Author Topic: Q6 TUT 0201  (Read 548 times)

Victor Ivrii

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Q6 TUT 0201
« on: November 17, 2018, 04:08:42 PM »
Find the Laurent series for the given function $f(z)$ about the indicated point. Also, give the residue of the function at the point.
$$
f(z)=\frac{z^2}{z^2-1};\qquad z_0=1.
$$

Junya Zhang

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Re: Q6 TUT 0201
« Reply #1 on: November 17, 2018, 04:20:00 PM »
This is question 8 from CH2.5 in the textbook.

Let $w = z - 1$. Then $z = w+ 1$.
$$\frac{z^2}{z^2 -1} = \frac{(w+1)^2}{(w+1)^2 -1} = \frac{w^2 + 2w + 1}{w^2 + 2w}= 1+ \frac{1}{w^2 + 2w}=1 + \frac{1}{2w} \cdot \frac{1}{1 + w/2} = 1 + \frac{1}{2w} \sum_{n = 0}^{\infty} \left(- \frac{w}{2}\right)^n$$ which is valid for $|w/2| < 1$, i.e. $|z-1|<2$. Then
$$
\frac{z^2}{z^2 -1} = 1 + \frac{1}{2w}\left(1 - \frac{w}{2} + \sum_{n = 2}^{\infty} (-1)^n 2^{-n}w^n \right)
= 1 +  \frac{1}{2w} - \frac{1}{4} + \sum_{n = 2}^{\infty} (-1)^n 2^{-n-1}w^{n-1}
= \frac{1}{2w} +\frac{3}{4}  + \sum_{n = 1}^{\infty} (-1)^{n+1} 2^{-n-2}w^{n}
= \frac{1}{2w} +\frac{3}{4}  + \sum_{n = 1}^{\infty} (-1)^{n+1} \frac{w^{n}}{2^{n+2}}
$$ Substitute $w = z-1$ back to the equation we get $$\frac{z^2}{z^2 -1} = \frac{1}{2(z-1)} +\frac{3}{4}  + \sum_{n = 1}^{\infty} (-1)^{n+1} \frac{(z-1)^{n}}{2^{n+2}}$$
Residue of the function at $z = 1$ is the coefficient of $\frac{1}{z-1}$ in the Laurent series, which is $\frac{1}{2}$.

Amy Zhao

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Re: Q6 TUT 0201
« Reply #2 on: November 17, 2018, 06:32:49 PM »
$\frac{z^2}{z^2-1}$$=\frac{z^2-1+1}{z^2-1}$
$=1+\frac{1}{z^2-1}$
$=1+\frac{1}{(z-1)(z+1)}$
$=1+\frac{1}{2}\frac{1}{z-1}-\frac{1}{z+1}$, by partial fractions
$=1+\frac{1}{2}(\frac{1}{z-1}-\frac{1}{2+(z-1)})$
$=1+\frac{1}{2}(\frac{1}{z-1}-\frac{1}{2}\frac{1}{1+\frac{z-1}{2}})$
$=1+\frac{1}{2}((z-1)^{-1}-\frac{1}{2}\frac{1}{1-\frac{1-z}{2}})$
$=1+\frac{1}{2}((z-1)^{-1}-\frac{1}{2}\sum_{0}^{\infty} (\frac{1-z}{2})^n)$
Residue=$\frac{1}{2}$

Victor Ivrii

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Re: Q6 TUT 0201
« Reply #3 on: November 28, 2018, 04:57:15 AM »
Two rather different solutions leadinfg to the same result.