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.
Link here: https://en.wikipedia.org/wiki/Bisection_method
My gut feeling is to say that it can't but I am not entirely sure how to justify it.
I know Newton's method can but am not sure about this one.
Any help would be highly appreciated!
Hint.
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