Finding a cubic equation from it's roots

For the cubic equation z^3+az^2+3z+9 where one zero is purely imaginary.

I factorized into (z-a)(z-bi)(z+bi)

And on to expand and equate coefficients to get a=3, and b = +/-sqrt(3)

Now to write back the cubic equation as a product of linear factors, how do I write b?

The answer is

(z-a)**(z+sqrt(3)i)**(z-sqrt(3)i)

But why?

Isn't it

(z-a)**(z+sqrt(3))**(z-sqrt(3)i) since one of the roots is a positive sqrt(3)?

Thank you so much!

Re: Finding a cubic equation from it's roots

Quote:

Originally Posted by

**Tutu** For the cubic equation z^3+az^2+3z+9 where one zero is purely imaginary.

I factorized into (z-a)(z-bi)(z+bi)

And on to expand and equate coefficients to get a=3, and b = +/-sqrt(3)

Now to write back the cubic equation as a product of linear factors, how do I write b?

The answer is

(z-a)**(z+sqrt(3)i)**(z-sqrt(3)i)

But why?

Isn't it

(z-a)**(z+sqrt(3))**(z-sqrt(3)i) since one of the roots is a positive sqrt(3)?

Thank you so much!

Nonreal roots always appear as complex conjugates. That means if $\displaystyle \displaystyle \begin{align*} z - \sqrt{3}\,i \end{align*}$ is a factor, so is $\displaystyle \displaystyle \begin{align*} z + \sqrt{3}\,i \end{align*}$.