Complex Numbers Question - None

Question

Let z = x + iy. If z satisfies $(2iy)^2 = 12(x^2 + y^2) - 4$, find the maximum value of $3\sqrt{3}\,\text{Re}(z) + 8\,\text{Im}(z)$.

Accepted Answer

Step 1: Expand the left side. Since (2iy)² = 4i²y² = -4y², the equation becomes:

-4y² = 12(x² + y²) - 4

Step 2: Simplify the constraint equation:

-4y² = 12x² + 12y² - 4

0 = 12x² + 16y² - 4

12x² + 16y² = 4

3x² + 4y² = 1

Step 3: This is an ellipse. We maximize f(x,y) = 3√3·x + 8·y subject to 3x² + 4y² = 1 using Lagrange multipliers:

∇f = λ∇g where g(x,y) = 3x² + 4y² - 1

3√3 = λ·6x and 8 = λ·8y

Step 4: From the second equation: λ = 1/y. From the first: λ = √3/(2x).

Therefore: 1/y = √3/(2x) ⟹ 2x = √3·y ⟹ x = (√3/2)y

Step 5: Substitute into the ellipse constraint:

3(√3y/2)² + 4y² = 1

3·(3y²/4) + 4y² = 1

(9y²/4) + 4y² = 1

(9y² + 16y²)/4 = 1

25y² = 4 ⟹ y = 2/5

Step 6: Then x = (√3/2)·(2/5) = √3/5

Step 7: Calculate the maximum value:

3√3·(√3/5) + 8·(2/5) = 9/5 + 16/5 = 25/5 = 5

∴ Answer: 5

Question: