Newton’s method is super attractive!

Suppose that f:\mathbb C \to \mathbb C is entire, f(z_0)=0, and f'(z_0)\neq0. Show that z_0 is a super-attractive fixed point of the corresponding Newton’s method iteration function.

Suppose that f:\mathbb{C} \rightarrow \mathbb{C} is entire, f(z_0) = 0 and f'(z_0) \neq 0.

Recall that Newton’s iteration function is defined as:

N(z) = z - \dfrac{f(z)}{f'(z)}

so let’s plug in z_0:

N(z_0) = z_0 - \dfrac{f(z_0)}{f'(z_0)} = z_0 - \dfrac{0}{f'(z_0)} = z_0

So this means that z_0 is a fixed point of Newton’s method iteration function, but we need to take the derivative to and check its value to see what properties it has:

Since N(z_0) = z_0, then N'(z_0) = 0, so |N(z_0)| = 0, therefore z_0 is a super attractive fixed point of Newton’s method iteration function.