Forming a Laurent Series for 4cos(z*pi) / (z-2i)

In summary, to form a Laurent series about the point z0= 2i for the function 4cos(z*pi)/(z-z0), one can first expand the cosine function using the power series formula. Then, by using the properties of power series, the function can be simplified to have only non-negative powers of (z-2i) and a single term of (z-2i)^{-1}. This approach allows for easier integration of the original function.
  • #1
Dissonance in E
71
0
How does one form a laurent series about the point z0 = 2i
for the function:

4cos(z*pi) / (z-z0).

Could one take advantage of the power series
1 / (z - z0)
1 / (z - 2i)

SUMMATION q^n
= SUMMATION (2i)^n
= 1 + 2i -4 -8i . . . . .

and somehow integrate the rest of the original function into the above results?
 
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  • #2
Well, first you have to handle that "cos(z)"!
I suspect it will be better to use cos(z)= cos(z- 2i+ 2i)= cos(z- 2i)cos(2i)- sin(z- 2i)sin(2i). Expand those as power series, then divide by z- 2i. That will give just a single [itex](z- 2i)^{-1}[/itex] and all the rest will be non-negative powers.
 

1. How do I determine the center and radius of convergence for a Laurent Series?

The center of convergence for a Laurent Series can be found by setting the denominator of the given function equal to zero and solving for the variable. In this case, the center is z=2i. The radius of convergence can be calculated by taking the absolute value of the difference between the center and the nearest singularity. In this case, the nearest singularity is at z=2i, so the radius of convergence is 2.

2. What is the general formula for a Laurent Series?

The general formula for a Laurent Series is:
f(z) = Σn=-∞∞ an (z-z0)n
where z0 is the center of convergence and the coefficients an can be calculated using the Cauchy Integral Formula.

3. How do I find the coefficients for a Laurent Series?

The coefficients for a Laurent Series can be found by using the Cauchy Integral Formula:
an = (1/2πi) ∫γ f(z) (z-z0)-n-1 dz
where γ is any closed contour that encloses the center of convergence z0. This integral can be evaluated using the Residue Theorem.

4. Can a Laurent Series have both positive and negative powers?

Yes, a Laurent Series can have both positive and negative powers. This is what differentiates it from a Taylor Series, which only has positive powers.

5. How do I determine the convergence or divergence of a Laurent Series?

The convergence or divergence of a Laurent Series can be determined by calculating the radius of convergence and examining the behavior of the function at the boundary of this radius. If the function is analytic at the boundary, it will converge. If the function has a singularity at the boundary, it will diverge.

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