Let $f(z) = z^2 + 1$ and let $N(z)$ be the corresponding Newton method iteration function. Show by direct computation that $i$ is a super-attractive fixed point for $N$.

An archived instance of discourse for discussion in undergraduate Complex Dynamics.

# A complex Newton method

mark

Rick

Let $f(z)=z^2+1$. Newton's method iteration function is given by $N(z)=z-\frac{f(z)}{f'(z)}.$ Then

\begin{align}

N'(z)&=1-\frac{f'(z)f'(z)+f''(z)f(z)}{f'(z)^2} \\

&=1-\frac{(2z)(2z)+(2)(z^2+1)}{(2z)^2} \\

&=1-\frac{4z^2+2z^2+2}{4z^2}=1-\frac{6z^2+2}{4z^2}.

\end{align}

Also, $$N'(i)=1-\frac{6(i)^2+2}{4(i)^2}=1-\frac{-4}{-4}=1-1=0,$$

Therefore, $i$ is a super attractive fixed point for N.

RedCrayon

@Rick I think the one thing missing from this is showing that $i$ is a fixed point of $N$.

Note that

$$N(z)=z-\frac{z^2+1}{2z}=\frac{z}{2}-\frac{1}{2z}$$ so

$$N(i)=\frac{i}{2}-\frac{1}{2i}=\frac{i}{2}+\frac{i}{2}=i$$

Therefore, $i$ is a fixed point of $N$.