Chapter 4, Problem 26RE

To determine

To calculate: The remaining zeros of f and a polynomial function with real coefficients that has the zeros.

Answer to Problem 26RE

The remaining zeros are $-i,1-i$ .

The polynomial function is $x^4-2x^3+3x^2-2x+2$ .

Explanation of Solution

Given information:

Degree 4; zeros: $i,1+i$ .

Formula used:

Conjugate Pairs Theorem:

The theorem states that if the coefficients of the polynomial $f(x)$ are real numbers, whose zeros are $r=a+bi$ then the complex conjugate $\overline{r}=a-bi$ is also a zero of f .

Linear Factorization Theorem:

If $r_1,r_2,\mathrm{....},r_n$ are zeros of polynomial function $f(x)$ , then $f(x)$ can be written as

  $f(x)=a(x-r_1)(x-r_2)\mathrm{.......}(x-r_n)$ where $a$ is a nonzero constant.

Calculation:

Consider the zeros: $i,1+i$

Since coefficients of f are real numbers, complex zeros appear as conjugate pairs.

This implies $-i,1-i$ is also a zeros of f .

Hence, the remaining zeros of f are $i,1+i,-i,1-i$ .

Using Linear Factorization Theorem $f(x)$ can be written as :

  $f(x)=a(x-i)(x-(-i))(x-(1+i))(x-(1-i))$

Here , $a=1$ .

  $\begin{array}{l}f(x)=a(x-i)(x+i)(x-(1+i))(x-(1-i))\\ =\left[x^2-i^2\right]\left\{x(x-(1-i))-(1+i)(x-(1-i))\right\}\\ =(x^2+1)\left\{x(x-1+i)-(1+i)(x-1+i)\right\}\end{array}$

  $\begin{array}{l}=(x^2+1)\left[(x^2-x+xi)-(x-1+i+ix-i+i^2)\right]\\ =(x^2+1)\left[(x^2-x+xi)-(x-1+ix-1)\right]\\ =(x^2+1)\left[(x^2-x+xi)-(x+ix-2)\right]\\ =(x^2+1)\left[x^2-x+xi-x-ix+2\right]\end{array}$ where $i^2=-1$

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

Therefore , the polynomial function is $x^4-2x^3+3x^2-2x+2$ .