Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Get immediate and reliable answers to your questions from a community of experienced experts on our platform. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
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]
We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
