(a) Prove that x² + 1 is an irreducible polynomial in Z₃[x]. Further show that the quotient ring (Z₃[x])/(⟨ x²+1 ⟩) is a field of…

(a) Irreducibility of x² + 1 and the field of 9 elements

A polynomial of degree 2 or 3 over a field is irreducible if and only if it has no root in the field. In Z₃, test roots:

• x = 0: 0² + 1 = 1 ≠ 0.
• x = 1: 1² + 1 = 2 ≠ 0.
• x = 2: 2² + 1 = 4 + 1 = 5 ≡ 2 ≠ 0.

Thus x² + 1 has no root in Z₃, so it is irreducible in Z₃[x].

Since Z₃[x] is a Euclidean domain, an irreducible element generates a maximal ideal. Hence ⟨x² + 1⟩ is maximal, and therefore Z₃[x]/⟨x² + 1⟩ is a field.

Every element of this quotient ring is of the form a + bx with a, b ∈ Z₃, because every polynomial can be reduced modulo x² + 1. There are 3 choices for a and 3 choices for b, hence 3² = 9 elements. So the quotient ring is a field of 9 elements.

Final: x² + 1 is irreducible in Z₃[x], and Z₃[x]/⟨x² + 1⟩ is a field of 9 elements.

(b) Harmonic function, conjugate and analytic function

For u(x, y) = eˣ(x cos y − y sin y), compute: ∂u/∂x = eˣ[(x + 1)cos y − y sin y], ∂²u/∂x² = eˣ[(x + 2)cos y − y sin y].

Also, ∂u/∂y = eˣ[−x sin y − sin y − y cos y] = eˣ[−(x + 1)sin y − y cos y], ∂²u/∂y² = eˣ[−(x + 2)cos y + y sin y].

Hence ∂²u/∂x² + ∂²u/∂y² = eˣ[(x + 2)cos y − y sin y − (x + 2)cos y + y sin y] = 0. So u is harmonic.

Using the Cauchy–Riemann equations: ∂u/∂x = ∂v/∂y, ∂u/∂y = −∂v/∂x. Thus ∂v/∂y = eˣ[(x + 1)cos y − y sin y]. Integrating with respect to y: v = eˣ[(x + 1)sin y + y cos y − sin y] + φ(x) = eˣ[x sin y + y cos y] + φ(x).

Then ∂v/∂x = eˣ[(x + 1)sin y + y cos y] + φ′(x). Since ∂u/∂y = eˣ[−(x + 1)sin y − y cos y], we get φ′(x) = 0, so φ(x) is constant. Taking it as 0, v(x, y) = eˣ(x sin y + y cos y).

Then f(z) = u + iv = eˣ[x cos y − y sin y + i(x sin y + y cos y)] = eˣ(x + iy)(cos y + i sin y) = z eᶻ.

Final: v(x, y) = eˣ(x sin y + y cos y), and f(z) = z eᶻ + iC.

(c) Big M method and uniqueness of the optimum

By the Big M method, introduce surplus variables S₁, S₂, slack S₃, and artificial variables A₁, A₂. Minimize: Z = 2x₁ + 3x₂ + M A₁ + M A₂, subject to x₁ + x₂ − S₁ + A₁ = 9, x₁ + 2x₂ − S₂ + A₂ = 15, 2x₁ − 3x₂ + S₃ = 9, all variables ≥ 0.

Initial basis: A₁ = 9, A₂ = 15, S₃ = 9. Since 3 − 3M < 2 − 2M for large M, x₂ enters. The minimum ratio is 15/2 = 7.5, so A₂ leaves.

After pivot: 0.5x₁ − S₁ + 0.5S₂ + A₁ − 0.5A₂ = 1.5 0.5x₁ + x₂ − 0.5S₂ + 0.5A₂ = 7.5 3.5x₁ − 1.5S₂ + S₃ + 1.5A₂ = 31.5

Now x₁ enters because its reduced cost 0.5 − 0.5M is more negative than that of S₂, namely 1.5 − 0.5M. The minimum ratio is 1.5/0.5 = 3, so A₁ leaves.

Final simplex rows: x₁ − 2S₁ + S₂ + 2A₁ − A₂ = 3 x₂ + S₁ − S₂ − A₁ + A₂ = 6 7S₁ − 5S₂ + S₃ − 7A₁ + 5A₂ = 21

At optimum, artificial variables are zero, so A₁ = A₂ = 0 and S₁ = S₂ = 0. Hence x₁ = 3, x₂ = 6, S₃ = 21.

Check constraints: 3 + 6 = 9, 3 + 12 = 15, 2(3) − 3(6) = −12 ≤ 9. Thus feasible.

Objective: Z = 2(3) + 3(6) = 6 + 18 = 24.

In the final reduced-cost row, the nonbasic variables S₁ and S₂ have reduced costs 1 and 1, while A₁ and A₂ have reduced costs M − 1 > 0. No nonbasic variable has zero reduced cost, so no alternative optimum exists.

Final: x₁ = 3, x₂ = 6, minimum Z = 24. The optimal solution is unique.