Find the Laurent series expansion of f(z) = 1/(z² - 3z + 2) valid in the annulus 1 < |z| < 2.
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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, z−1−1, we need an expansion valid for ∣z∣>1. We factor out z from the denominator and use the geometric series formula 1−x1=∑n=0∞xn for ∣x∣<1. Here, x=z1, and since ∣z∣>1, we have ∣z1∣<1.
Step 5: Expand the Second Term for ∣z∣<2
For the second term, z−21, we need an expansion valid for ∣z∣<2. We factor out −2 from the denominator and again use the geometric series formula. Here, x=2z, and since ∣z∣<2, we have ∣2z∣<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.