Answer
$f(z)=(z-1+i)(z-1-i)$
The zeros:
$z=1-i$
$z=1+i$
Work Step by Step
$f(z)=z^2-2z+2$
$f(z)=z^2-2z+1+1$
$f(z)=(z-1)^2+1$
$f(z)=(z-1)^2-i^2$
$f(z)=(z-1+i)(z-1-i)$
The zeros:
$z-1+i=0$
$z=1-i$
$z-1-i=0$
$z=1+i$
Algebra and Trigonometry 10th Edition
by Larson, Ron
Published by
Cengage Learning
ISBN 10:
9781337271172
ISBN 13:
978-1-33727-117-2
Chapter 3 - 3.4 - Zeroes of Polynomial Functions - 3.4 Exercises - Page 284: 67
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