Advanced Topics (Elite Level)

Question 1 (10 marks)

(a) Poles where denominators vanish: $z^2+1=0\Rightarrow z=\pm i$; $z^2+4=0\Rightarrow z=\pm 2i$. In the upper half-plane: $z=i$ and $z=2i$, both simple poles (denominator factors are distinct linear factors, numerator nonzero there). (3)

(b) For a simple pole at $z_0$: $\operatorname{Res} = \dfrac{z_0^2}{\text{derivative of denom / factored limit}}$.

At $z=i$: $\operatorname{Res}_{z=i} = \frac{z^2}{(z+i)(z^2+4)}\Big|_{z=i} = \frac{-1}{(2i)(3)} = \frac{-1}{6i} = \frac{i}{6}.$

At $z=2i$: $\operatorname{Res}_{z=2i} = \frac{z^2}{(z^2+1)(z+2i)}\Big|_{z=2i} = \frac{-4}{(-3)(4i)} = \frac{-4}{-12i} = \frac{1}{3i} = -\frac{i}{3}.$ (4) (2 each)

(c) On the arc $|z|=R$, $|f|\sim R^2/R^4 = 1/R^2$, length $\pi R$, so contribution $\sim \pi/R \to 0$. Thus $I = 2\pi i\left(\tfrac{i}{6} - \tfrac{i}{3}\right) = 2\pi i\left(-\tfrac{i}{6}\right) = \frac{2\pi}{6} = \frac{\pi}{3}.$ (3) — arc justification (1), sum of residues (1), final $I=\pi/3$ (1).

Question 2 (10 marks)

(a) Partial fractions: $\dfrac{1}{z(z-2)} = \dfrac{-1/2}{z} + \dfrac{1/2}{z-2}$. For $|z|<2$: $\dfrac{1/2}{z-2} = \dfrac{-1/4}{1-z/2} = -\tfrac14\sum_{n\ge0}(z/2)^n$. So $g(z) = -\frac{1}{2z} - \frac14\sum_{n=0}^{\infty}\frac{z^n}{2^n} = -\frac{1}{2z} - \sum_{n=0}^\infty \frac{z^n}{2^{n+2}}.$ (4)

(b) For $|z|>2$: $\dfrac{1/2}{z-2} = \dfrac{1}{2z}\cdot\dfrac{1}{1-2/z} = \dfrac{1}{2z}\sum_{n\ge0}(2/z)^n$. So $g(z) = -\frac{1}{2z} + \sum_{n=0}^\infty \frac{2^n}{2\,z^{n+1}} = -\frac{1}{2z} + \frac{1}{2z}\sum_{n=0}^\infty\frac{2^n}{z^n}.$ Combining, the $1/z$ terms give $-\tfrac12+\tfrac12 = 0$, so $g(z)=\sum_{n=1}^{\infty}\dfrac{2^{n-1}}{z^{n+1}} = \dfrac{1}{z^2}+\dfrac{2}{z^3}+\cdots$. (4)

(c) Residue at $z=0$ is the coefficient of $1/z$ in (a): $-\tfrac12$. The annulus matters because the residue is read from the expansion valid on a punctured disc around the pole ($0<|z|<2$); the $|z|>2$ expansion has no $1/z$ term and cannot be used to read $\operatorname{Res}_{z=0}$. (2)

Question 3 (10 marks)

(a) Energy conservation from rest: $\tfrac12 v^2 = g y \Rightarrow v=\sqrt{2gy}$. Time $=\int ds/v$ with $ds=\sqrt{1+y'^2}\,dx$, giving the stated $T[y]$. (2)

(b) With $F=\sqrt{\dfrac{1+y'^2}{2gy}}$, drop constant $1/\sqrt{2g}$; use $F-y'F_{y'}=C$. $F_{y'} = \dfrac{y'}{\sqrt{y}\sqrt{1+y'^2}}$ (up to constant). Then $F - y'F_{y'} = \frac{\sqrt{1+y'^2}}{\sqrt y} - \frac{y'^2}{\sqrt y\sqrt{1+y'^2}} = \frac{(1+y'^2)-y'^2}{\sqrt y\sqrt{1+y'^2}} = \frac{1}{\sqrt y\sqrt{1+y'^2}} = C.$ Square: $y(1+y'^2)=1/C^2 \equiv k$. (5)

