Answer
$z = (-1+i)~~$ 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.
