- #1
Yura
- 39
- 0
im not sure i can fully remember the rules for complex numbers but i have to solve an equation that has 3 solutions to equal zero. so far i have (x=2), (x=i) and i think there was a + rule for a solution with complex numbers that there would always be a conjugate solution of it. so i figured that the last solution would be (x=-i) but when i try and solve it i keep getting different answers and they are all non zero.
i think i may be doing something wrong with the powers of the (-i) or it might be that i thought wrong and (x=-i) is not a solution.
could someone please confirm ths for me?
here is the equation:
x^3 - (2-i)*x^2 + (2-2*i)*x - 4 = 0
im trying to solve for:
(-i)^3 - (2-i)*(-i)^2 + (2-2*i)*(-i) - 4 = 0
thanks in advance ^^
i think i may be doing something wrong with the powers of the (-i) or it might be that i thought wrong and (x=-i) is not a solution.
could someone please confirm ths for me?
here is the equation:
x^3 - (2-i)*x^2 + (2-2*i)*x - 4 = 0
im trying to solve for:
(-i)^3 - (2-i)*(-i)^2 + (2-2*i)*(-i) - 4 = 0
thanks in advance ^^