Question

If$\mathit{a}\mathbf{+}\mathit{i}\mathit{b}$ is a root of ${\mathit{x}}^{\mathbf{3}}\mathbf{-}\mathbf{3}{\mathit{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathit{i}\mathit{x}\mathbf{+}\mathit{i}\mathbf{-}\mathbf{1}\mathbf{\in }\mathbf{\square }\left[x\right]$, then it is true that$\mathit{a}\mathbf{-}\mathit{i}\mathit{b}$ is also a root?

Short answer

No, $a-ib$ is not a root of $\underline{the}$given polynomial

Step-by-step solution

Use method of contradiction.

Suppose $a+ib$and $a-ib$are both roots of given polynomial.

Then, $x^2-2ax+a^2+b^2\in \square \left[x\right]$is a factor of $x^3-3x^2+2ix+i-1$.

Therefore, there must be some$c+id\in \square$ such that

$$\begin{gathered} x^3-3x^2+2ix+i-1=\left(x+c+id\right)\left(x^2-ax+a^2+b^2\right) \\ =x^3+\left(c+id-2a\right)x^2-2a\left(c+id\right)x+\left(a^2+b^2\right)\left(c+id\right) \end{gathered}$$

Compare coefficients of both side

So, we have a system of equations:

$\left\{\begin{array}{l}c-2a+di=-3\\ a^2+b^2-2ac-2adi=2i\\ \left(a^2+b^2\right)\left(c+di\right)=i-1\end{array}\right.$

From the first equation, we have $\sout{that}$, but that implies the left-hand of the last two equation$\underline{s}$are real, while the right-hand side is not, $\underline{w}$hich is $\underline{a}$contradiction.

Therefore, $x^3-3x^2+2ix+i-1$does not have a pair of conjugate roots.