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Algebra, analysis and differential equations II course of the University of Oxford

Past exam of the algebra, analysis and differential equations II course of the University of Oxford

2012

Inside the unit circle, use the <geometric series> for $1/(1-z)$:
$$
\boxed{\frac1{z^2(z-1)}=-\sum_{n=0}^{\infty}z^{n-2}=-\frac1{z^2}-\frac1z-1-z-z^2-\cdots,\qquad0<|z|<1.}
$$
Outside the unit circle, expand in powers of $1/z$ instead:
$$
\boxed{\frac1{z^2(z-1)}=\frac1{z^3}\frac1{1-z^{-1}}=\sum_{n=0}^{\infty}z^{-n-3},\qquad |z|>1.}
$$
The different <Laurent series> reflect different annuli separated by the <pole> at one. Both follow from absolutely convergent <geometric series> in their stated regions.

Solution

Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 5 / c / Solution

Past exam of the algebra, analysis and differential equations II course of the University of Oxford / 2012 / 5 / c / Solution

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