Laurent Series of 1/[(z-i)(z-2)] at z0=i

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In summary, the conversation discusses finding the Laurent series at z0=i, which is convergent in the given annulus, for the function 1/[(z-i)(z-2)]. The method of using geometric series is suggested to find the Taylor's series for 1/(z-2) and then dividing each term by z-i.
  • #1
CTID17
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Homework Statement



Find the Laurent series at z0=i, which is convergent in the annulus A ={z:0<|z-i|<51/2 } of

1/[(z-i)(z-2)]


Homework Equations





The Attempt at a Solution


|z-i|/51/2 <1

i make
1/[(z-i)(z-2)] = 1/[51/2 (z-i)((i-2)/51/2 + (z-i)/51/2 )
now how do i make it so that i have 1-(z-i)/51/2 to use the binomial series?
Is this a right approach to this type of questions? I could use partial factions to get

1/(2-i)*[1/(z-2)-1/(z-i)] but again i don't know how i can take advantage of |z-i|/51/2 <1 to use in the binomial series.

Thanks in advance.
 
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  • #2
CTID17 said:

Homework Statement



Find the Laurent series at z0=i, which is convergent in the annulus A ={z:0<|z-i|<51/2 } of

1/[(z-i)(z-2)]


Homework Equations





The Attempt at a Solution


|z-i|/51/2 <1

i make
1/[(z-i)(z-2)] = 1/[51/2 (z-i)((i-2)/51/2 + (z-i)/51/2 )
now how do i make it so that i have 1-(z-i)/51/2 to use the binomial series?
Is this a right approach to this type of questions? I could use partial factions to get

1/(2-i)*[1/(z-2)-1/(z-i)] but again i don't know how i can take advantage of |z-i|/51/2 <1 to use in the binomial series.

Thanks in advance.
[tex]\frac{1}{(z-i)(z-2)}= \frac{1}{z-i}\left(\frac{1}{z-2}\right)[/tex]

[tex]\frac{1}{z-2}[/tex]
is analytic in the annulus given so it has a Taylor's series about z= i. Find that Taylor's series and divide each term by z-i.

(Hint: write 1/(z-2) as
[tex]\frac{1}{z-2}= \frac{1}{z- i+ i- 2}= -\frac{1}{(i-2)+ (z-i)}= -\frac{\frac{1}{i-2}}{1- \frac{z-i}{i-2}}[/tex]
and think of geometric series.)
 
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  • #3
Thanks a lot . I was doing the same thing, except I've forgotten about geometric series :) . Got it now :) .
 

What is a Laurent Series?

A Laurent Series is a mathematical representation of a complex function in the form of an infinite sum of terms. It is used to approximate functions that cannot be easily expressed as a polynomial.

What is the specific Laurent Series for 1/[(z-i)(z-2)] at z0=i?

The Laurent Series for 1/[(z-i)(z-2)] at z0=i is given by the formula:
1/[(z-i)(z-2)] = -1/(z-i) + 1/(z-2) + 1/[(z-i)^2] + 1/[(z-i)^3] + ...

How is the Laurent Series different from a Taylor Series?

The main difference between a Laurent Series and a Taylor Series is that a Laurent Series includes both positive and negative powers of the variable, while a Taylor Series only includes positive powers. This makes the Laurent Series useful for representing functions with poles or singularities, while the Taylor Series is better for representing smooth functions.

What is the significance of z0=i in the Laurent Series for 1/[(z-i)(z-2)] at z0=i?

The value of z0=i in the Laurent Series represents the center of the series. This means that the series is valid for all complex values of z as long as the distance between z and z0 is within the radius of convergence. In this case, the radius of convergence is the distance between z0=i and the nearest pole, which is z=2.

Can the Laurent Series for 1/[(z-i)(z-2)] at z0=i be used to calculate the value of the function at other points?

Yes, the Laurent Series can be used to approximate the value of the function at other points within the radius of convergence. However, outside of the radius of convergence, the series may not accurately represent the function. Additionally, the series may not converge at points on the boundary of the radius of convergence.

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