to calculate it at the singularities you calculated. Just pick one to start.

 

[cbarker1]

ok. I would pick the z=ipi/2.

 

[cbarker1]

So far, the Taylor series is the following:
$f(z)=i\sum_{n=1}^\infty \frac{(z-\frac{i\pi}{2})^n}{n!}$

 

[Office_Shredder]

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.

 

[cbarker1]

when n=0, the $a_0=0$, because $f(\frac{i\pi}{2})=e^{\frac{i\pi}{2}}-i=0$.

 

[Office_Shredder]

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.

 

[cbarker1]

I am thinking that I am doing the long division wrong.

1|i(z-ipi/2)/1!+...

 

[cbarker1]

Here is the function's series so far: $$\frac{1}{f(z)}=\frac{i}{(z-\frac{i\pi}{2})}-\frac{i}{2!}-\frac{i(z-\frac{i\pi}{2})}{2!^2}+\dots$$

 

[Office_Shredder]

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?

$$\frac{z}{e^z-i} =\frac{1}{z-\frac{i\pi}{2}} \frac{z}{i\sum_{n=1}^\infty \frac{(z-\frac{i\pi}{2})^{n-1}}{n!}}$$

 

[cbarker1]

I would say the summation part of the function, right?

 

[Office_Shredder]

Can you say what the $f(z)$ in cauchy's integral formula is? Can you evaluate it at $\frac{i\pi}{2}$?

 

[cbarker1]

F(z) is the z/ the sum.

 

[Office_Shredder]

Ok. So can you evaluate the integral?

 

[cbarker1]

$f(z_0)=i\pi/4$, I think.

 

[Fred Wright]

$$f(z_0)=\frac{1}{2i\pi}\int_C\frac{f(z)}{z-z_0}dz$$,

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.

 

[cbarker1]

Your function doesn't resonate; it's just a short circuit.

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.

 

[Fred Wright]

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 $f(0)=0$.

 

[cbarker1]

So how do we get the value for the Laurent series as given in the wolfromalpha?

 

[jasonRF]

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