Chapter 8.6, Problem 23WE

a.

To determine

To show that the given number is a root of the equation.

Explanation of Solution

Given information:

The given equation is $x^3+2x^2+x-1+i=0$ , and the given number is $-1+i$

Concept used:

If a number is the root of a certain equation then it must satisfy the equation.

Proof:

Substitute the given number in the left hand side of equation

  $\begin{array}{l}{x}^3+2x^2+x-1+i\\ {\left(-1+i\right)}^3+2{\left(-1+i\right)}^2+\left(-1+i\right)-1+i\\ \left(2+2i\right)-4i-1+i-1+i\\ 2-2-4i+4i\\ 0\end{array}$

The number satisfies the equation hence it is a root of the equation.

b.

To determine

To show that the conjugate of given number is not a root of the equation.

Explanation of Solution

Given information:

The given equation is $x^3+2x^2+x-1+i=0$ , and the given number is $-1+i$

Proof:

Conjugate of $-1+i$ is $-1-i$

Substitute the conjugate in the left hand side of equation

  $\begin{array}{l}{x}^3+2x^2+x-1+i\\ {\left(-1-i\right)}^3+2{\left(-1-i\right)}^2+\left(-1-i\right)-1+i\\ \left(2-2i\right)+4i-1-i-1+i\\ 2-2-2i+4i\\ 2i\end{array}$

2i is not equals to zero hence is not satisfies the equation.

The conjugate of the number does not satisfies the equation hence it is not a root of the equation.