Noesis
Level 2 — RecallAdvanced Topics (Elite Level)

Advanced Topics (Elite Level)

⏱ 30 minutes40 marksprintable — key stays hidden on paper

Level: 2 — Recall (definitions, standard problems, short derivations) Time Limit: 30 minutes Total Marks: 40

Q1. State the Cauchy–Riemann equations for f(z)=u(x,y)+iv(x,y)f(z)=u(x,y)+iv(x,y), and verify that f(z)=z2f(z)=z^2 is analytic. (4 marks)

Q2. Evaluate the contour integral ∮∣z∣=1dzz\displaystyle \oint_{|z|=1} \frac{dz}{z} where the contour is traversed once counterclockwise. (4 marks)

Q3. Using Cauchy's integral formula, evaluate ∮∣z∣=2ezz−1 dz\displaystyle \oint_{|z|=2} \frac{e^{z}}{z-1}\,dz. (4 marks)

Q4. Find the residue of f(z)=1z2+1\displaystyle f(z)=\frac{1}{z^2+1} at the pole z=iz=i. (4 marks)

Q5. Using the residue theorem, evaluate the real integral ∫−∞∞dxx2+1.\int_{-\infty}^{\infty}\frac{dx}{x^2+1}. (5 marks)

Q6. State the Einstein summation convention. Given the metric tensor gijg_{ij} and a contravariant vector vjv^j, write the expression for the covariant component viv_i. (4 marks)

Q7. State the Euler–Lagrange equation for a functional J[y]=∫abF(x,y,y′) dxJ[y]=\int_a^b F(x,y,y')\,dx. Apply it to F=(y′)2F=(y')^2 and find the general solution y(x)y(x). (5 marks)

Q8. Define a convex function on an interval. Determine, with justification, whether f(x)=x2f(x)=x^2 is convex. (4 marks)

Q9. A Markov chain has transition matrix P=(0.50.50.20.8)P=\begin{pmatrix} 0.5 & 0.5 \\ 0.2 & 0.8 \end{pmatrix}. Find the steady-state distribution π=(π1,π2)\pi=(\pi_1,\pi_2). (6 marks)

End of paper.

Answer keyMark scheme & solutions

Q1. (4 marks) Cauchy–Riemann equations: ux=vy,uy=−vx.(2 marks)u_x = v_y, \qquad u_y = -v_x. \quad \text{(2 marks)} For f(z)=z2=(x+iy)2=(x2−y2)+i(2xy)f(z)=z^2=(x+iy)^2 = (x^2-y^2)+i(2xy), so u=x2−y2u=x^2-y^2, v=2xyv=2xy.

  • ux=2xu_x = 2x, vy=2xv_y=2x → equal. (1 mark)
  • uy=−2yu_y=-2y, vx=2yv_x=2y, so uy=−vxu_y=-v_x. (1 mark) CR equations satisfied everywhere → ff analytic.

Q2. (4 marks) By Cauchy's integral formula / standard result, ∮∣z∣=1dzz=2πi\oint_{|z|=1}\frac{dz}{z} = 2\pi i.

  • Parametrize z=eiθz=e^{i\theta}, dz=ieiθdθdz=ie^{i\theta}d\theta (1 mark)
  • Integral =∫02πieiθeiθdθ=∫02πi dθ=2πi=\int_0^{2\pi} \frac{ie^{i\theta}}{e^{i\theta}}d\theta = \int_0^{2\pi} i\,d\theta = 2\pi i. (3 marks)

Q3. (4 marks) Cauchy's integral formula: ∮Cf(z)z−adz=2πi f(a)\oint_C \frac{f(z)}{z-a}dz = 2\pi i\, f(a) for aa inside CC. (1 mark) Here f(z)=ezf(z)=e^z, a=1a=1, which lies inside ∣z∣=2|z|=2. (1 mark) ∮=2πi e1=2πie.(2 marks)\oint = 2\pi i\, e^{1} = 2\pi i e. \quad \text{(2 marks)}

Q4. (4 marks) z2+1=(z−i)(z+i)z^2+1=(z-i)(z+i); simple pole at z=iz=i. (1 mark) Resz=if=lim⁡z→i(z−i)1(z−i)(z+i)=12i.(3 marks)\text{Res}_{z=i}f = \lim_{z\to i}(z-i)\frac{1}{(z-i)(z+i)} = \frac{1}{2i}. \quad \text{(3 marks)} (Equivalently −i2-\tfrac{i}{2}.)

