Question

given that p and q are real & that 1+2i are aroot of the equation z^2+(p+5i)z +q(2-i) determine the value of p& q and the other root of the equation

Answer 1

Given that $p$ and $q$ are real and that $1+2i$ are a root of the equation $z^2+(p+5i)z +q(2-i)$ let us determine the value of $p$ and $q$ and the other root of the equation.

Let $a+bi$ be the other root of the equation. Then

$z^2+(p+5i)z +q(2-i)=(z-1-2i)(z-a-bi)\\ =z^2+(-1-2i-a-bi)z+(1+2i)(a+bi).$

It follows that

$p+5i=-1-2i-a-bi\text{ and}\\ q(2-i)=(1+2i)(a+bi)=(a-2b)+(2a+b)i.$

Therefore, we get the following system:

$\begin{cases} p=-1-a\\ 5=-2-b\\ 2q=a-2b\\ -q=2a+b \end{cases}$

which is equivalent to the following systems

$\begin{cases} p=-1-a\\ b=-7\\ 2q=a+14\\ -q=2a-7 \end{cases}$

| let us myltiply the last equation by 2 and add to the third equation |

$\begin{cases} p=-1-a\\ b=-7\\ 2q=a+14\\ 0=5a \end{cases}$

$\begin{cases} p=-1\\ b=-7\\ q=7\\ a=0 \end{cases}$

We conclude that $p=-1,\ q=7$ and the other root of the equation is $-7i.$