Find the third iterate of the function f(t) = 9t - 3 when t0 = 1/3 + i/9.
To find the third iterate of the function f(t) = 9t - 3, we need to apply the function three times consecutively.
Given that t0 = 1/3 + i/9, let's denote it as t0 = a + ib, where a = 1/3 and b = 1/9.
The first iterate, f(t1), is obtained by substituting t0 into the function f(t):
f(t1) = 9 * t0 - 3
= 9 * (a + ib) - 3
= 9a + 9ib - 3
= 9a - 3 + 9ib
= 9 * (1/3) - 3 + 9ib
= 3 - 3 + 9ib
= 9ib
Note that the real part a disappears in the first iterate, leaving only the imaginary part b, multiplied by 9i.
For the second iterate, f(t2), we substitute t1 (9ib) back into the function:
f(t2) = 9 * t1 - 3
= 9 * (9ib) - 3
= 81ib - 3
= -3 + 81ib
Again, the real part disappears, leaving only the imaginary part multiplied by 81i.
For the third iterate, f(t3), we substitute t2 (-3 + 81ib) back into the function:
f(t3) = 9 * t2 - 3
= 9 * (-3 + 81ib) - 3
= -27 + 729ib - 3
= -30 + 729ib
Therefore, the third iterate of the function f(t) = 9t - 3, when t0 = 1/3 + i/9, is -30 + 729ib.