## Trigonometry (11th Edition) Clone

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