Cauchy's integral formula

In summary: So, you don't really need to worry about that unless you are interested. So, in summary, use Cauchy's integral formula with n=2 for f(z)=sinh(z) and w=0 to find that the integral is 0.In summary, the conversation discusses how to use Cauchy's integral formula to compute the formula for a circle of radius greater than zero around the origin. The formula is derived using the derivative of Cauchy's integral formula with respect to z_0. It is then discussed how the integral of sinh(z)/z^3 can be written in the form of the derivative of Cauchy's formula. The conversation also clarifies the meaning of a removable singularity
  • #1
sara_87
763
0

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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  • #2
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.
 
  • #3
I didnt understand what you said about the derivative...why are we differentiating?

what do you mean 'removable singularity'?

thank you
 
  • #4
Just use the Laurent series for [tex]\sinh(z)[/itex], and the Cauchy integral formula for simple poles of order>1...
 
  • #5
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]
 
  • #6
f(0)=sihn(0)=0
?

why is it entire?
 
  • #7
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?
 
  • #8
yes, i also should have put the formula you posted in my post.

thanks to all
:)
 
  • #9
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.
 

What is Cauchy's integral formula?

Cauchy's integral formula is a fundamental theorem in complex analysis that allows for the computation of integrals of complex functions over a closed contour.

What is the significance of Cauchy's integral formula?

Cauchy's integral formula is significant because it provides a powerful tool for solving complex integrals, which are important in many areas of mathematics, physics, and engineering.

What is the formula for Cauchy's integral formula?

The formula is given by: ∫𝛾 f(z) dz = (1/2𝜋i) ∫𝛾 f(z) / (z-z0) dzwhere f(z) is a complex-valued function, 𝛾 is a closed contour, and z0 is a point inside the contour.

How is Cauchy's integral formula related to Cauchy's theorem?

Cauchy's integral formula is a special case of Cauchy's theorem, which states that for a simply connected region in the complex plane, the integral of a holomorphic function along a closed contour is equal to zero.

What is the use of Cauchy's integral formula in solving complex integrals?

Cauchy's integral formula allows for the computation of complex integrals through the use of residues, which are the values of a function at its singularities. This makes it a powerful tool in solving integrals that would otherwise be difficult or impossible to solve using traditional methods.

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