If $8iz^3 +12z^2 -18z +27i =0$ find the value of $4|z|^2$

Question

Problem :

If $8iz^3 +12z^2 -18z +27i =0$ find the value of $4|z|^2$ where $z$ is a complex number.

Working :

Now let $z = x+iy$ then $8i (x+iy)^3 +12(x+iy)^2 -18(x+iy) +27i =0$

$\Rightarrow 8i\{x^3 +(iy)^3 +3xiy (x+iy)\} +12(x^2-y^2+2xiy) -18(x+iy) +27i=0$

$\Rightarrow 8i\{ x^3 -iy^3 +3x^2iy -3xy^2\} +12(x^2-y^2+2xiy)-18(x+iy) +27i=0$

Please suggest whether this is the correct way of approaching such problems or if there is any other method please suggest thanks..

Answer 1

To start,

$$ 8iz^3 + 12z^2 - 18z + 27i = 4z^2(2iz+3) + 9i(2iz+3). $$

Answer 2

The product of the complex roots of $az^3 + bz^2 + cz + d$ is given by $-d/a$. In your case, the product of the roots is $z_1z_2z_3 = (27i)/(8i) = 27/8$. Now, assuming that each of the roots has the same modulus $|z_1| = |z_2| = |z_3| = |z|$, we get $|z|^3 = 27/8$, hence $|z| = 3/2$ and so $\boxed{4|z|^2 = 9}$.

The reason I assumed that $|z_1| = |z_2| = |z_3|$ was that the question asked for the value of $4|z|^2$, which implies that this value doesn't depend on which root you choose. It is not true in general that the modulus of all the roots of a cubic equation are equal.