(c) $\dfrac{dy}{d\theta}=a\sin\theta$, $\dfrac{dx}{d\theta}=a(1-\cos\theta)$, so $y'=\dfrac{\sin\theta}{1-\cos\theta}$. $1+y'^2 = 1+\frac{\sin^2\theta}{(1-\cos\theta)^2} = \frac{(1-\cos\theta)^2+\sin^2\theta}{(1-\cos\theta)^2} = \frac{2-2\cos\theta}{(1-\cos\theta)^2} = \frac{2}{1-\cos\theta}.$ Then $y(1+y'^2)= a(1-\cos\theta)\cdot\dfrac{2}{1-\cos\theta} = 2a =$ const. ✓ (3)

Question 4 (10 marks)

(a) Lagrangian $L = x^2+y^2 + \mu(2-x-y)$. KKT:

• Stationarity: $2x-\mu=0$, $2y-\mu=0$.
• Primal feasibility: $2-x-y\le0$.
• Dual feasibility: $\mu\ge0$.
• Complementary slackness: $\mu(2-x-y)=0$. (4)

(b) From stationarity $x=y=\mu/2$. If $\mu=0$: $x=y=0$, but then $2-0=2>0$ violates feasibility. So $\mu>0$, forcing $2-x-y=0\Rightarrow x+y=2\Rightarrow x=y=1$, $\mu=2$. Minimiser $(1,1)$, minimum value $f=1+1=2$. (4)

(c) Constraint is active ($x+y=2$). Justification: complementary slackness with $\mu=2>0$ requires $2-x-y=0$; the unconstrained minimum $(0,0)$ is infeasible, so the optimum lies on the boundary. (2)

Question 5 (10 marks)

(a) From 1 you can reach 2, from 2 reach 1,2,3, from 3 reach 2 — all states communicate, so irreducible. State 2 has a self-loop ($P_{22}>0$) giving period 1; irreducibility spreads aperiodicity, so the chain is aperiodic. (3)

(b) $\pi P=\pi$:

• $\pi_1 = \tfrac14\pi_2$
• $\pi_2 = \pi_1 + \tfrac12\pi_2 + \pi_3$
• $\pi_3 = \tfrac14\pi_2$

From eq 1 and 3: $\pi_1=\pi_3=\tfrac14\pi_2$. Check eq 2: $\pi_2 = \tfrac14\pi_2 + \tfrac12\pi_2 + \tfrac14\pi_2 = \pi_2$ ✓. Normalise: $\pi_1+\pi_2+\pi_3 = \tfrac14\pi_2+\pi_2+\tfrac14\pi_2 = \tfrac32\pi_2 =1 \Rightarrow \pi_2=\tfrac23$, $\pi_1=\pi_3=\tfrac16$. $\pi = \left(\tfrac16,\ \tfrac23,\ \tfrac16\right).$ (5)

(c) By the 1↔3 symmetry, $\pi_1=\pi_3=\tfrac16$, consistent with the solved values; sum $=\tfrac16+\tfrac23+\tfrac16=1$. (2)

[
  {"claim":"Q1 integral equals pi/3","code":"x=symbols('x',real=True); I=integrate(x**2/((x**2+1)*(x**2+4)),(x,-oo,oo)); result = simplify(I - pi/3)==0"},
  {"claim":"Q1 residues sum: 2*pi*i*(i/6 - i/3) = pi/3","code":"result = simplify(2*pi*I*(I/Rational(1,1)/6 - I/3) - pi/3)==0"},
  {"claim":"Q4 minimum value is 2 at (1,1)","code":"result = (1**2+1**2)==2"},
  {"claim":"Q5 stationary distribution pi P = pi","code":"P=Matrix([[0,1,0],[Rational(1,4),Rational(1,2),Rational(1,4)],[0,1,0]]); pv=Matrix([[Rational(1,6),Rational(2,3),Rational(1,6)]]); result = (pv*P == pv) and (sum(pv)==1)"}
]