Substituting x=i in the given equation, we have:
LHS = 2i-(-i)= 3i= RHS
Thus, we have x=i as a root of the equation.
Then, dividing (2x-x^3 - 3i) by (x-i) we have:
-x^2-ix-3
Then, considering x^2-ix-3 =0, we have:
x = (i+/-sqrt(-1+12))/2 = (I +/- sqrt(11))/2
Edited on April 28, 2023, 12:08 pm