Solution
Core Logic
Let's expand the terms by grouping real and imaginary parts explicitly:
ω₁ = (8 θ + 7 θ) + i( θ + 4 θ) ω₂ = ( θ + 4 θ) + i(8 θ + 7 θ)Notice that if we let u = 8 θ + 7 θ and v = θ + 4 θ, then:
ω₁ = u + iv and ω₂ = v + iuStep 1: Calculating the Product
Multiplying ω₁ and \omega_2:
ω₁ω₂ = (u + iv)(v + iu) = uv + iu² + iv² - uv = i(u² + v²)Since the product is given as α + iβ:
α = 0
β = u² + v² = (8 θ + 7 θ)² + ( θ + 4 θ)²Step 2: Simplifying the expression for alpha + beta
Expanding the terms for β:
β = (64 ²θ + 49 ²θ + 112 θ θ) + ( ²θ + 16 ²θ + 8 θ θ) α + β = 0 + β = 65 ²θ + 65 ²θ + 120 θ θUsing the identity ²θ + ²θ = 1 and 2 θ θ = 2θ:
α + β = 65 + 60 2θStep 3: Max and Min Extrema Analysis
Since -1 ≤ 2θ ≤ 1:
Maximum value p = 65 + 60(1) = 125 Minimum value q = 65 + 60(-1) = 5Sum of maximum and minimum bounds equals:
p + q = 125 + 5 = 130Pattern Recognition
Observe the symmetric structure in complex variables: (u+iv) and (v+iu). Their product structurally completely cancels out the real component, saving you from a highly messy component expansion.
Chapter Mix
Class 11 Mathematics: Complex Numbers Class 11 Mathematics: Trigonometric Functions