CVS3

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Find the Laurent series for $\frac{1}{z^2(z-1)}\,$ about all its singular points.

The singularities are $z=0,1\,$ .

Centered at the singular point $z_0=0\,$ there are two possible Laurent series. One valid for $0<|z|<1\,$ and one for $|z|>1\,$ .

In the domain $0<|z|<1\,$ , $\frac{1}{z^2(z-1)}=-\frac{1}{z^2}\sum_{n=0}^\infty z^n\,$

$= -\sum_{n=0}^\infty z^{n-2}=-\sum_{n=-2}^\infty z^n\,$

For the domain $|z|>1\,$ , $\frac{1}{z^2(z-1)}=\frac{1/z}{z^2(1-1/z)}\,$

$=\frac{1}{z^3}\sum_{n=0}^\infty \left(\frac{1}{z}\right)^n=\sum_{n=0}^\infty \left(\frac{1}{z}\right)^{n+3}=\sum_{n=0}^\infty \left( \frac{1}{z} \right)^{n+3}\,$

Let $n\to -n-3\,$ .

$=\sum_{n=-\infty}^{-3} z^{n}, |z|>1\,$

For the singular point $z=1\,$ make a change of variables $u=z-1\,$ .

$\frac{1}{z^2(z-1)}=\frac{1}{(u+1)^2u}\,$

Now when $0<|u|<1\,$

$\frac{1}{(u+1)^2u} = \frac{1}{u}\sum_{n=0}^\infty (-1)^n (n+1) u^n \Bigg|_{u=z-1}\,$

$= \frac{1}{z-1} \sum_{n=0}^\infty (-1)^n (n+1)(z-1)^{n-1} \,$

$= \sum_{n=0}^\infty (-1)^n (n+1)(z-1)^{n-2} \,$

With a change of index $n\to n+1\,$

$\sum_{n=-1}^\infty (-1)^{n+1} (n+2)(z-1)^n, 0<|z-1|<z\,$

When $|u|>1\,$

$\frac{1}{(u+1)^2u} = \frac{1}{u^3(1+1/u)^2} = \frac{1}{u^3} \sum_{n=0}^\infty (-1)^n (n+1) u^{-n} \Bigg|_{u=z-1}\,$

$=\frac{1}{(z-1)^3}\sum_{n=0}^\infty (-1)^n (n+1) (z-1)^{-n}\,$

$=\sum_{n=0}^\infty (-1)^n (n+1) (z-1)^{-n-3}\,$

$=\sum_{n=-\infty}^{-3} (-1)^n (n+2) (z-1)^n, |z-1|>1\,$

Main Page : Complex Variables : Series

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