Newton’s method is not super attractive - an example

Suppose that we apply Newton’s method to f(z)=z^3. Show that the origin in attractive, but not super-attractive.

As f(z)=z^3 and f'(z)=3z^2, then the Newton’s method equation for this function is N(z)=z-\frac{z^3}{3z^2} or N(z)=\frac{2}{3}z. A fixed point z_{0} is attractive if |N'(z_{0})|<1 and super attractive if |N'(z_{0})|=0.

N'(z_{0})=\frac{2}{3}

The only fixed point of N(z) is the origin z_{0}=0.

N'(0)=\frac{2}{3} and because 0<|\frac{2}{3}|<1, the origin is an attractive fixed point of N(z), but not a super-attractive fixed point.