Chapter 2.5, Problem 69E

To determine

To write : the polynomial as the product of linear factors and list all the zeros of function

Answer to Problem 69E

The function as product of linear factors: $\boxed{\boxed{f\left(z\right)=\left(z-1-i\right)\left(z-1+i\right)}}$

The zeros of the function are: $\boxed{\boxed{z=1+i;1-i}}$

Explanation of Solution

Given information: $f\left(z\right)=z^2-2z+2$

Concept Involved:

Linear Factorization Theorem:If $f\left(x\right)$ is a polynomial of degree $n$ , where $n>0$ , then $f\left(x\right)$ has precisely $n$ linearfactors $f\left(x\right)=a_n\left(x-c_1\right)\left(x-c_2\right)...\left(x-c_n\right)$ where $c_1,c_2,...,c_n$ are complex numbers.

Complex Zeros Occur in Conjugate Pairs: Let f be a polynomial function that has real coefficients. If $a+bi$ , where $b\ne 0$ , is a zero of the function, then the complex conjugates $a-bi$ is alsoa zero of the function.

Calculation:

To find the zeros of the function we need to set the function to zero $f\left(z\right)=0$

  $z^2-2z+2=0$

We can solve this equation using completing the square method by subtracting2 on both sides of the equation

  $z^2-2z+2-2=0-2$

Simplify on both sides of the equation

  $z^2-2z=-2$

In order to make the left side expression as perfect square trinomial, we need to add square of half of coefficient of x on both sides

  $z^2-2z+1=-2+1$

Simplify on right side of the equation

  $z^2-2z+1=-1$

Write the left side as a perfect square

  ${\left(z-1\right)}^2=-1$

Take square root on both sides

  $\sqrt{{\left(z-1\right)}^2}=\sqrt{-1}$

Simplifying square root on both sides of the equation

  $z-1=\pm \sqrt{-1}$

Replace $i$ for $\sqrt{-1}$

  $z-1=\pm i$

Add 1 on both sides of the equation

  $z-1+1=1\pm i$

Simplify in left side of the equation

  $z=1\pm i$

List the zeros of the functions given:

  $z=1+i;z=1-i$

If $z=k$ is zero, then $z-k$ will be the factor of the function.

So $z-\left(1+i\right)$ & $z-\left(1-i\right)$ are the factors of the function

Write the function $f\left(z\right)=z^2-2z+2$ as product of linear factors:

  $f\left(z\right)=\left(z-1-i\right)\left(z-1+i\right)$

Conclusion:

The zeros of the given function $f\left(z\right)=z^2-2z+2$ are: $z=1+i;1-i$

The function written as product of linear factors: $f\left(z\right)=\left(z-1-i\right)\left(z-1+i\right)$