# Laurent series expansion

In attachment

## The Attempt at a Solution

f(z) = -1/3[3(z+1)] + 4/3[z+4]

Now im stuck.

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Hi!

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

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

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

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

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

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

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

vela
Staff Emeritus
Homework Helper
If a Laurent series contains only non-negative powers, it will be identical to the Taylor series for the function.

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

vela
Staff Emeritus
Homework Helper
What do you mean by "largest pole"?

Note that

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

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

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

has an infinite number of negative-power terms.

What you wrote clarified my problem. tks