find the first three iterates of the function f(z) = z^2+z-i if the initial value is i.
My answers were: z=i; z1= -1, z2 = i, and z3 = i
Is this right?
sorry, not in my area of expertise.
Perhaps drwls, Damon, bobpursley or DrBob or others can help out here.
start with f(z) = i as directed
then
f(i) = -1 +i - i = -1 check
f(-1) =+1 -1 - i = -i do not check
f(-i) =-1 -i - i = -1 -2i
To find the first three iterates of the function f(z) = z^2+z-i with the initial value i, we can use the iteration formula z(n+1) = f(z(n)).
Starting with the initial value z = i:
- Plug in z = i into f(z) to find f(i) = i^2 + i - i = i^2.
- Therefore, z1 = i^2.
Now, use z1 as the next initial value:
- Plug in z = z1 into f(z) to find f(z1) = z1^2 + z1 - i.
- Therefore, z2 = z1^2 + z1 - i.
Lastly, use z2 as the next initial value:
- Plug in z = z2 into f(z) to find f(z2) = z2^2 + z2 - i.
- Therefore, z3 = z2^2 + z2 - i.
Now let's compute these iterates step by step:
z1 = i^2 = -1
z2 = (-1)^2 + (-1) - i = 0 - 1 - i = -1 - i
z3 = (-1 - i)^2 + (-1 - i) - i = (1 - 2i - i^2) - 1 - i - i = 1 - 2i + 1 - 2i - 1 - i - i = -2 - 6i
Therefore, the first three iterates with the initial value i are z1 = -1, z2 = -1 - i, and z3 = -2 - 6i.
Your answer of z1 = -1, z2 = i, and z3 = i is incorrect. Please recheck your calculations using the iteration formula f(z(n)).