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Problem 2

# Expand $$f(z)=\frac{3 z-1}{z^{2}-2 z-3}$$ in Laurent series valid for (i) $$1<|z|<3$$ (ii) $$|z|>3$$ (iii) $$|z|<1$$.

Expert verified
The Laurent series of $$f(z)$$ are $$-\frac{3}{z} + \frac{7}{z^2}$$ for $$|z|>3$$, $$\frac{1}{z} + \frac{1}{z^2}$$ for $$1<|z|<3$$, and $$\sum_{n=0}^{\infty} (-z)^n + \sum_{n=0}^{\infty} (\frac{2}{3}z)^n$$ for $$|z|<1$$.
See the step by step solution

## Step 1: Factorize the Denominator

First, factorize the denominator of the fraction. The quadratic $$z^{2}-2z-3$$ can be factored into $$(z-3)(z+1)$$. So the function can be rewritten as $$f(z) = \frac{3z - 1}{(z-3)(z+1)}$$.

## Step 2: Partial Fraction Decomposition

Perform a partial fraction decomposition for the above expression, you get $$f(z) = \frac{1}{z+1} - \frac{2}{z-3}$$. This is the function we will need to expand.

## Step 3: Laurent Expansion for |z|>3

For $$|z|>3$$, expand the function as a power series in $$1/z$$. The first factor becomes $$\frac{1}{z(-1 + 1/z)}$$ and the second becomes $$-\frac{2}{z(1 - 3/z)}$$. Expanding both expressions you get $$\frac{-1}{z} + \frac{1}{z^2} - \frac{2}{z} + \frac{6}{z^2}$$, so the Laurent series for $$|z|>3$$ is $$-\frac{3}{z} + \frac{7}{z^2}$$.

## Step 4: Laurent Expansion for 1

For $$1<|z|<3$$, express the function as a power series in $$(z-3)$$ and $$z$$. The corresponding expression for the first factor would be $$-\frac1{z}(\frac{1}{1 + \frac{1}{z}})$$ and for the second factor $$2\frac{1}{1 - \frac{3}{z}}$$. Expanding both we get $$-\frac{1}{z} + \frac{1}{z^2} + \frac{2}{z}$$, so the Laurent series for $$1<|z|<3$$ is $$\frac{1}{z} + \frac{1}{z^2}$$.

## Step 5: Laurent Expansion for |z|

For $$|z|<1$$, express the function as a power series in $$z$$. Express the first factor as $$\frac{1}{1 + z}$$ and the second one as $$-\frac{2}{3-z}$$. The power series expansions will be $$\sum_{n=0}^{\infty} (-z)^n$$ and $$\sum_{n=0}^{\infty} (\frac{2}{3}z)^n$$. So the Laurent series for $$|z|<1$$ are these two sums.

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