How to Use Cauchy's Integral Formula for Computing Circles Around the Origin

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Homework Help Overview

The discussion revolves around the application of Cauchy's integral formula to compute the integral of sinh(z)/z^3 over a circle around the origin. Participants are exploring the implications of the formula and the nature of the singularities involved.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to differentiate Cauchy's integral formula and how this relates to the integral of sinh(z)/z^3. There are questions about the concept of removable singularities and the nature of entire functions. Some suggest using the Laurent series and the general form of Cauchy's integral formula for higher-order poles.

Discussion Status

The discussion is active, with participants offering various insights and clarifications regarding the application of Cauchy's integral formula. There is an exchange of ideas about the nature of the function sinh(z) and its properties, as well as the relevance of derivatives in this context. No consensus has been reached yet.

Contextual Notes

Some participants express uncertainty about the concepts of singularities and Laurent series, indicating that these may not have been covered in their course yet. There is also mention of needing to evaluate derivatives at specific points, which adds to the complexity of the discussion.

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Homework Statement



Use the Cauchy integral formula to compute the formula:

[tex]\int_\gamma\frac{sinh(z)}{z^3}[/tex] for [tex]\gamma[/tex] any circle of radius >0 around the origin

Homework Equations



cauchy's integral formula:

[tex]f(z_0)=\frac{1}{2i\pi}\int\frac{f(z)}{z-z_0}dz[/tex]

The Attempt at a Solution



I don't know how to start
?
 
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What you need here is the derivative of the Cauchy's integral formula w.r.t. z_0.

If you differentiate the right hand side, you easily find that you need to differentiate the integrand w.r.t. z_0, i.e. you can bring the derivative inside the integral. (and if )

Now, what you then need to do is recognize that the integral of
sinh(z)/z^3 can be written in the form of that derivative of Cauchy's formula:

sinh(z)/z * 1/(z-z_0)^2

with z_0 = 0

sinh(z)/z only has a removable singularity at z = 0, you can define this function to be 1 at z = 0 to make it a complex differentiable function everywhere.
 
I didnt understand what you said about the derivative...why are we differentiating?

what do you mean 'removable singularity'?

thank you
 
Just use the Laurent series for [tex]\sinh(z)[/itex], and the Cauchy integral formula for simple poles of order>1...[/tex]
 
There should be the general form of Cauchy's integral formula located in your text. It is
[tex]f^{(n)}(w) = \frac{n!}{2\pi i} \int_\gamma \frac{f(z)}{(z-w)^{n+1}} \,dz[/tex]

In your case, f(z)=sinh(z), w=0, n=2. Note that f is then entire, i.e. analytic everywhere. Then
[tex]\int_\gamma \frac{\sinh z}{(z-0)^3} \,dz = \frac{2\pi i}{2!} f^{(2)}(0)[/tex]
 
f(0)=sihn(0)=0
?

why is it entire?
 
sara_87 said:
f(0)=sihn(0)=0
?

why is it entire?

Because
[tex]\sinh z = \frac{e^z - e^{-z}}{2},[/tex]
so it is the sum of two entire functions.

And technically you should be evaluating the second derivative at z=0, not just f, but in this case, the second derivative is just f(2)(z)=f(z)=sinh(z). If you don't know what singularities or Laurent series are, then don't worry about them. It seems that if you are just now doing Cauchy's integral formula, then your course hasn't arrived at those other topics yet. Does your text have the general formula I posted?
 
yes, i also should have put the formula you posted in my post.

thanks to all
:)
 
sara_87 said:
yes, i also should have put the formula you posted in my post.

thanks to all
:)

Okay, that's good and no problem for not posting it. What Count Iblis was talking about when they were referring to taking the derivative, they were basically deriving the formula I posted.
 

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