- #1

- Thread starter j-lee00
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Hi!

I will help you with a full list of steps for one of the parts:

[tex]

\frac{4}{3(z + 4)} = \frac{4}{3(4(\frac{z}{4} + 1))} = \frac{1}{3(\frac{z}{4} + 1)} = \frac{1}{3(1 + \frac{z}{4})} = \frac{1}{3}\cdot \frac{1}{(1 + \frac{z}{4})} = \frac{1}{3}\cdot\frac{1}{(1-(-\frac{z}{4}))} = \frac{1}{3}\left ( 1 + (-\frac{z}{4}) + (-\frac{z}{4})^{2} + (-\frac{z}{4})^{3} +\cdots \right ) =

[/tex]

[tex]

= \frac{1}{3}\left ( 1 - \frac{z}{4} + \frac{z^{2}}{4^{2}} - \frac{z^{3}}{4^{3}} +\cdots \right ) = \frac{1}{3} - \frac{z}{3\cdot 4} + \frac{z^{2}}{3\cdot 4^{2}} - \frac{z^{3}}{3\cdot 4^{3}} +\cdots

[/tex]

Remember that for a given geometric series to converge |x| has to be less than 1. which in your case means:

[tex]

\left |\frac{z}{4} \right |<1

[/tex]

I hope you will understand all the steps otherwise do ask.

P.S. I used the sum of the geometric series (which is also the Taylor and Laurent series): 1/(1 - x) = 1 + x + x^2 + x^3 + ...

|x|<1

I will help you with a full list of steps for one of the parts:

[tex]

\frac{4}{3(z + 4)} = \frac{4}{3(4(\frac{z}{4} + 1))} = \frac{1}{3(\frac{z}{4} + 1)} = \frac{1}{3(1 + \frac{z}{4})} = \frac{1}{3}\cdot \frac{1}{(1 + \frac{z}{4})} = \frac{1}{3}\cdot\frac{1}{(1-(-\frac{z}{4}))} = \frac{1}{3}\left ( 1 + (-\frac{z}{4}) + (-\frac{z}{4})^{2} + (-\frac{z}{4})^{3} +\cdots \right ) =

[/tex]

[tex]

= \frac{1}{3}\left ( 1 - \frac{z}{4} + \frac{z^{2}}{4^{2}} - \frac{z^{3}}{4^{3}} +\cdots \right ) = \frac{1}{3} - \frac{z}{3\cdot 4} + \frac{z^{2}}{3\cdot 4^{2}} - \frac{z^{3}}{3\cdot 4^{3}} +\cdots

[/tex]

Remember that for a given geometric series to converge |x| has to be less than 1. which in your case means:

[tex]

\left |\frac{z}{4} \right |<1

[/tex]

I hope you will understand all the steps otherwise do ask.

P.S. I used the sum of the geometric series (which is also the Taylor and Laurent series): 1/(1 - x) = 1 + x + x^2 + x^3 + ...

|x|<1

Last edited:

- #3

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With the expansion that you made, how does this differ to the taylor expansion? or it the same?

- #4

vela

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Are the negative powers of the Laurent expansion only as big as the largest pole?

- #6

vela

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Note that

[tex]\frac{4}{3(z + 4)} = \frac{1}{3} \left( 1 - \frac{z}{4} + \frac{z^2}{4^2} - \frac{z^3}{4^3} + \cdots \right)[/tex]

has a pole at z=-4 yet no negative-power terms. The Laurent series about z=0 of the function for |z|>4 is

[tex]\frac{4}{3(z + 4)} = \frac{4}{3z}\left( 1 - \frac{4}{z} + \frac{4^2}{z^2} - \frac{4^3}{z^3} + \cdots \right)[/tex]

has an infinite number of negative-power terms.

- #7

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What you wrote clarified my problem. tks

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