Find the Laurent series expansion of f(z) = 1/(z² - $3z + 2$ ) valid in the annulus 1 < |z| < 2.

Answer: $f(z) = -\sum_{n=0}^{\infty} \frac{1}{z^{n+1}} - \sum_{n=0}^{\infty} \frac{z^n}{2^{n+1}}$

Step-by-step solution

Step 1: Factor the Denominator and Perform Partial Fraction Decomposition

First, we factor the denominator of the given function $f(z)$ . Then, we use partial fraction decomposition to express $f(z)$ as a sum of simpler fractions. This makes it easier to find the Laurent series.

Step 2: Solve for A and B

To find the constants $A$ and $B$ , we set the numerators equal. By substituting the roots of the denominator, $z=1$ and $z=2$ , we can easily solve for $A$ and $B$ .

Step 3: Rewrite f(z) using Partial Fractions

Now that we have found the values of $A$ and $B$ , we can substitute them back into the partial fraction decomposition to get the simplified form of $f(z)$ .

Step 4: Expand the First Term for $|z| > 1$

For the first term, $\frac{-1}{z-1}$ , we need an expansion valid for $|z| > 1$ . We factor out $z$ from the denominator and use the geometric series formula $\frac{1}{1-x} = \sum_{n=0}^{\infty} x^n$ for $|x| < 1$ . Here, $x = \frac{1}{z}$ , and since $|z| > 1$ , we have $|\frac{1}{z}| < 1$ .

Step 5: Expand the Second Term for $|z| < 2$

For the second term, $\frac{1}{z-2}$ , we need an expansion valid for $|z| < 2$ . We factor out $-2$ from the denominator and again use the geometric series formula. Here, $x = \frac{z}{2}$ , and since $|z| < 2$ , we have $|\frac{z}{2}| < 1$ .

Step 6: Combine the Series Expansions

Finally, we combine the two series expansions obtained for each term. This combined series is the Laurent series expansion of $f(z)$ valid in the annulus $1 < |z| < 2$ .

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Find the Laurent series expansion of f(z) = 1/(z² - 3z+23z + 23z+2) valid in the annulus 1 < |z| < 2.