# Roots of polynomial with imaginary coefficients

This is the first time I see this kind of problem, so it might be trivial but I am just not used to it.

What are the roots of $x^3-6ix^2-11x+6i$

I am not sure If I should ignore the imaginary numbers and simply compute the polynomial or factor the imaginary part out separately.

I tried to use the rational polynomial root test but it has no rational roots when I ignore the Imaginary coefficients.

When I factor them out as $x^3-11x-i(6x^2-6)$ I get $i$ and $-i$ as a root which is definitely wrong.

All I ask for here is to provide me advice on what method should I use to solve this type of problems. Thanks in advance.

• @abiessu yes I sorry, I corrected it – Fred Dec 19 '17 at 14:50
• Hint: search for pure complex roots. – lulu Dec 19 '17 at 14:53
• Hint: define $u$ such that $x=iu$ – Martigan Dec 19 '17 at 14:54
• If you plug in $x=iy$, you get $-iy^3+6iy^2-11iy+6i$, which should have at least one real solution in $y$... This approach is not available in general, but is reasonable here where the even and odd degree terms all match in terms of coefficients being complex (or not). – abiessu Dec 19 '17 at 14:54
• @abiessu I was going to put exactly the same - the alternating real and imaginary coefficients indicate this approach. – Mark Bennet Dec 19 '17 at 14:58

$x=iy$
Now if $f(y)=y^3-6y^2+11y-6$
$f(1)=?$