Complex Polynomial, Input That Also Is A Solution

Question

I'm trying to solve this problem, but I don't understand how the z can be a solution to the equation when it already is an input.

Let z = a + ib be a complex number. Find real numbers p and q such that z is a solution of $$z^2 + pz + q = 0$$

I have tried to use the quadratic formula, but that does not lead me anywhere since p and q is yet defined.

There second part of the problem asks for

Find all complex numbers w such that w and i are solutions of a polynomial $$z^2 + pz + q$$ with p and q real.

Is it correct to let $w = z_1$ and $i = z_2$ from the quadratic formula?

Answer 1

For the first part, note that when there is a complex root, its complex conjugate is also a root. Then, if $z=a+ib$, then its complex conjugate will be $\bar{z}=a-ib$ and they will be solutions for the equation $z^2 + pz + q = 0$. Then, you can use Vietta's Formulas as in usual quadratics to determine them.

For the second part, since it demands that $w$ and $i$ are solutions of the quadratic polynomial, recall that you can write it as in a factorized way :

$$z^2 + pz + q = (z-i)(z-w)$$

You can figure out $w$ from the expression above, when $p,q \in \mathbb R$.

Answer 2

Hint: choose $p$ and $q$ so that $z=a+ib$ and $\overline{z}=a-ib$ are the solutions of $z^2+pz+q=0$. Use Vieta's formulas to find $p$ and $q$.

2nd part: take $w=\overline{i}=-i$.

Answer 3

If $z=a+bi$, then$$z^2+pz+q=(a+bi)^2+p(a+bi)+q=a^2-b^2+pa+q+i(2ab+pb).$$Therefore, take $p=-2a$ and $q=-a^2+b^2-pa=-a^2+b^2+2a^2=a^2+b^2$.

For the second part: if $i$ is a root of a polynomial with real coefficients, then $-i(=\overline i)$ is also a root.