Question #00dfc

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Answer 1 · 2017-10-17T09:42:44

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

$i^{2} = - 1$ $i^2=-1$

Use $z = 1$ $z=1$ as the first input value

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

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

Then $z = - 2 + 2 i$ $z=-2+2i$

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

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

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

$= - 3 - 6 i$ $=-3-6i$

And finally $z = - 3 - 6 i$ $z=-3-6i$

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

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

$= - 30 + 38 i$ $=-30+38i$