Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 8 - Test - Page 412: 16


$z = (-1+i)~~$ is not in the Julia set since the absolute value of $z$ exceeds 2.

Work Step by Step

$z = -1+i$ We can find the value of $z^2-1$: $z^2-1 = (-1+i)(-1+i)-1$ $z^2-1 = (-2i)-1$ $z^2-1 = -1-2i$ We can find the magnitude of $z^2-1$: $\sqrt{(-1)^2+(-2)^2} = \sqrt{5} \gt 2$ Therefore, $~~z = (-1+i)~~$ is not in the Julia set.
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