Question #00dfc

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Answer 1 · 2017-10-17T09:42:44

The first three output values are $(-2+2i)$ , $(-3+6i)$ , and $(-30+38i)$

$i^2=-1$

Use $z=1$ as the first input value

$F(z)=z^2-3+2i$

$F(1)=1-3+2i=-2+2i$

Then $z=-2+2i$

$F(-2+2i)=(-2+2i)^2-3+2i=(2-2i)^2-3+2i$

$=4-8i+4i^2-3+2i$

$=4-8i-4-3+2i$

$=-3-6i$

And finally $z=-3-6i$

$F(-3-6i)=(-3-6i)^2-3+2i$

$=9+36i^2+36i-3+2i$

$=-30+38i$