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UPSC 2022 Maths Optional Paper 2 Q1d — Step-by-Step Solution

10 marks · Section A

Laurent's series in an annulus · Complex Analysis · asked 8× in 14 yrs · Read the full method →

Question

Expand $f(z)=\dfrac{1}{(z-1)^2(z-3)}$ in Laurent series valid for (i) $0<|z-1|<2$ and (ii) $0<|z-3|<2$.

Technique

Partial fractions; expand each term as a Laurent (negative powers from singularity) or Taylor (positive powers, geometric series) in the appropriate region.

Solution

Setup. Singularities at $z=1$ (double pole) and $z=3$ (simple pole). The regions specified are around each singularity.

Step 1 — Partial fractions

$\dfrac{1}{(z-1)^2(z-3)}=\dfrac{A}{z-1}+\dfrac{B}{(z-1)^2}+\dfrac{C}{z-3}$.

$1=A(z-1)(z-3)+B(z-3)+C(z-1)^2$.

• $z=1$: $1=B(-2)\Rightarrow B=-1/2$.
• $z=3$: $1=C(4)\Rightarrow C=1/4$.
• $z=0$: $1=A(-1)(-3)+B(-3)+C(1)=3A-3(-1/2)+1/4=3A+3/2+1/4$. So $3A=1-3/2-1/4=-3/4$, $A=-1/4$.

So $f(z)=-\dfrac{1/4}{z-1}-\dfrac{1/2}{(z-1)^2}+\dfrac{1/4}{z-3}$.

Step 2 — Region (i): $0<|z-1|<2$

Let $w=z-1$, so $0<|w|<2$. Substitute $z=w+1$, $z-3=w-2$.

$f=-\dfrac{1/4}{w}-\dfrac{1/2}{w^2}+\dfrac{1/4}{w-2}$.

The first two terms are already in Laurent form in $w$. The third: $\dfrac{1/4}{w-2}=-\dfrac{1/4}{2-w}=-\dfrac{1/8}{1-w/2}=-\dfrac{1}{8}\sum_{n=0}^\infty(w/2)^n=-\sum_{n=0}^\infty\dfrac{w^n}{2^{n+3}}$.

(Valid for $|w/2|<1$, i.e. $|w|<2$ ✓.)

Combine (replacing $w=z-1$):

Step 3 — Region (ii): $0<|z-3|<2$

Let $w=z-3$, so $0<|w|<2$. Substitute $z=w+3$, $z-1=w+2$.

$f=-\dfrac{1/4}{w+2}-\dfrac{1/2}{(w+2)^2}+\dfrac{1/4}{w}$.

The third term is already in Laurent form. The first two need expansion in $w$ around $0$ (so as positive powers of $w$, valid for $|w|<2$).

$\dfrac{1/4}{w+2}=\dfrac{1/8}{1+w/2}=\dfrac{1}{8}\sum_{n=0}^\infty(-w/2)^n=\dfrac{1}{8}\sum_{n=0}^\infty\dfrac{(-1)^n w^n}{2^n}=\sum_{n=0}^\infty\dfrac{(-1)^n w^n}{2^{n+3}}$.

$\dfrac{1/2}{(w+2)^2}=\dfrac{1}{2}\cdot\dfrac{1}{4}\cdot\dfrac{1}{(1+w/2)^2}=\dfrac{1}{8}\sum_{n=0}^\infty(-1)^n(n+1)\dfrac{w^n}{2^n}=\sum_{n=0}^\infty\dfrac{(-1)^n(n+1)w^n}{2^{n+3}}$.

(Using $\dfrac{1}{(1-u)^2}=\sum_{n=0}^\infty(n+1)u^n$, so $\dfrac{1}{(1+u)^2}=\sum(-1)^n(n+1)u^n$.)

Combine:

$f=\dfrac{1/4}{w}-\sum_{n=0}^\infty\dfrac{(-1)^n w^n}{2^{n+3}}-\sum_{n=0}^\infty\dfrac{(-1)^n(n+1)w^n}{2^{n+3}}$

$=\dfrac{1/4}{w}-\sum_{n=0}^\infty\dfrac{(-1)^n(n+2)w^n}{2^{n+3}}$.

In terms of $z-3$:

Answer