



 
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.
simplifies to
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
Divide z + i into z^{3} + (5i  6)z^{2} + (9  24i)z + 13i + 18 to find p(z). Solve p(z) = 0. Harley  


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