Chapter 8.6, Problem 29WE

To determine

To justify the following statements of conjugate root theorem proof.

1. if $S\left(x\right)=\left[x-\left(a+bi\right)\right]\left[x-\left(a-bi\right)\right]=x^2-2ax+\left(a^2+b^2\right)$ , then $S\left(a+bi\right)=0\text{ and }S\left(a-bi\right)=0$
2. There are real numbers c and d and a polynomial $Q\left(x\right)$ for which $P\left(x\right)=Q\left(x\right)S\left(x\right)+cx+d$ .
3. Setting $x=a+bi$ , you obtain $0=Q\left(a+bi\right)\cdot 0+c\left(a+bi\right)+d$ .
4. $c\left(a+bi\right)+d=0$ , or $\left(ac+d\right)+bci=0$ .
5. $bc=0,$ and since $b\ne 0$ , $c=0$
6. $ac+d=0,$ and since $c=0,d=0$
7. $P\left(x\right)=Q\left(x\right)S\left(x\right)$
8. $P\left(a-bi\right)=Q\left(a-bi\right)S\left(a-bi\right)=Q\left(a-bi\right)\cdot 0=0$

Explanation of Solution

Given information:

The given statement is that $P\left(x\right)$ be a polynomial with real coefficients, and $a+bi\left(b\ne 0\right)$ be imaginary root of $P\left(x\right)=0$ .

Proof:

1.  $S\left(x\right)=\left[x-\left(a+bi\right)\right]\left[x-\left(a-bi\right)\right]=x^2-2ax+\left(a^2+b^2\right)$

Substitute $x=a+bi$

  $\begin{array}{l}{\left(a+bi\right)}^2-2a\left(a+bi\right)+\left(a^2+b^2\right)\\ =a^2+2abi-b^2-2a^2-2abi+a^2+b^2\\ =0\end{array}$

Substitute $x=a-bi$

  $\begin{array}{l}{\left(a-bi\right)}^2-2a\left(a-bi\right)+\left(a^2+b^2\right)\\ =a^2-2abi-b^2-2a^2+2abi+a^2+b^2\\ =0\end{array}$

Hence, the statement is justified.

2. There are real numbers c and d and a polynomial $Q\left(x\right)$ for which $P\left(x\right)=Q\left(x\right)S\left(x\right)+cx+d$

The above statement is justifiable because the degree of $S\left(x\right)$ is 2 and $cx+d$ is linear that means its degree is less than that of $S\left(x\right)$ .

3. Setting $x=a+bi$ , you obtain $0=Q\left(a+bi\right)\cdot 0+c\left(a+bi\right)+d$ .

Substitute $x=a+bi$ in $P\left(x\right)=Q\left(x\right)S\left(x\right)+cx+d$

Since $a+bi\left(b\ne 0\right)$ is root of $P\left(x\right)=0$ therefore $P\left(a+bi\right)=0$ and $S\left(a+bi\right)=0$ is already proved in statement 1.

Therefore,

  $0=Q\left(a+bi\right)\cdot 0+c\left(a+bi\right)+d$

4. $c\left(a+bi\right)+d=0$ , or $\left(ac+d\right)+bci=0$ .

First write the equation of statement 3 and then simplify

  $\begin{array}{l}0=Q\left(a+bi\right)\cdot 0+c\left(a+bi\right)+d\\ 0=c\left(a+bi\right)+d\\ 0=ac+bci+d\\ 0=\left(ac+d\right)+bci\end{array}$

The equation is justified.

5. $bc=0$ , and since $b\ne 0$ , $c=0$

Since $S\left(a+bi\right)=0\text{ and }S\left(a-bi\right)=0$

  $bc=0$ and $b\ne 0$

Therefore $c=0$

6. $ac+d=0$ , and since $c=0,d=0$

Since $0=\left(ac+d\right)+bci$ and $c=0$

Therefore $d=0$

7. $P\left(x\right)=Q\left(x\right)S\left(x\right)$

Substitute $d=0$ and $c=0$ in $P\left(x\right)=Q\left(x\right)S\left(x\right)+cx+d$

  $\begin{array}{l}P\left(x\right)=Q\left(x\right)S\left(x\right)+0\cdot x+0\\ P\left(x\right)=Q\left(x\right)S\left(x\right)\end{array}$

8. $P\left(a-bi\right)=Q\left(a-bi\right)S\left(a-bi\right)=Q\left(a-bi\right)\cdot 0=0$

Since $P\left(x\right)=Q\left(x\right)S\left(x\right)$ substitute $x=a-bi$

  $\begin{array}{l}P\left(a-bi\right)=Q\left(a-bi\right)S\left(a-bi\right)\\ P\left(a-bi\right)=Q\left(a-bi\right)\cdot 0=0\end{array}$

The above statement is justified

Hence, the conjugate root theorem is proved.