Math Problem Statement

evaluate 1/(z-1)(z-2) is valid for the region mod(z-1)<1 using laurents series

Solution

To evaluate the function 1(z1)(z2)\frac{1}{(z-1)(z-2)} using a Laurent series that is valid in the region z1<1|z-1| < 1, follow these steps:

Step 1: Express the Function as a Sum of Partial Fractions

The given function can be decomposed into partial fractions: 1(z1)(z2)=Az1+Bz2\frac{1}{(z-1)(z-2)} = \frac{A}{z-1} + \frac{B}{z-2} To find the constants AA and BB, multiply both sides by (z1)(z2)(z-1)(z-2) and set the equation equal to the numerator: 1=A(z2)+B(z1)1 = A(z-2) + B(z-1) Expanding and equating coefficients: 1=A(z)2A+B(z)B1 = A(z) - 2A + B(z) - B 1=(A+B)z(2A+B)1 = (A + B)z - (2A + B) From this, we equate the coefficients of zz and the constant term: A+B=0A + B = 0 2AB=1-2A - B = 1

Step 2: Solve for AA and BB

From A+B=0A + B = 0, we have B=AB = -A.

Substitute B=AB = -A into the second equation: 2A(A)=12A+A=1A=1A=1-2A - (-A) = 1 \quad \Rightarrow \quad -2A + A = 1 \quad \Rightarrow \quad -A = 1 \quad \Rightarrow \quad A = -1 Then, B=1B = 1.

So, the partial fraction decomposition is: 1(z1)(z2)=1z1+1z2\frac{1}{(z-1)(z-2)} = \frac{-1}{z-1} + \frac{1}{z-2}

Step 3: Expand Each Term in the Region z1<1|z-1| < 1

We can now expand each term in a Laurent series. Since z1<1|z-1| < 1, we consider the series expansion around z=1z = 1:

For the first term: 1z1\frac{-1}{z-1} This is already in a simple form and is the first term of the Laurent series.

For the second term 1z2\frac{1}{z-2}, rewrite it by factoring out (z1)(z-1): 1z2=1(z1)1\frac{1}{z-2} = \frac{1}{(z-1) - 1} =1(1(z1))=11(z1)= \frac{1}{-(1 - (z-1))} = -\frac{1}{1-(z-1)} Expand this using the geometric series for z1<1|z-1| < 1: 11(z1)=n=0(z1)n-\frac{1}{1-(z-1)} = -\sum_{n=0}^{\infty} (z-1)^n This is valid for z1<1|z-1| < 1.

Step 4: Combine the Series

Now combine the two series: 1(z1)(z2)=1z1n=0(z1)n\frac{1}{(z-1)(z-2)} = -\frac{1}{z-1} - \sum_{n=0}^{\infty} (z-1)^n This is the Laurent series expansion of the function 1(z1)(z2)\frac{1}{(z-1)(z-2)} valid in the region z1<1|z-1| < 1.

Final Answer

1(z1)(z2)=1z11(z1)(z1)2(z1)3\frac{1}{(z-1)(z-2)} = -\frac{1}{z-1} - 1 - (z-1) - (z-1)^2 - (z-1)^3 - \cdots

Would you like more details or have any questions?


  1. How would the Laurent series change for a different region, such as z2<1|z-2| < 1?
  2. What are the general steps to perform partial fraction decomposition?
  3. How can the geometric series expansion be generalized for other functions?
  4. What are the conditions for the convergence of the Laurent series in this region?
  5. How would the series expansion look if the region were 1<z1<z21 < |z-1| < |z-2|?

Tip: When working with Laurent series, always carefully consider the region of convergence to determine the correct expansion.

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Math Problem Analysis

Mathematical Concepts

Complex Analysis
Laurent Series
Partial Fraction Decomposition

Formulas

Partial fraction decomposition formula
Geometric series expansion formula

Theorems

Laurent series theorem

Suitable Grade Level

Advanced Undergraduate