It looks like we're given
z₁ = 1 - (2 - √3) i
z₂ = -2 - 2i
and we want to find
(i z₁ z₂)²
z₁ lies in the fourth quadrant of the complex plane, so
|z₁| = √(1² + (-2 + √3)²) = 2√(2 - √3)
arg(z₁) = arctan(-2 + √3)
while z₂ lies in the third quadrant, so
|z₂| = √((-2)² + (-2)²) = 2√2
arg(z₂) = arctan(-2 / -2) - π = arctan(1) - π = -3π/4
so their polar forms are
z₁ = 2√(2 - √3) exp(i arctan(-2 + √3))
z₂ = 2√2 exp(-i 3π/4)
Before we continue, we can actually simplify the argument to z₁. Let θ = arg(z₁). From its polar form, it's evident that
2√(2 - √3) cos(θ) = 1
which reduces to
cos(θ) = 1 / (2√(2 - √3))
Recall the half-angle identity,
cos²(x) = (1 + cos(2x))/2
By taking squares on both sides, we have
cos²(θ) = 1 / (8 - 4√3)
(1 + cos(2θ))/2 = 1 / (8 - 4√3)
1 + cos(2θ) = 1 / (4 - 2√3)
cos(2θ) = (-3 + 2√3) / (4 - 2√3)
cos(2θ) = √3/2
2θ = arccos(√3/2) + 2nπ or 2θ = -arccos(√3/2) + 2nπ
(where n is any integer)
2θ = π/6 + 2nπ or 2θ = -π/6 + 2nπ
θ = π/12 + nπ or θ = -π/12 + nπ
We know that z₁ lies in the fourth quadrant, so θ = -π/12.
All this to say
z₁ = 2√(2 - √3) exp(-i π/12)
z₂ = 2√2 exp(-i 3π/4)
Since i = exp(i π/2), we then have
i z₁ z₂ = 4 √(4 - 2√3) exp(i (π/2 - π/12 - 3π/4))
… = 4 √(4 - 2√3) exp(-i π/3)
and squaring both sides gives
(i z₁ z₂)² = 16 (4 - 2√3) exp(-i 2π/3)
