Complex Numbers and Quadratic Equations

Assuming the equation is (x + iy)² = I (x² + y²)
x² - y² + 2ixy = I (x² + y²)
Compare real and imaginary parts
x² - y² = 0 ⇒ x = y or x = -y
2xy = x² + y²
If x=y, then 2x² = x² + x², which is true for all x.
If x=-y, then -2y² = y² + y² = 2y², which implies 4y²=0, so y=0 and x=0.
The non-trivial solution is x = y.

Given equation is 2x (2x + 1) = 1 ⇒ 4x² + 2x - 1 = 0. Roots of the equation are α and β.
∴ α + β = -2/4 = -1/2 ⇒ β = -1/2 - α
and
4α² + 2α - 1 = 0 ⇒ α² = (1-2α)/4 = 1/4 - α/2
Now
α = 1/2 - 2α²
Substituting into the expression for β:
β = -1/2 - (1/2 - 2α²) = -1 + 2α²

Given: 3α² – 10α + 27λ = 0
3α² – 3α + 6λ = 0
Subtract -7α + 21λ = 0
3λ = 0
By (ii) 9λ² – 3λ + 2λ = 0
⇒ λ = 0, 1/9
∴ α = 1/3, β = 2/3, α = 1/3, γ = 3
∴ (βγ)/λ = (2/3) * 3) / (1/9) = 18

= [√3e^ (iπ/3)]^4
= 9 (cos (2π/3) + isin (2π/3)
= -9/2 + 9√3i/2
⇒ 0 + 9 (-1 + i√3)/2)
∴ a = 0, b = 9

u = (2z+i)/ (z-ki)
= (2x² + (2y+1) (y-k)/ (x²+ (y-k)²) + I (x (2y+1) - 2x (y-k)/ (x²+ (y-k)²)
Since Re (u) + Im (u) = 1
⇒ 2x² + (2y+1) (y-k) + x (2y+1) - 2x (y-k) = x² + (y-k)²
P (0, y? )
Q (0, y? )
⇒ y² + y - k - k² = 0 {y? + y? = -1, y? = -k-k²}
∴ PQ = 5
⇒ |y? - y? | = 5 ⇒ k² + k - 6 = 0
⇒ k = -3, 2
So, k = 2 (k > 0)

x² - 3x + p = 0
α, β, γ, δ in G.P.
α + αr = 3
x² - 6x + q = 0
ar² + ar³ = 6
(2) ÷ (1) ⇒ r² = 2
So, 2q+p/2q-p = (2r? +r)/ (2r? -r) = (2r? +1)/ (2r? -1) = 9/7