# Bisection method complex roots

Can the bisection method of root finding be used to find the complex roots of a polynomial, i.e, complex values of $$x$$ such that $$f\left(x\right)\:=\:0$$?

In the bisection method we arbitrarily choose two starting points with opposite signs and then see if the value of the function at the midpoint of these two points is a root. If it is, we stop, if it is not we halve the interval depending on which side of the interval gives us a closer result.

My gut feeling is to say that it can't but I am not entirely sure how to justify it.

Any help would be highly appreciated!

• There are no signs in complex numbers...and there are no intervals, either. It's like working with the plane $\;\Bbb R^2\;$ for that matter Commented Oct 28, 2020 at 8:01
• @DonAntonio I see. So because of this, the bisection can't be used to find complex roots? Commented Oct 28, 2020 at 8:12
• There is a winding number variation on the bisection method. Here is 3blue1brown's video on the subject. I don't know to what degree it's actually usable in practice, though. (This winding number idea is the core of my favourite proof of the fundamental theorem of arithmetic.) Commented Oct 28, 2020 at 8:17

If $$f(z) = 0$$ can be represented as
$$\cases{ f_r(x,y) = 0\\ f_i(x,y) = 0 }$$
we can use a bintree to determine into a plane region the trace of $$f_r=0, f_i=0$$ and then we will have within a required precision the set of points which verify the equation. The binthree for $$\mathbb{R}^2$$ or octree for $$\mathbb{R}^3$$ can be processed recursively in a fashion very akin to the binary search process to $$\mathbb{R}$$ problems