Can we find the complex roots by using Newton's method?

Question

Is it possible to find the complex roots of a function by using Newton's method? If the answer is yes, how to do it?

Answer 1

Newton's method works for complex differentiable functions too. In fact, we do exactly the same thing as in the real case, namely repeat the following operation: $$ z_n = z_{n+1} - \frac{f(z_n)}{f'(z_n)} $$ The only difference is that this time the fraction may have complex numerator and denominator.

(Note that for complex functions, the requirement of being differrentiable is a lot stricter than it is for real functions.)

Answer 2

You can but, as usual, you typically need to be close to one root for that to work well.

Understanding how Newton's method converges and where it converges to depending on the initial value is very hard. See the famous Newton fractal.

A nontrivial study of how to choose a finite set of initial points that guarantee the convergence for Newton's method for complex polynomials is given in the paper

How to find all roots of complex polynomials by Newton's method by Hubbard et al.
Invent. Math. 146 (2001), no. 1, 1–33. pdf

Answer 3

The two answers given up to now are correct and complete; however, I would like to add this simple practical remark: if $f$ maps real values to real values (like, for instance, a polynomial or rational function with real coefficients), then Newton's method started from a real $x_0$ produces real iterates, and hence it cannot converge to complex (non-real) roots. So you need to start from a non-real $x_0$ to find non-real roots.

(And, if there are no real roots, Newton's method with a real $x_0$ just won't converge: see what happens for instance with $f(x)=x^2+1$).