← 2018 Paper 2

UPSC 2018 Maths Optional Paper 2 Q4b — Step-by-Step Solution

15 marks · Section A

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

Question

Find the Laurent’s series which represent the function $\dfrac{1}{(1+z^2)(z+2)}$ when

• (i) $|z|<1$
• (ii) $1<|z|<2$
• (iii) $|z|>2$

Technique

Partial fractions, then expand each factor as a geometric series in $z$ (when its singularity radius exceeds $|z|$) or in $1/z$ (when smaller); the three annuli split at the singular radii $|z|=1$ and $|z|=2$.

Solution

Step 1 — Partial fractions

The singularities are at $z=\pm i$ (from $1+z^2$, both with $|z|=1$) and $z=-2$ (with $|z|=2$). Decompose

Clearing: $1=(Az+B)(z+2)+C(z^2+1)$. At $z=-2$: $1=C(5)\Rightarrow C=\tfrac15$. Matching $z^2$: $A+C=0\Rightarrow A=-\tfrac15$. Matching constants: $2B+C=1\Rightarrow 2B=\tfrac45\Rightarrow B=\tfrac25$. Thus

The two natural radii are $|z|=1$ (poles $\pm i$) and $|z|=2$ (pole $-2$), giving three annuli.

Step 2 — Region (i): $|z|<1$ (both factors expanded in powers of $z$)

Here $|z|<1<2$, so both $\dfrac{1}{z+2}$ and $\dfrac{1}{z^2+1}$ are expanded as ordinary (Taylor) power series.

So $\dfrac{2-z}{z^2+1}=(2-z)\displaystyle\sum_{n=0}^{\infty}(-1)^n z^{2n}$. Combining via (PF):

This is a pure Taylor series (analytic at $0$). Leading terms: $f(z)=\tfrac15\big[\tfrac12+2\big]+\cdots=\tfrac12+\cdots$ — and indeed $f(0)=\tfrac{1}{1\cdot2}=\tfrac12$.

Step 3 — Region (ii): $1<|z|<2$ ($\frac{1}{z+2}$ in $z$; $\frac{1}{z^2+1}$ in $1/z$)

Now $|z|<2$ still, so $\dfrac{1}{z+2}$ keeps its power series; but $|z|>1$ forces the $z^2+1$ part into negative powers:

Therefore

This genuine Laurent series has both positive powers (from $1/(z+2)$) and negative powers (from $1/(z^2+1)$).

Step 4 — Region (iii): $|z|>2$ (both factors in $1/z$)

Now $|z|>2>1$, so both factors expand in negative powers:

Hence

Answer