(a) Show that the multiplicative group G = {1, -1, i, -i}, where i = √(-1), is isomorphic to the group G′ = ({0, 1, 2, 3}, +₄)…

(a) Isomorphism of G and G′

Define φ: G → G′ by φ(1) = 0, φ(i) = 1, φ(-1) = 2, φ(-i) = 3. Since G = {i⁰, i¹, i², i³} and G′ = {0,1,2,3}, the map is φ(i^k) = k mod 4, k = 0,1,2,3. It is clearly bijective. For any i^m, i^n ∈ G, φ(i^m i^n) = φ(i^(m+n)) = (m+n) mod 4 = (m mod 4) +₄ (n mod 4) = φ(i^m) +₄ φ(i^n). Thus φ is a homomorphism. Since it is bijective, it is an isomorphism. Hence G ≅ G′.

(b) Analytic function from u − v (Milne-Thomson)

Let F(z) = (1+i)f(z). Then F is analytic and Re F = u − v = (cos x + sin x − e⁻ʸ)/(2 cos x − eʸ − e⁻ʸ).

Put w(x,y) = (cos x + sin x − e⁻ʸ)/(2 cos x − eʸ − e⁻ʸ). By Milne-Thomson method, if F = U + iV and U = w, then F′(z) = w_x(z,0) − i w_y(z,0).

At y = 0, w(x,0) = (cos x + sin x − 1)/(2 cos x − 2) = 1/2 − 1/2 cot(x/2). Hence w_x(x,0) = 1/4 csc²(x/2). Also, w_y(x,0) = 1/(2 cos x − 2) = −1/4 csc²(x/2). Therefore F′(z) = 1/4 csc²(z/2) − i(−1/4 csc²(z/2)) = (1+i)/4 csc²(z/2).

Integrating, F(z) = (1+i)/4 · (−2 cot(z/2)) + C = −(1+i)/2 cot(z/2) + C. Since F(z) = (1+i)f(z), f(z) = −1/2 cot(z/2) + C/(1+i). Let K = C/(1+i). Given f(π/2) = 0, 0 = −1/2 cot(π/4) + K = −1/2 + K, so K = 1/2. Thus f(z) = 1/2 − 1/2 cot(z/2) = (1 − cot(z/2))/2.

(c) Convergence of the improper integral

Consider ∫₀^∞ (cos x)/(1+x²) dx. Split at x = 1: ∫₀^∞ = ∫₀¹ + ∫₁^∞. The first integral is finite because the integrand is continuous on [0,1]. For x ≥ 1,

|cos x|/(1+x²) ≤ 1/(1+x²). Now ∫₁^∞ dx/(1+x²) = [arctan x]₁^∞ = π/2 − π/4 = π/4, which is finite. By the comparison test for improper integrals, ∫₁^∞ |cos x|/(1+x²) dx converges. Therefore the original integral converges absolutely, and hence it converges. (Its value is π/(2e), by contour integration.)

(d) Laurent expansions

Given f(z) = 1/((z−1)²(z−3)).

(d)(i) Region 0 < |z−1| < 2

For 0 < |z−1| < 2, put u = z−1. Then z−3 = u−2, so f(z) = 1/(u²(u−2)) = −1/(2u²(1−u/2)). Since |u/2| < 1, 1/(1−u/2) = Σ_(n=0)^∞ (u/2)^n. Hence f(z) = −1/(2u²) Σ_(n=0)^∞ (u/2)^n = −Σ_(n=0)^∞ u^(n−2)/2^(n+1). Thus f(z) = −Σ_(n=0)^∞ (z−1)^(n−2)/2^(n+1), 0 < |z−1| < 2. Equivalently, f(z) = −1/[2(z−1)²] − 1/[4(z−1)] − 1/8 − (z−1)/16 − (z−1)²/32 − ...

(d)(ii) Region 0 < |z−3| < 2

For 0 < |z−3| < 2, put v = z−3. Then z−1 = v+2, so f(z) = 1/[v(v+2)²] = 1/(4v) · (1+v/2)^(−2). Using (1+t)^(−2) = Σ_(n=0)^∞ (−1)^n (n+1)t^n, |t| < 1, with t = v/2, we get f(z) = 1/(4v) Σ_(n=0)^∞ (−1)^n (n+1)(v/2)^n = Σ_(n=0)^∞ (−1)^n (n+1) v^(n−1)/2^(n+2). Therefore f(z) = Σ_(n=0)^∞ (−1)^n (n+1)(z−3)^(n−1)/2^(n+2), 0 < |z−3| < 2. Equivalently, f(z) = 1/[4(z−3)] − 1/4 + 3(z−3)/16 − (z−3)²/8 + 5(z−3)³/64 − ...

(e) Two-phase simplex method

Minimize Z = x₁ + x₂ subject to 2x₁ + x₂ ≥ 4, x₁ + 7x₂ ≥ 7, x₁, x₂ ≥ 0.

Introduce surplus variables S₁, S₂ and artificial variables A₁, A₂: 2x₁ + x₂ − S₁ + A₁ = 4, x₁ + 7x₂ − S₂ + A₂ = 7, all variables ≥ 0.

Phase I: Minimize W = A₁ + A₂. Initial basis: A₁ = 4, A₂ = 7, so W = 11. From the constraints, W = 11 − 3x₁ − 8x₂ + S₁ + S₂. The most negative coefficient is −8 for x₂, so x₂ enters. Ratios: 4/1 = 4 for A₁, 7/7 = 1 for A₂. Minimum is 1, so A₂ leaves.

After pivoting on x₂ in the A₂ row: (1/7)x₁ + x₂ − (1/7)S₂ + (1/7)A₂ = 1, (13/7)x₁ − S₁ + (1/7)S₂ + A₁ − (1/7)A₂ = 3, W = 3 − (13/7)x₁ + S₁ − (1/7)S₂ + (8/7)A₂. Now x₁ enters. Ratios: 3/(13/7) = 21/13 and 1/(1/7) = 7. Minimum is 21/13, so A₁ leaves.

After pivoting: x₁ − (7/13)S₁ + (1/13)S₂ + (7/13)A₁ − (1/13)A₂ = 21/13, x₂ + (1/13)S₁ − (2/13)S₂ − (1/13)A₁ + (2/13)A₂ = 10/13. Now W = A₁ + A₂. Since A₁ and A₂ are nonbasic, W = 0. Hence a feasible solution is x₁ = 21/13, x₂ = 10/13, S₁ = S₂ = 0.

Phase II: Use original objective Z = x₁ + x₂. From the Phase I rows, x₁ = 21/13 + (7/13)S₁ − (1/13)S₂ − (7/13)A₁ + (1/13)A₂, x₂ = 10/13 − (1/13)S₁ + (2/13)S₂ + (1/13)A₁ − (2/13)A₂. Thus Z = 31/13 + (6/13)S₁ + (1/13)S₂ − (6/13)A₁ − (1/13)A₂. Setting artificial variables to zero, the coefficients of S₁ and S₂ are positive. Hence Z is minimized at S₁ = S₂ = 0.

Therefore x₁ = 21/13, x₂ = 10/13, and Z_min = 31/13.