Question: Newton-Raphson formula

Assume that |T(z)| < 1. Why does it follow that (T(z))^r → 0 as r → ∞?

(Hint |(T(z))^r| = |T(z)|^r.) Deduce from the above that, as k→∞, we have T(N^k(z)) → 0. Why does it follow that N^k(z)→i ?

What is the corresponding result when |T(z)| > 1 ?

(Hint: Let l/T(z) = U(z)so that |U(z)| < 1, then apply the same argument (check U(N(z)) = (U(z))², noting that U^-1(0) = -i).

Now consider the two regions of the plane:

R₊ = {z : N^k{z) → i as k → ∞}; R_- = {z : N^k(z) → -i as k → ∞}.

Draw a diagram to illustrate these regions, the line L and the roots i and -i. We call R₊ the basin of attraction for the root +i, and similarly R_-is the basin of attraction for the root -i.

Show that if z is on the set L (the common boundary of the two regions R₊ and R_{_}, then N^k(z) stays on L for all values of k. (This is easy once you identify what L is). So in this case iteration does not produce a root at all.

It would be appreciated if any one could help me with clear up my confusion with questions.

Thanks a lot.