Answer :
Certainly! Let's tackle these two problems step by step.
### (a) Laurent Series Expansion of [tex]\( f(z) = \frac{1}{z^2 + 2z} \)[/tex] for [tex]\( 1 < |z-1| < 3 \)[/tex]
First, let's rewrite the function to make the singularities more apparent:
[tex]\[ f(z) = \frac{1}{z(z+2)} \][/tex]
This function has poles at [tex]\( z = 0 \)[/tex] and [tex]\( z = -2 \)[/tex]. To find the Laurent series for [tex]\( 1 < |z - 1| < 3 \)[/tex], we need to express [tex]\( f(z) \)[/tex] in a form that separates the singularities inside and outside the annulus.
We use partial fraction decomposition to rewrite [tex]\( f(z) \)[/tex]:
[tex]\[ f(z) = \frac{1}{z(z+2)} = \frac{A}{z} + \frac{B}{z+2} \][/tex]
Solving for [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ 1 = A(z+2) + Bz \][/tex]
Setting [tex]\( z = 0 \)[/tex]:
[tex]\[ 1 = 2A \implies A = \frac{1}{2} \][/tex]
Setting [tex]\( z = -2 \)[/tex]:
[tex]\[ 1 = -2B \implies B = -\frac{1}{2} \][/tex]
So the decomposition is:
[tex]\[ f(z) = \frac{1}{2z} - \frac{1}{2(z+2)} \][/tex]
Now, in the given annulus [tex]\( 1 < |z-1| < 3 \)[/tex], we need to work with the term [tex]\( \frac{1}{2(z+2)} \)[/tex] by shifting the origin to [tex]\( z = 1 \)[/tex] using [tex]\( z+2 = (z-1) + 3 \)[/tex]:
[tex]\[ \frac{1}{z+2} = \frac{1}{(z-1)+3} = \frac{1}{3\left(1 + \frac{z-1}{3}\right)} \][/tex]
Using the geometric series expansion [tex]\( \frac{1}{1+w} = \sum_{n=0}^{\infty} (-w)^n \)[/tex], where [tex]\( w = \frac{z-1}{3} \)[/tex]:
[tex]\[ \frac{1}{z+2} = \frac{1}{3} \sum_{n=0}^{\infty} \left( -\frac{z-1}{3} \right)^n = \frac{1}{3} \sum_{n=0}^{\infty} \left( -1 \right)^n \left( \frac{z-1}{3} \right)^n \][/tex]
Combining the partial fractions:
[tex]\[ f(z) = \frac{1}{2z} - \frac{1}{2} \left( \frac{1}{3} \sum_{n=0}^{\infty} \left( -1 \right)^n \left( \frac{z-1}{3} \right)^n \right) \][/tex]
Simplifying:
[tex]\[ f(z) = \frac{1}{2z} - \frac{1}{6} \sum_{n=0}^{\infty} \left( -1 \right)^n \left( \frac{z-1}{3} \right)^n \][/tex]
[tex]\[ f(z) = \frac{1}{2z} - \frac{1}{6} \sum_{n=0}^{\infty} \left( -\frac{z-1}{3} \right)^n \][/tex]
### (b) Laurent Series Expansion of [tex]\( f(z) = z^2 \sin \left( \frac{1}{z^2} \right) \)[/tex] in the domain [tex]\( 0<|z|<\infty \)[/tex]
Firstly, recall the Maclaurin series expansion for [tex]\( \sin x \)[/tex]:
[tex]\[ \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \][/tex]
Substitute [tex]\( x = \frac{1}{z^2} \)[/tex]:
[tex]\[ \sin \left( \frac{1}{z^2} \right) = \sum_{n=0}^{\infty} \frac{(-1)^n \left( \frac{1}{z^2} \right)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{4n+2}} \][/tex]
Now, multiply by [tex]\( z^2 \)[/tex]:
[tex]\[ f(z) = z^2 \sin \left( \frac{1}{z^2} \right) = z^2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{4n+2}} \][/tex]
[tex]\[ f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{z^2}{z^{4n+2}} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{2-4n-2} \][/tex]
[tex]\[ f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{-4n} \][/tex]
Therefore, the Laurent series of [tex]\( f(z) \)[/tex] in the domain [tex]\( 0 < |z| < \infty \)[/tex] is:
[tex]\[ f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{-4n} \][/tex]
### (a) Laurent Series Expansion of [tex]\( f(z) = \frac{1}{z^2 + 2z} \)[/tex] for [tex]\( 1 < |z-1| < 3 \)[/tex]
First, let's rewrite the function to make the singularities more apparent:
[tex]\[ f(z) = \frac{1}{z(z+2)} \][/tex]
This function has poles at [tex]\( z = 0 \)[/tex] and [tex]\( z = -2 \)[/tex]. To find the Laurent series for [tex]\( 1 < |z - 1| < 3 \)[/tex], we need to express [tex]\( f(z) \)[/tex] in a form that separates the singularities inside and outside the annulus.
