We all know the graphical proof of the Newton Raphson method that draws a tangent to the graph in question and then finds where this intersects the x-axis, giving better approximations. How could I adapt this proof to explain why the Newton Raphson method works for complex roots (provided that you put complex starting points in).
Thanks!!!

Suppose that you are trying to find a root of a function $f(x)$, and that root is $\alpha$. Let your initial guess be $X=\alpha +\epsilon$, where $\epsilon$ is quite small.

Then we can expand $f(x)$ in a Taylor series at $X$, so we have $f(\alpha )=f(X-\epsilon )=f(X)-\epsilon f\text{'}(X)+\mathrm{something}\mathrm{very}\mathrm{small}$.

However, since $\alpha$ is a root, $f(\alpha )=0$. So we have $f(X)-\epsilon f\text{'}(X)=0$, or $\epsilon =f(X)/f\text{'}(X)$.

Thus a better estimate for $\alpha$ is $X-\epsilon =X-f(X)/f\text{'}(X)$, which is the Newton-Raphson formula.

(However, we have assumed that differentiation still exists for complex functions. This is not trivial, but relatively safe for well behaved functions like polynomials and rational functions.)

David

Well,
if we can represent complex numbers on a graph then why can we not give the same graphical explanation for complex numbers as well.
Arun

Kerwin