Answer
$z = -0.2i$ is in the Julia set
Work Step by Step
For $z = -0.2i$,
$z^2 - 1$
= $(-0.2i)^2 - 1$
= $-0.04 - 1$ (since $i^2$ = $-1$)
= $-1.04$
$(z^2 - 1)^2 - 1$
= $(-1.04)^2 - 1$
= $0.0816$
$[(z^2 - 1)^2 - 1]^2 - 1$
= $0.0816^2 - 1$
= $-0.993$ (correct to 3 sig. fig.)
$\{[(z^2 - 1)^2 - 1]^2 - 1\}^2 - 1$
= $(-0.993)^2 - 1$
= $-0.014$ (correct to 3 sig. fig.)
and so on.
Since the respective absolute values do not exceed 2, so $z = -0.2i$ is in the Julia set and the point $(0, -0.2)$ is part of the graph.