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 Question from wael, a student: how do we solve in C the following equation: z^3 + (5i-6)z^2 + (9-24i)z + 13i + 18 = 0 if it admits a pure imaginary root?

Wael,

I can help with the first step. If this equation has a purely imaginary root then this root is of the form ki where k is a real number, thus substitution of z = ki into the left side of the equation should yield a value of zero.

(ki)3 + (5i - 6)(ki)2 + (9 - 24i)(ki) + 13i + 18

simplifies to

(-k3 - 5k2 + 9k + 13)i + 6(k2 + 4k + 3)

If this is zero then both the real and imaginary parts must be zero. Setting the real part to zero yields k = -1 and -3. Substitution of these values into the imaginary part shows that k must be -1. Thus z = -i is a root of the original cubic equation and thus there is a complex polynomial p(z) so that

z3 + (5i - 6)z2 + (9 - 24i)z + 13i + 18 = (z + i) p(z)

Divide z + i into z3 + (5i - 6)z2 + (9 - 24i)z + 13i + 18 to find p(z). Solve p(z) = 0.

Harley

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.