UPSC 2014 Maths Optional Paper 2 Q1d — Step-by-Step Solution

10 marks · Section A

Laurent's series in an annulus · Complex Analysis · asked 7× in 13 yrs · Read the full method →

Question

Expand in Laurent series the function $f(z)=\dfrac{1}{z^{2}(z-1)}$ about $z=0$ and $z=1$ .

Standard Laurent expansion by writing $1/(z-a)$ as a geometric series in the appropriate variable for the annulus.

The function has poles at $z=0$ (double) and $z=1$ (simple). Expansion about each singular point gives a Laurent series.

Use $\dfrac{1}{z-1}=-\dfrac{1}{1-z}=-\sum_{n=0}^\infty z^{n}$ for $|z|<1$ .

f(z)=\dfrac{1}{z^{2}}\cdot\dfrac{1}{z-1}=-\dfrac{1}{z^{2}}\sum_{n=0}^\infty z^{n}=-\sum_{n=0}^\infty z^{n-2}.

Splitting into principal part and analytic part:

\boxed{\;f(z)=-\dfrac{1}{z^{2}}-\dfrac{1}{z}-1-z-z^{2}-\cdots,\;0<|z|<1.\;}

For the annulus $|z|>1$ , use $\dfrac{1}{z-1}=\dfrac{1}{z}\cdot\dfrac{1}{1-1/z}=\dfrac{1}{z}\sum_{n=0}^\infty z^{-n}$ :

f(z)=\dfrac{1}{z^{2}}\cdot\dfrac{1}{z}\sum_{n=0}^\infty z^{-n}=\sum_{n=0}^\infty z^{-(n+3)}=\dfrac{1}{z^{3}}+\dfrac{1}{z^{4}}+\dfrac{1}{z^{5}}+\cdots,\;|z|>1.

Substitute $w=z-1$ . Then $z=w+1$ , and

f(z)=\dfrac{1}{w(w+1)^{2}}.

Use $\dfrac{1}{(w+1)^{2}}=\sum_{n=0}^\infty(-1)^{n}(n+1)w^{n}=1-2w+3w^{2}-4w^{3}+\cdots$ for $|w|<1$ .

Therefore

f(z)=\dfrac{1}{w}(1-2w+3w^{2}-4w^{3}+\cdots)=\dfrac{1}{w}-2+3w-4w^{2}+\cdots.

In terms of $z$ :

\boxed{\;f(z)=\dfrac{1}{z-1}-2+3(z-1)-4(z-1)^{2}+5(z-1)^{3}-\cdots,\;0<|z-1|<1.\;}