x̄ -> Cauchy equation for Laurent series is a fundamental concept in complex analysis example questions along with their solutions

The Cauchy equation for Laurent series is a fundamental concept in complex analysis. 

Here are four example questions along with their solutions: 

Question 1: Find the Laurent series expansion for the function f(z)=1/z^2(z−1) in the annulus 1<|z|<2. 

Solution 1:To find the Laurent series expansion, we decompose f(z) into partial fractions: 

f(z)=A/z+B/z^2+C/z−1 where A, B, and C are constants to be determined.

By finding common denominators and equating coefficients, we can solve for A, B, and C. 

After finding the values of A, B, and C, we get: f(z)=1/z−1/z^2−1/z−1 Now, we can express each term in 

the Laurent series expansion: f(z)=1/z−1/z^2−1/1−z=1/z−1/z^2  - ∑n=0,∞ zn 

This series converges for |z|>1 and |z−1|>1, which is the annulus 1<|z|<2. 

Question 2: Find the Laurent series expansion for the function f(z)=1/z^2−z in the annulus 1<|z−1|<2. 

Solution 2: We can rewrite the function as: f(z)=1/z(z−1) This is already in a form where we can apply 

the geometric series expansion: 1/1−z=∑n=0,∞zn Now, we have: f(z)=1/z+1/1−z So, the Laurent series 

expansion for f(z) is: f(z)=1/z+∑n=0,∞zn This series converges for |z|>1 and |z−1|<2, which is the 

annulus 1<|z−1|<2. 

Question 3: Find the Laurent series expansion for the function f(z)=1/z(z−1)(z−2) in the annulus 

1<|z−1|<2. 

Solution 3: Similarly, we can decompose f(z) into partial fractions and find the Laurent series 

expansion. After decomposing and finding the values of the constants, we obtain: f(z)=1/z−2/z−1+1/z−2 

Thus, the Laurent series expansion for f(z) is: f(z)=1z−∑n=0,∞2(z−1)n+∑n=0,∞(z−2)n This series 

converges for |z|>2 and |z−1|<1, which is the annulus 1<|z−1|<2. 

Question 4: Find the Laurent series expansion for the function f(z)=z^2/z^3−1 in the annulus 0<|z|<1. 

Solution 4: We can rewrite f(z) as: f(z)=z^2/(z−1)(z−ω)(z−ω^2) where ω=e2πi3 is a cube root of unity. 

Now, we express f(z) in partial fractions: f(z)=A/z−1+B/z−ω+C/z−ω2 After finding the values of A, B, 

 and C, we obtain: f(z)=1/z−1−1/z−ω+1/z−ω2 Hence, the Laurent series expansion for f(z) is: 

f(z)=1/z+∑n=1∞,ω2nz−n−∑n=1∞ωnz−n This series converges for 0<|z|<1.

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