← 2019 Paper 2

UPSC 2019 Maths Optional Paper 2 Q2d — Step-by-Step Solution

15 marks · Section A

Residues: computation at poles of various orders · Complex Analysis · asked 3× in 13 yrs · Read the full method →

Question

Show that an isolated singular point z0z_0 of a function f(z)f(z) is a pole of order mm if and only if f(z)f(z) can be written in the form f(z)=ϕ(z)(z−z0)mf(z)=\dfrac{\phi(z)}{(z-z_0)^m} where ϕ(z)\phi(z) is analytic and non zero at z0z_0. Moreover Res⁡z=z0f(z)=ϕ(m−1)(z0)(m−1)!\displaystyle\operatorname*{Res}_{z=z_0} f(z)=\frac{\phi^{(m-1)}(z_0)}{(m-1)!} if m≥1m\ge 1.

Technique

Laurent series at an isolated singularity; multiply/divide by (z−z0)m(z-z_0)^m to pass between the pole and the analytic-numerator form; read a−1a_{-1} off as the k=m−1k=m-1 Taylor coefficient of ϕ\phi.

Solution

Let z0z_0 be an isolated singularity: ff is analytic on a punctured disc 0<∣z−z0∣<R0<|z-z_0|<R, with Laurent expansion

f(z)=∑n=−∞∞an(z−z0)n.(L)f(z)=\sum_{n=-\infty}^{\infty} a_n (z-z_0)^n.\tag{L}

By definition, z0z_0 is a pole of order mm (m≥1m\ge1) iff the principal part has exactly mm terms, i.e. a−m≠0a_{-m}\ne0 and a−n=0a_{-n}=0 for all n>mn>m.

Step 1 — (⇒\Rightarrow) Pole of order mm implies the form ϕ/(z−z0)m\phi/(z-z_0)^m

If z0z_0 is a pole of order mm, the Laurent series (L) starts at n=−mn=-m:

f(z)=a−m(z−z0)m+⋯+a−1z−z0+∑n≥0an(z−z0)n,a−m≠0.f(z)=\frac{a_{-m}}{(z-z_0)^m}+\cdots+\frac{a_{-1}}{z-z_0}+\sum_{n\ge0}a_n(z-z_0)^n,\qquad a_{-m}\ne0.

Multiply by (z−z0)m(z-z_0)^m:

ϕ(z):=(z−z0)mf(z)=a−m+a−m+1(z−z0)+⋯=∑k≥0ak−m(z−z0)k.\phi(z):=(z-z_0)^m f(z)=a_{-m}+a_{-m+1}(z-z_0)+\cdots=\sum_{k\ge0}a_{k-m}(z-z_0)^k.

This is a convergent power series on 0<∣z−z0∣<R0<|z-z_0|<R; it defines an analytic function on the full disc ∣z−z0∣<R|z-z_0|<R (the singularity is removed), with

ϕ(z0)=a−m≠0.\phi(z_0)=a_{-m}\ne0.

Thus f(z)=ϕ(z)(z−z0)mf(z)=\dfrac{\phi(z)}{(z-z_0)^m} with ϕ\phi analytic at z0z_0 and ϕ(z0)≠0\phi(z_0)\ne0.

Step 2 — (⇐\Leftarrow) The form ϕ/(z−z0)m\phi/(z-z_0)^m implies a pole of order mm

Suppose f(z)=ϕ(z)(z−z0)mf(z)=\dfrac{\phi(z)}{(z-z_0)^m} with ϕ\phi analytic at z0z_0 and ϕ(z0)≠0\phi(z_0)\ne0. Expand ϕ\phi in its Taylor series (valid since ϕ\phi is analytic at z0z_0):

ϕ(z)=∑k≥0ϕ(k)(z0)k!(z−z0)k,ϕ(z0)≠0.\phi(z)=\sum_{k\ge0}\frac{\phi^{(k)}(z_0)}{k!}(z-z_0)^k,\qquad \phi(z_0)\ne0.

Then

f(z)=1(z−z0)m∑k≥0ϕ(k)(z0)k!(z−z0)k=∑k≥0ϕ(k)(z0)k!(z−z0)k−m.f(z)=\frac{1}{(z-z_0)^m}\sum_{k\ge0}\frac{\phi^{(k)}(z_0)}{k!}(z-z_0)^k=\sum_{k\ge0}\frac{\phi^{(k)}(z_0)}{k!}(z-z_0)^{k-m}.

The lowest power is (z−z0)−m(z-z_0)^{-m} with coefficient ϕ(z0)≠0\phi(z_0)\ne0, and there is no power below −m-m. So the principal part has exactly mm terms: z0z_0 is a pole of order mm.

This proves the equivalence. ■\blacksquare

Step 3 — Residue formula

The residue is the coefficient a−1a_{-1} in (L), i.e. the coefficient of (z−z0)−1(z-z_0)^{-1} in the last display, which occurs at k−m=−1k-m=-1, i.e. k=m−1k=m-1:

Res⁡z=z0f=a−1=ϕ(m−1)(z0)(m−1)!.\operatorname*{Res}_{z=z_0}f=a_{-1}=\frac{\phi^{(m-1)}(z_0)}{(m-1)!}.

Answer

  z0 pole of order m  ⟺  f=ϕ(z)(z−z0)m, ϕ analytic, ϕ(z0)≠0;Res⁡z=z0f=ϕ(m−1)(z0)(m−1)!.  \boxed{\;z_0\text{ pole of order }m\iff f=\frac{\phi(z)}{(z-z_0)^m},\ \phi\text{ analytic},\ \phi(z_0)\ne0;\quad \operatorname*{Res}_{z=z_0}f=\frac{\phi^{(m-1)}(z_0)}{(m-1)!}.\;}
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