Question

Find the first three output values of the function f(z) = z2 – 1 + 2i. Use z = 0 as the first input value.

–1 + 2i , –4 – 2i , 19 – 14i
–1 + 2i , 2i , 3 + 2i
–1 + 2i , –2 –+ 2i , –1 + 4i
–1 + 2i , –4 – 2i , 11 + 18i

Answer 1

f(0) = 0^2 -1 + 2i = -1 + 2i

now use the output as the next input

f(-1 + 2i) = (-1 + 2i ) ^2 -1 + 2i

Answer 2

you can do this nicely with a TI 84/ TI 83 calculator

Answer 3

console: 0
screen output: 0

type :
ANS^2 -1 + 2i
press enter , keep pressing enter

Answer 4

i dont see the ANS button

Answer 5

wait found it

Answer 6

2nd (-)

Answer 7

also make sure you can use i, which is 2nd . (next to 0 button)

Answer 8

ok i clicked enter 4 times
should i keep going? im still confused

Answer 9

@perl ?

Answer 10

you should see your sequence

Answer 11

so first press 0.
then type
Ans^2 -1 + 2i

Enter:
-1 + 2i
Enter :
-4 -2i
Enter :
11 + 18i

Answer 12

if you keep pressing enter you see that it diverges , gets bigger and bigger

Answer 13

Enter :
-204 + 398i
Enter:
-116789 - 162382i

etc

Answer 14

this is a recursive sequence

Answer 15

@Michele_Laino
z0 = z0
z1 = f(z0)
z2 = f(z1) = f(f(z0))
z3 = f(z2) = f(f(f(z0)))
...

Answer 16

ok! I understand!

Answer 17

no problem :) the directions were not clear

Answer 18

I believe this is called a 'Julia set'

Answer 19

if I remember correctly
given a complex number c, and a seed z0
f(z0)=z0
f(zn) = (z_n-1)^2 + c ,

iterating the function above produces a Julia set or Julia sequence

or more simply
f(z) = z^2 + c is the Julia set