We use partial fraction decomposition to rewrite [tex]\( f(z) \)[/tex]:
[tex]\[ f(z) = \frac{1}{z(z+2)} = \frac{A}{z} + \frac{B}{z+2} \][/tex]
Solving for [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ 1 = A(z+2) + Bz \][/tex]
Setting [tex]\( z = 0 \)[/tex]:
[tex]\[ 1 = 2A \implies A = \frac{1}{2} \][/tex]
Setting [tex]\( z = -2 \)[/tex]:
[tex]\[ 1 = -2B \implies B = -\frac{1}{2} \][/tex]
So the decomposition is:
[tex]\[ f(z) = \frac{1}{2z} - \frac{1}{2(z+2)} \][/tex]
Now, in the given annulus [tex]\( 1 < |z-1| < 3 \)[/tex], we need to work with the term [tex]\( \frac{1}{2(z+2)} \)[/tex] by shifting the origin to [tex]\( z = 1 \)[/tex] using [tex]\( z+2 = (z-1) + 3 \)[/tex]:
[tex]\[ \frac{1}{z+2} = \frac{1}{(z-1)+3} = \frac{1}{3\left(1 + \frac{z-1}{3}\right)} \][/tex]
Using the geometric series expansion [tex]\( \frac{1}{1+w} = \sum_{n=0}^{\infty} (-w)^n \)[/tex], where [tex]\( w = \frac{z-1}{3} \)[/tex]:
[tex]\[ \frac{1}{z+2} = \frac{1}{3} \sum_{n=0}^{\infty} \left( -\frac{z-1}{3} \right)^n = \frac{1}{3} \sum_{n=0}^{\infty} \left( -1 \right)^n \left( \frac{z-1}{3} \right)^n \][/tex]
Combining the partial fractions:
[tex]\[ f(z) = \frac{1}{2z} - \frac{1}{2} \left( \frac{1}{3} \sum_{n=0}^{\infty} \left( -1 \right)^n \left( \frac{z-1}{3} \right)^n \right) \][/tex]
Simplifying:
[tex]\[ f(z) = \frac{1}{2z} - \frac{1}{6} \sum_{n=0}^{\infty} \left( -1 \right)^n \left( \frac{z-1}{3} \right)^n \][/tex]
[tex]\[ f(z) = \frac{1}{2z} - \frac{1}{6} \sum_{n=0}^{\infty} \left( -\frac{z-1}{3} \right)^n \][/tex]
### (b) Laurent Series Expansion of [tex]\( f(z) = z^2 \sin \left( \frac{1}{z^2} \right) \)[/tex] in the domain [tex]\( 0<|z|<\infty \)[/tex]
Firstly, recall the Maclaurin series expansion for [tex]\( \sin x \)[/tex]:
[tex]\[ \sin x = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n+1}}{(2n+1)!} \][/tex]
Substitute [tex]\( x = \frac{1}{z^2} \)[/tex]:
[tex]\[ \sin \left( \frac{1}{z^2} \right) = \sum_{n=0}^{\infty} \frac{(-1)^n \left( \frac{1}{z^2} \right)^{2n+1}}{(2n+1)!} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{4n+2}} \][/tex]
Now, multiply by [tex]\( z^2 \)[/tex]:
[tex]\[ f(z) = z^2 \sin \left( \frac{1}{z^2} \right) = z^2 \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{1}{z^{4n+2}} \][/tex]
[tex]\[ f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \frac{z^2}{z^{4n+2}} = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{2-4n-2} \][/tex]
[tex]\[ f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{-4n} \][/tex]
Therefore, the Laurent series of [tex]\( f(z) \)[/tex] in the domain [tex]\( 0 < |z| < \infty \)[/tex] is:
[tex]\[ f(z) = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} z^{-4n} \][/tex]