Q5. (5 marks) Close contour in upper half-plane; only pole enclosed is z=iz=i. (1 mark) Residue at z=iz=i is 12i\frac{1}{2i} (from Q4). (1 mark) By residue theorem, integral =2πi⋅12i=π=2\pi i \cdot \frac{1}{2i} = \pi. (2 marks) Semicircle contribution → 0 as R→∞R\to\infty. (1 mark) ∫−∞∞dxx2+1=π.\int_{-\infty}^{\infty}\frac{dx}{x^2+1}=\pi.

Q6. (4 marks) Einstein summation convention: repeated indices (one upper, one lower) are summed over their range; the summation symbol is omitted. (2 marks) Lowering an index with the metric: vi=gijvj(summed over j).(2 marks)v_i = g_{ij}v^j \quad(\text{summed over }j). \quad \text{(2 marks)}

Q7. (5 marks) Euler–Lagrange equation: ∂F∂y−ddx(∂F∂y′)=0.(2 marks)\frac{\partial F}{\partial y} - \frac{d}{dx}\left(\frac{\partial F}{\partial y'}\right)=0. \quad \text{(2 marks)} For F=(y′)2F=(y')^2: ∂F∂y=0\frac{\partial F}{\partial y}=0, ∂F∂y′=2y′\frac{\partial F}{\partial y'}=2y'. (1 mark) So −ddx(2y′)=0⇒y′′=0-\frac{d}{dx}(2y')=0 \Rightarrow y''=0. (1 mark) General solution: y(x)=Ax+By(x)=Ax+B (straight line). (1 mark)

Q8. (4 marks) Definition: ff is convex on II if for all x1,x2∈Ix_1,x_2\in I and λ∈[0,1]\lambda\in[0,1]: f(λx1+(1−λ)x2)≤λf(x1)+(1−λ)f(x2).(2 marks)f(\lambda x_1+(1-\lambda)x_2)\le \lambda f(x_1)+(1-\lambda)f(x_2). \quad \text{(2 marks)} For f(x)=x2f(x)=x^2: f′′(x)=2>0f''(x)=2>0 for all xx, so ff is convex. (2 marks)

Q9. (6 marks) Steady state: πP=π\pi P=\pi, π1+π2=1\pi_1+\pi_2=1. (1 mark) Equations: 0.5π1+0.2π2=π1⇒−0.5π1+0.2π2=0⇒π2=2.5π10.5\pi_1+0.2\pi_2=\pi_1 \Rightarrow -0.5\pi_1+0.2\pi_2=0 \Rightarrow \pi_2=2.5\pi_1. (2 marks) Normalize: π1+2.5π1=1⇒π1=1/3.5=2/7\pi_1+2.5\pi_1=1 \Rightarrow \pi_1=1/3.5=2/7. (2 marks) π=(27,57).(1 mark)\pi=\left(\tfrac{2}{7},\tfrac{5}{7}\right). \quad \text{(1 mark)}

[
  {"claim":"Integral of dz/z over unit circle = 2*pi*i", "code":"theta=symbols('theta'); z=exp(I*theta); val=integrate(diff(z,theta)/z,(theta,0,2*pi)); result = simplify(val-2*pi*I)==0"},
  {"claim":"Residue of 1/(z^2+1) at z=i is 1/(2*I)", "code":"z=symbols('z'); res=limit((z-I)*1/(z**2+1),z,I); result = simplify(res-1/(2*I))==0"},
  {"claim":"Integral of 1/(x^2+1) over reals = pi", "code":"x=symbols('x'); val=integrate(1/(x**2+1),(x,-oo,oo)); result = simplify(val-pi)==0"},
  {"claim":"Steady state of Markov chain is (2/7,5/7)", "code":"p1,p2=symbols('p1 p2'); sol=solve([0.5*p1+0.2*p2-p1, p1+p2-1],[p1,p2]); result = abs(sol[p1]-Rational(2,7))<1e-9 and abs(sol[p2]-Rational(5,7))<1e-9"}
]