It is well known to experts that the probability that a complex number will be a root of a random polynomial is . Here is a simulation of the values of for :

There are many ways to prove such a statement, one is using the following theorem of Erdős:

Theorem: Let be a sequence of nonzero complex numbers and a complex number. ThenHere the implied constant is absolute.

The simplest case of this theorem is when all the -s are equal and —a simple random walk on . Then the probability to hit is either if is odd, or

,

if is even. Here means that there exists an absolute constant such that . The last inequality is elementary, see here, and using Stirling’s formula (or Wallis formula) can be made into the asymptotic formula .

The goal of this post is to describe an easy “circle-method-argument” that reduces the proof of the theorem to the case of a simple random walk on . I’ve learnt this argument from James Maynard.

**Proof.**

By conditioning on the last sign, one sees that it suffices to prove the statement for even . Indeed, had we knew the assertion for even -s, then for odd we have

We start by considering the case where all the -s are rational. Since if and only if with a common denominator for the -s, we may assume w.l.o.g. that the are integers. We may also assume that is an integer, otherwise there is no solution and so the probability is . We put to be the number of choice of signs so that so that . To compute we use the trivial identity for integers

where . Thus, by expanding the product, we have

We now apply the AM-GM inequality to get that

Since is real and is even, we may remove the absolute value in the integral, and then by the same argument as before

So we are back to the simple random walk, that is,

.

Averaging over we conclude that

as needed. This finishes the proof for rational -s.

Next we consider the general setting, namely, when the -s are nonzero complex numbers. Let be the -vector space that is spanned by the -s and proceed by induction on . When , then we are back in the setting of rational -s, so we may assume that . Take of dimension . So the -s divide into two nonempty sets and such that and .

Fix some so that . Any breaks as with rational and . We may assume that (otherwise the probability is zero) and hence can be decompose as with rational and . Then if and only if and with . Denote the former event by and for any choice of signs for the elements in denote the latter event by .

Putting , we get by the induction assumption that and conditioning on the choice of signs for the elements in , the probability of is . Thus by conditioning on we get that

This conclude the proof since clearly (separate to the cases that for some and that both ). QED