UPSC Maths 2025 Paper 2 Q1d — Solution

10 marks · Section A

Question

Expand $f(z) = \dfrac{1}{(z+1)(z+3)}$ in a Laurent series valid for $1 < |z| < 3$ .

Technique

Split by partial fractions, then expand each term in the annulus: for $|z|>1$ expand in powers of $1/z$ , and for $|z|<3$ expand in powers of $z$ .

Solution

Step 1 — Partial fractions.

$\frac{1}{(z+1)(z+3)} = \frac{A}{z+1} + \frac{B}{z+3}.$ Then $1 = A(z+3) + B(z+1)$ . Put $z=-1$ : $1 = A(2) \Rightarrow A=\tfrac12$ . Put $z=-3$ : $1 = B(-2) \Rightarrow B=-\tfrac12$ . Hence $f(z) = \frac{1}{2}\,\frac{1}{z+1} - \frac{1}{2}\,\frac{1}{z+3}.$

Step 2 — Expand each term for the annulus $1<|z|<3$ .

Term 1: $\dfrac{1}{z+1}$ with $|z|>1$ . Factor out $z$ so the ratio has modulus $<1$ : $\frac{1}{z+1} = \frac{1}{z}\cdot\frac{1}{1+\frac1z} = \frac{1}{z}\sum_{n=0}^{\infty}\left(-\frac1z\right)^n = \sum_{n=0}^{\infty} \frac{(-1)^n}{z^{n+1}}, \qquad |z|>1.$ This is the principal part (negative powers of $z$ ).

Term 2: $\dfrac{1}{z+3}$ with $|z|<3$ . Factor out $3$ : $\frac{1}{z+3} = \frac{1}{3}\cdot\frac{1}{1+\frac z3} = \frac{1}{3}\sum_{n=0}^{\infty}\left(-\frac z3\right)^n = \sum_{n=0}^{\infty} \frac{(-1)^n z^n}{3^{n+1}}, \qquad |z|<3.$ This is the analytic part (non-negative powers of $z$ ).

Step 3 — Combine.

$f(z) = \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^n}{z^{n+1}} - \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^n z^n}{3^{n+1}}, \qquad 1<|z|<3.$

Writing out the first few terms: $f(z) = \cdots - \frac{1}{2z^3} + \frac{1}{2z^2} - \frac{1}{2z} \;-\; \frac{1}{6} + \frac{z}{18} - \frac{z^2}{54} + \cdots$

Answer

$\boxed{\;f(z) = \frac{1}{(z+1)(z+3)} = \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^n}{z^{n+1}} \;-\; \frac{1}{2}\sum_{n=0}^{\infty} \frac{(-1)^n}{3^{n+1}}\,z^n,\qquad 1<|z|<3.\;}$

Equivalently $\displaystyle f(z)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{2\,z^{n}}-\sum_{n=0}^{\infty}\frac{(-1)^n}{2\cdot 3^{n+1}}z^n.$