If z=2+2i is the solution to a quartic equation of the form z^n+w=0 z,w ∈ C. Determine the quartic equation.
Let's denote the quartic equation as z^4 + az^3 + bz^2 + cz + d = 0, where a, b, c, and d are complex constants.
We are given that z = 2 + 2i is a solution to the quartic equation. So, we have:
(2 + 2i)^4 + a(2 + 2i)^3 + b(2 + 2i)^2 + c(2 + 2i) + d = 0
Expanding the terms, we get:
16 + 96i + 192i^2 + 128i^3 + 16a + 48ai + 48ai^2 + 8a^2 + 12a^2i + 6a^2i^2 + ai^3 + 2ai^3 + bi^2 + 2bi + ci + d = 0
Note that i^2 = -1 and i^3 = -i. Substituting these values, we simplify the equation as follows:
16 + 96i - 192 - 128i + 16a - 48a - 48a - 8a^2 - 12a^2 - 6a^2 - a - 2a + b - 2b + ci + d = 0
Combining like terms and simplifying further, we get:
-192 + (16a - 48a - 48a - 8a^2 - 12a^2 - 6a^2 - a - 2a) + (96i - 128i + b - 2b + c)i + d = 0
-192 - 60a - 27a^2 + (96 - 128 + b - 2b + c)i + d = 0
To satisfy the equation, we must have the individual real and imaginary parts equal to zero. Therefore, we have:
-192 - 60a - 27a^2 = 0 (Equation for real parts)
96 - 128 + b - 2b + c = 0 (Equation for imaginary parts)
d = 0 (Since there is no constant term in z)
Simplifying the real part equation, we have:
192 + 60a + 27a^2 = 0
Dividing through by 3, we get:
64 + 20a + 9a^2 = 0
Factoring the quadratic equation, we have:
(3a + 4)(3a + 5) = 0
So, we have two possible values for a:
a = -4/3 or a = -5/3
Substituting these values back into the imaginary part equation and simplifying, we have:
b - c = -64
Therefore, we have two possible solutions for the quartic equation:
z^4 - (4/3)z^3 - (5/3)z^2 + 64z - 0 = 0
or
z^4 - (5/3)z^3 - (4/3)z^2 + 64z - 0 = 0