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Sagot :
To find the Laurent series expansion of [tex]\( f(z) = \frac{1}{z^2 + 2z} \)[/tex] for [tex]\( 1 < |z - 1| < 3 \)[/tex], we start with partial fraction decomposition.
First, we can rewrite [tex]\( f(z) \)[/tex] as:
[tex]\[ f(z) = \frac{1}{z^2 + 2z} = \frac{1}{z(z + 2)}. \][/tex]
To decompose [tex]\( \frac{1}{z(z + 2)} \)[/tex] into partial fractions, we write:
[tex]\[ \frac{1}{z(z + 2)} = \frac{A}{z} + \frac{B}{z + 2}. \][/tex]
Solving for [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ 1 = A(z + 2) + Bz. \][/tex]
Setting [tex]\( z = 0 \)[/tex]:
[tex]\[ 1 = A(2), \][/tex]
[tex]\[ A = \frac{1}{2}. \][/tex]
Setting [tex]\( z = -2 \)[/tex]:
[tex]\[ 1 = B(-2), \][/tex]
[tex]\[ B = -\frac{1}{2}. \][/tex]
So, the partial fractions decomposition is:
[tex]\[ \frac{1}{z(z + 2)} = \frac{1}{2z} - \frac{1}{2(z + 2)}. \][/tex]
We now need to express this in a form suitable for the Laurent series in the given annulus [tex]\( 1 < |z - 1| < 3 \)[/tex].
The expression [tex]\( \frac{1}{2z} \)[/tex] is already in a suitable form.
For [tex]\( \frac{1}{2(z + 2)} \)[/tex], we note that [tex]\( z + 2 = (z - 1) + 3 \)[/tex].
We will write this expression in terms of [tex]\( \frac{1}{z-1} \)[/tex]:
[tex]\[ \frac{1}{z + 2} = \frac{1}{(z - 1) + 3}. \][/tex]
Using the series expansion for [tex]\( \frac{1}{a+b} \)[/tex], where [tex]\( |b| < |a| \)[/tex],
[tex]\[ \frac{1}{a+b} = \frac{1}{a} \cdot \frac{1}{1 - \left(-\frac{b}{a}\right)} = \frac{1}{a} \sum_{n=0}^\infty \left(-\frac{b}{a}\right)^n. \][/tex]
Here [tex]\( a = 3 \)[/tex] and [tex]\( b = z-1 \)[/tex], so:
[tex]\[ \frac{1}{(z - 1) + 3} = \frac{1}{3} \cdot \frac{1}{1 - \left(-\frac{z-1}{3}\right)} = \frac{1}{3} \sum_{n=0}^\infty \left( \frac{z-1}{3} \right)^n. \][/tex]
Therefore,
[tex]\[ \frac{1}{z + 2} = \frac{1}{3} \sum_{n=0}^\infty \left( \frac{z-1}{3} \right)^n = \sum_{n=0}^\infty \frac{(z-1)^n}{3^{n+1}}. \][/tex]
Now multiply by [tex]\(-\frac{1}{2}\)[/tex]:
[tex]\[ -\frac{1}{2(z + 2)} = -\frac{1}{2} \sum_{n=0}^\infty \frac{(z-1)^n}{3^{n+1}} = - \sum_{n=0}^\infty \frac{(z-1)^n}{2 \cdot 3^{n+1}} = - \sum_{n=0}^\infty \frac{(z-1)^n}{2^{(n+1)} \cdot 3}. \][/tex]
Thus, our Laurent Series in the given annulus is:
[tex]\[ \frac{1}{2z} - \sum_{n=0}^\infty \frac{(z-1)^n}{2 \cdot 3^{n+1}}. \][/tex]
It can be written as:
[tex]\[ f(z) = \frac{1}{2z} - \sum_{n=0}^\infty \frac{(z-1)^n}{6 \cdot 3^n}. \][/tex]
This is the required Laurent series expansion for the given function in the specified region.
First, we can rewrite [tex]\( f(z) \)[/tex] as:
[tex]\[ f(z) = \frac{1}{z^2 + 2z} = \frac{1}{z(z + 2)}. \][/tex]
To decompose [tex]\( \frac{1}{z(z + 2)} \)[/tex] into partial fractions, we write:
[tex]\[ \frac{1}{z(z + 2)} = \frac{A}{z} + \frac{B}{z + 2}. \][/tex]
Solving for [tex]\( A \)[/tex] and [tex]\( B \)[/tex]:
[tex]\[ 1 = A(z + 2) + Bz. \][/tex]
Setting [tex]\( z = 0 \)[/tex]:
[tex]\[ 1 = A(2), \][/tex]
[tex]\[ A = \frac{1}{2}. \][/tex]
Setting [tex]\( z = -2 \)[/tex]:
[tex]\[ 1 = B(-2), \][/tex]
[tex]\[ B = -\frac{1}{2}. \][/tex]
So, the partial fractions decomposition is:
[tex]\[ \frac{1}{z(z + 2)} = \frac{1}{2z} - \frac{1}{2(z + 2)}. \][/tex]
We now need to express this in a form suitable for the Laurent series in the given annulus [tex]\( 1 < |z - 1| < 3 \)[/tex].
The expression [tex]\( \frac{1}{2z} \)[/tex] is already in a suitable form.
For [tex]\( \frac{1}{2(z + 2)} \)[/tex], we note that [tex]\( z + 2 = (z - 1) + 3 \)[/tex].
We will write this expression in terms of [tex]\( \frac{1}{z-1} \)[/tex]:
[tex]\[ \frac{1}{z + 2} = \frac{1}{(z - 1) + 3}. \][/tex]
Using the series expansion for [tex]\( \frac{1}{a+b} \)[/tex], where [tex]\( |b| < |a| \)[/tex],
[tex]\[ \frac{1}{a+b} = \frac{1}{a} \cdot \frac{1}{1 - \left(-\frac{b}{a}\right)} = \frac{1}{a} \sum_{n=0}^\infty \left(-\frac{b}{a}\right)^n. \][/tex]
Here [tex]\( a = 3 \)[/tex] and [tex]\( b = z-1 \)[/tex], so:
[tex]\[ \frac{1}{(z - 1) + 3} = \frac{1}{3} \cdot \frac{1}{1 - \left(-\frac{z-1}{3}\right)} = \frac{1}{3} \sum_{n=0}^\infty \left( \frac{z-1}{3} \right)^n. \][/tex]
Therefore,
[tex]\[ \frac{1}{z + 2} = \frac{1}{3} \sum_{n=0}^\infty \left( \frac{z-1}{3} \right)^n = \sum_{n=0}^\infty \frac{(z-1)^n}{3^{n+1}}. \][/tex]
Now multiply by [tex]\(-\frac{1}{2}\)[/tex]:
[tex]\[ -\frac{1}{2(z + 2)} = -\frac{1}{2} \sum_{n=0}^\infty \frac{(z-1)^n}{3^{n+1}} = - \sum_{n=0}^\infty \frac{(z-1)^n}{2 \cdot 3^{n+1}} = - \sum_{n=0}^\infty \frac{(z-1)^n}{2^{(n+1)} \cdot 3}. \][/tex]
Thus, our Laurent Series in the given annulus is:
[tex]\[ \frac{1}{2z} - \sum_{n=0}^\infty \frac{(z-1)^n}{2 \cdot 3^{n+1}}. \][/tex]
It can be written as:
[tex]\[ f(z) = \frac{1}{2z} - \sum_{n=0}^\infty \frac{(z-1)^n}{6 \cdot 3^n}. \][/tex]
This is the required Laurent series expansion for the given function in the specified region.
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