← 2021 Paper 2

UPSC 2021 Maths Optional Paper 2 Q2c — Step-by-Step Solution

20 marks · Section A

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

Question

Find Laurent series of $f(z)=\dfrac{z^2-z+1}{z(z^2-3z+2)}$ in powers of $(z+1)$ in region $|z+1|>3$.

Technique

Partial fractions; substitute $w=z+1$; expand each term in negative powers of $w$ (since $|w|>3$).

Solution

Step 1 — Partial fractions

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

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

$z^2-z+1=A(z-1)(z-2)+Bz(z-2)+Cz(z-1)$.

• $z=0$: $1=A\cdot 2$, $A=1/2$.
• $z=1$: $1=B\cdot(-1)$, $B=-1$.
• $z=2$: $3=C\cdot 2$, $C=3/2$.

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

Step 2 — Substitute $w=z+1$, so $z=w-1$

$\dfrac{1/2}{z}=\dfrac{1/2}{w-1}$. $\dfrac{-1}{z-1}=\dfrac{-1}{w-2}$. $\dfrac{3/2}{z-2}=\dfrac{3/2}{w-3}$.

Region: $|w|>3$, i.e., $|w|$ greater than each of $\{1,2,3\}$. So expand each term in negative powers of $w$.

Step 3 — Expand each term in $1/w$

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

$\dfrac{1}{w-2}=\dfrac{1}{w}\cdot\dfrac{1}{1-2/w}=\sum_{n=0}^\infty\dfrac{2^n}{w^{n+1}}=\sum_{n=1}^\infty\dfrac{2^{n-1}}{w^n}$.

$\dfrac{1}{w-3}=\dfrac{1}{w}\cdot\dfrac{1}{1-3/w}=\sum_{n=1}^\infty\dfrac{3^{n-1}}{w^n}$.

Step 4 — Combine

$f=\dfrac{1}{2}\sum_{n=1}^\infty\dfrac{1}{w^n}-\sum_{n=1}^\infty\dfrac{2^{n-1}}{w^n}+\dfrac{3}{2}\sum_{n=1}^\infty\dfrac{3^{n-1}}{w^n}$.

Coefficient of $1/w^n$: $\dfrac{1}{2}-2^{n-1}+\dfrac{3^n}{2}=\dfrac{1-2\cdot 2^{n-1}+3^n}{2}=\dfrac{1-2^n+3^n}{2}$.

Answer