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#