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Math 3.3

The document outlines the process of finding the Laurent series expansion of the function f(z) = (z^2 - 1) / ((z + 2)(1 + z^2)) for two annular regions: 2 < |z| < 3 and 1 < |z| < 2. It details the steps of partial fraction decomposition and the subsequent series expansion for each term in the specified regions. The final Laurent series for each region is presented, showing the coefficients and structure of the expansions.

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17 views3 pages

Math 3.3

The document outlines the process of finding the Laurent series expansion of the function f(z) = (z^2 - 1) / ((z + 2)(1 + z^2)) for two annular regions: 2 < |z| < 3 and 1 < |z| < 2. It details the steps of partial fraction decomposition and the subsequent series expansion for each term in the specified regions. The final Laurent series for each region is presented, showing the coefficients and structure of the expansions.

Uploaded by

daviddikejesus
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Solution

We are tasked with finding the Laurent series expansion of the function:

z2 − 1
f (z) =
(z + 2)(1 + z 2 )

for two different annular regions:


1. 2 < |z| < 3
2. 1 < |z| < 2

Step 1: Factorize the Denominator and Perform Partial


Fraction Decomposition
First, factorize the denominator:

(z + 2)(1 + z 2 ) = (z + 2)(z + i)(z − i)

Perform partial fraction decomposition on f (z):

z2 − 1 A B C
= + +
(z + 2)(z + i)(z − i) z+2 z+i z−i

Multiply through by the denominator to solve for A, B, and C:

z 2 − 1 = A(z + i)(z − i) + B(z + 2)(z − i) + C(z + 2)(z + i)

Substitute z = −2:

(−2)2 − 1 = A(−2 + i)(−2 − i)

3 = A(4 − i2 ) = A(4 + 1) = 5A
3
A=
5
Substitute z = −i:

(−i)2 − 1 = B(−i + 2)(−i − i)

−1 − 1 = B(2 − i)(−2i)
−2 = B(−4i + 2i2 ) = B(−4i − 2)
−2 = B(−2 − 4i)
−2 1 1 − 2i 1 − 2i 1 − 2i
B= = · = =
−2 − 4i 1 + 2i 1 − 2i 1+4 5
Substitute z = i:
(i)2 − 1 = C(i + 2)(i + i)

1
−1 − 1 = C(2 + i)(2i)
−2 = C(4i + 2i2 ) = C(4i − 2)
−2 = C(−2 + 4i)
−2 1 1 + 2i 1 + 2i 1 + 2i
C= = · = =
−2 + 4i 1 − 2i 1 + 2i 1+4 5
Thus, the partial fraction decomposition is:

3/5 (1 − 2i)/5 (1 + 2i)/5


f (z) = + +
z+2 z+i z−i

Step 2: Rewrite Each Term for Series Expansion


We will expand each term in the partial fraction decomposition as a Laurent
series in the given annular regions.

(i) For 2 < |z| < 3:


3/5
1. First term: z+2

∞  n ∞
3/5 1 3 X 2 3X 2n
· 2 = − = (−1)n n+1
z 1+ z
5z n=0 z 5 n=0 z

(1−2i)/5
2. Second term: z+i

∞  n ∞
(1 − 2i)/5 1 1 − 2i X i 1 − 2i X 1
· i
= − = (−i)n n+1
z 1+ z
5z n=0 z 5 n=0 z

(1+2i)/5
3. Third term: z−i

∞  n ∞
(1 + 2i)/5 1 1 + 2i X i 1 + 2i X n 1
· i
= = i
z 1− z
5z n=0 z 5 n=0 z n+1

Combine the series:


∞ ∞ ∞
3X 2n 1 − 2i X 1 1 + 2i X n 1
f (z) = (−1)n n+1 + (−i)n n+1 + i
5 n=0 z 5 n=0 z 5 n=0 z n+1

Simplify the coefficients:


∞  
X 3 1 − 2i
n n n 1 + 2i n 1
f (z) = (−1) 2 + (−i) + i n+1
n=0
5 5 5 z

2
(ii) For 1 < |z| < 2:
3/5
1. First term: z+2

∞ ∞
3/5 1 3 X  z n 3 X zn
· z = − = (−1)n n
2 1+ 2 10 n=0 2 10 n=0 2

(1−2i)/5
2. Second term: z+i

∞
1 − 2i X 1
(−i)n n
5z n=0 z

(1+2i)/5
3. Third term: z−i

∞
1 + 2i X n 1
i
5z n=0 z n

Combine the series:


∞ ∞ 
z n X 1 − 2i

3 X 1 + 2i n 1
f (z) = (−1)n n + (−i)n + i n+1
10 n=0 2 n=0
5 5 z

Final Answer:
1. Laurent series for 2 < |z| < 3:
∞  
X 3 1 − 2i 1 + 2i n 1
f (z) = (−1)n 2n + (−i)n + i
n=0
5 5 5 z n+1

2. Laurent series for 1 < |z| < 2:


∞ n ∞  
3 X nz
X 1 − 2i n 1 + 2i n 1
f (z) = (−1) n + (−i) + i n+1
10 n=0 2 n=0
5 5 z

These series represent the Laurent expansions of f (z) in the specified annular
regions. The coefficients can be further simplified if desired, but the above forms
are complete and correct.

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