Question

Find the first four iterates of the function f(z)=z^2+2+i with the initial value of z0=2+2i

Answer 1

If this is supposed to be a sequence, with z representing the previous term, what happens when you apply the function to your initial value?

Answer 2

I appreciate your trying to help . . . but, my way of learning doesn't involve being asked questions in order to get it out of me.

Answer 3

I'm just trying to figure out what's going on here, actually. Is this supposed to be a sequence? (honest question)

Answer 4

Sorry. I honestly, have no clue. I'm so confused with it myself :/

Answer 5

I think it is, so I'll rewrite it as if it were to make it clear to myself. Then, we can both pretend I'm right. \[f(z_n)=z _{n-1}^{2}+2-i\]

Answer 6

Do you know how to square complex numbers?

Answer 7

Lol, alright. & nope, I don't.

Answer 8

f( z) = z^2 + 2 + i , z0 = 2+2i

Remember the formulas: (a ± b)^2 = a^2 ± 2ab + b^2
Now what if we replace "b" with "bi" then:
(a ± bi)^2 = a^2 ± 2abi + b^2 i^2 >> but i^2 = - 1 by definition so:
(a ± bi)^2 = a^2 ± 2abi - b2 ----> This is the key to solve this problem

z1 = f(z0) = (2 + 2i)^2 + 2 + i = 4 + 8i +(4i^2) + 2 + i = 4 - 4 + 2 + 9i = 2 + 9i

z2 = f(z1) = (2 + 9i)^2 + 2 + i = 4 + 36i - 81 + 2 + i = -75 + 37i

z3 = f(z2) = (-75 + 37i)^2 + 2 + i = 5625 - 5550i - 1369 + 2 + i = 4258 - 5549i

z4= f(z3) = (4258 - 5549i)^2 + 2 + i = 18130564 - 47255284i - 30791401 + 2 + i =

-12660835 - 47255283i

Answer 9

Thank you :)
Both, of you.

Answer 10

Anytime.