Answer:
We can simplify this expression using Euler's formula, which states that for any real number x,
e^(ix) = cos(x) + i sin(x)
Let's start by writing (-1 + i√3) as a complex number in polar form:
-1 + i√3 = 2e^(i(2π/3))
Similarly, we can write (-1 - i√3) as:
-1 - i√3 = 2e^(-i(2π/3))
Now we can raise each of these complex numbers to the fourth and fifth powers, respectively:
(-1 + i√3)⁴ = (2e^(i(2π/3)))⁴ = 16e^(i(8π/3)) = 16e^(i(2π/3))
(-1 - i√3)⁵ = (2e^(-i(2π/3)))⁵ = 32e^(-i(10π/3)) = 32e^(i(2π/3))
Multiplying these two expressions together, we get:
(-1 + i√3)⁴(−1 - i√3)⁵ = 16e^(i(2π/3)) * 32e^(i(2π/3)) = 512e^(i(4π/3))
Now, we can use Euler's formula again to convert this expression back to rectangular form:
512e^(i(4π/3)) = 512(cos(4π/3) + i sin(4π/3)) = 512(-1/2 + i(-√3/2)) = 512ω²
where ω = e^(iπ/3) is a primitive cube root of unity. Therefore, we have shown that:
(-1 + i√3)⁴(−1 - i√3)⁵ = 512ω²
as desired.
Explanation: