Answer
Ans. Proved $|z̅|=|z|$
Work Step by Step
Given:-let $z=a+ib$
$|Z|=\sqrt {a^2+b^2}$
Conjugate of Z is given by
$z̅=a-ib$
$|z̅|=\sqrt {a^2+(-b)^2}$
$|z̅|=\sqrt {a^2+(b)^2}$ $ (-b)^2=b^2$
$|z̅|=|z|$
Proved.
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
Published by
Pearson
ISBN 10:
0-32187-896-5
ISBN 13:
978-0-32187-896-0
Appendices - Section A.7 - Complex Numbers - Exercises A.7 - Page AP-35: 28
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