← 2015 Paper 2

UPSC 2015 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 all possible Taylor’s and Laurent’s series expansions of the function $f(z)=\dfrac{2z-3}{z^2-3z+2}$ about the point $z=0$.

Technique

Partial fractions to separate the two singularities; then in each region of the $|z|$-plane (bounded by the singularities), expand each piece as a power series (Taylor, in $z$) or a Laurent series (in $1/z$) depending on which converges in that annulus.

Solution

Step 1 — Partial fractions

$z^2-3z+2=(z-1)(z-2)$. Decompose:

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

• $z=1$: $-1=-A\Rightarrow A=1$.
• $z=2$: $1=B$.

Step 2 — Singularities

Simple poles at $z=1$ and $z=2$. About $z=0$, the relevant radii are $|z|=1$ and $|z|=2$, dividing the plane into three regions:

• Region 1: $|z|<1$ (Taylor — analytic inside the smaller disc).
• Region 2: $1<|z|<2$ (Laurent — annular).
• Region 3: $|z|>2$ (Laurent — outside both poles).

Step 3 — Region 1: $|z|<1$ (Taylor)

In this disc, $|z|<1$ and $|z/2|<1/2<1$. Use geometric series:

Sum:

Step 4 — Region 2: $1<|z|<2$ (Laurent)

Here $|1/z|<1$ (so expand $1/(z-1)$ in negative powers) and $|z/2|<1$ (so expand $1/(z-2)$ in positive powers as before).

Sum:

Step 5 — Region 3: $|z|>2$ (Laurent)

Here both $|1/z|<1$ and $|2/z|<1$. Expand both in negative powers:

Sum:

Answer