## Trigonometry (11th Edition) Clone

$z = -0.2i$ is in the Julia set
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.