3.3 Zeros of a Holomorphic Function
For a region and a holomorphic function , a point is a zero of iff . Furthermore, if has the Taylor expansion at of
then the zero at has multiplicity .
We will introduce a fundamental application of Liouville’s Theorem (Theorem 3.2.3) below.
Theorem 3.3.1 (Fundamental Theorem of Algebra).
Every non-constant polynomial with complex coefficients has at least one complex zero.
Proof.
For the sake of contradiction, suppose that has no complex zeros. Then the function is continuous and entire, because has no zeros in . Moreover, as , , so , and thus is bounded. By Liouville’s Theorem (Theorem 3.2.3), every bounded entire function is constant. Thus, is constant, and so must also be constant. By contradiction, has at least one complex zero.
Theorem 3.3.2.
Let be open and connected, and be holomorphic over . Then if the set defined by
has an accumulation point in , then over .
Proof.
Let be a subset of and assume it has an accumulation point in . Since is holomorphic over , such that is holomorphic over . Then over this disk, has the Taylor expansion
By Definition 1.1.3, such that , . Since is a zero of , . Then, by the continuity of ,
Using this result in comparison to (3.3.1), we get that .
The function
has a Taylor expansion over of
Let for some . Then vanishes, leaving
Letting , , and . Define
Then,
Similarly, . Letting
the sequence vanishes, and on .
Let
For all , since locally vanishes (and has vanishing derivatives as a consequence),
Furthermore, for all , such that has a convergent Taylor series with vanishing coefficients on . Then on . Then for all , since is constant at , it also has vanishing derivatives. It follows that
Since every point in has an open neighborhood also in , is open.
It is evident that for all , is continuous in by the holomorphy of . Let
For any sequence converging to some , by the continuity of ,
and therefore . Thus, contains all of its accumulation points in and is therefore closed in (if , then it is no longer relevant; we are concerned about it being closed within ). Since
and each of is closed in , is the intersection of closed sets and consequently closed.
Since is nonempty and clopen in the connected set , (by Theorem 3.2.9). It follows that on .
Remark.
This is a trivial property of holomorphic functions that allows for the uniqueness of analytic continuations. It is oftentimes stated in the form below:
Theorem 3.3.3 (Identity Theorem).
Let be open and connected, and define and to be two holomorphic functions on . For a set with an accumulation point in , if on , then on .
Proof.
Let be holomorphic over . Since has an accumulation point in , and over , then by Theorem 3.3.2, over .
Theorem 3.3.4 (Holomorphic Argument Principle).
Let be a region and be holomorphic. Let be a simple, closed, positively oriented curve that is null-homotopic in . If has no zeros on , then has finitely many zeros in the region bounded by , and this number, counting multiplicities, is given by
Let be the image of under the map . Then
where denotes the total change in argument of as it traverses .
Proof.
Let be the distinct zeros of enclosed by with the respective multiplicities . Choose disjoint disks centered at each with radii , each contained in the interior of and avoiding . The function
is holomorphic on the domain
where denotes the interior relative to . The oriented boundary of this domain is
By Cauchy–Goursat (Theorem 3.1.7),
which rearranges to
Near each , express
where is holomorphic and non-vanishing on . Differentiation yields
and thus
Since is holomorphic and non-vanishing on , the function is holomorphic there. By the Cauchy–Goursat Theorem,
The Cauchy–Goursat Formula (Theorem 3.1.8) gives
Combining results,
Finally, parameterize by . Then , and
which proves the result.
Thus, one defines the winding index () to quantify how many times a closed curve winds counterclockwise around a given point in the complex plane. Formally, if is a counterclockwise-oriented closed curve and is a point satisfying , then
Theorem 3.3.5.
Let be a sequence of holomorphic functions on the open set that uniformly converges to on every compact subset of . If , has no zeros in , then is either identically or has no zeros in .
Proof.
By the holomorphy of , for any simple closed rectifiable curve (whose interior is a subset of ), by the Cauchy–Goursat Theorem (Theorem 3.1.7),
Since is a subset of any compact subset of , uniformly converges on , and by Theorem 2.3.6,
Then by Morera’s Theorem (Theorem 3.2.4), is holomorphic, and is holomorphic. We aim to show that .
Let be arbitrary and compact and be open and relatively compact in . Since is holomorphic, by Corollary 3.2.1, there exists a finite constant such that
By the definition of uniform convergence, the right-hand side approaches , and is then uniformly convergent to by the same reasoning.
Through the proof of Theorem 3.3.2, if over , then the zeros of do not have an accumulation point in and are therefore discrete. In this case, let be a curve that does not pass through the zeros of . Since each function in the sequence does not contain zeros in , by the Argument Principle (Theorem 3.3.4),
Since and are holomorphic over , by Theorem 1.2.13, there exists a finite value such that , .
Since does not pass through the zeros of , such that , . By the uniform convergence of , such that
Then on . Hence, and its limit are uniformly bounded;
Therefore, is uniformly convergent on . By Theorem 2.3.6, we can pass the limit through the integral in (3.3.3). Then,
By the Argument Principle (Theorem 3.3.4), has no zeros in the interior of . Since was arbitrarily chosen, either on or has no zeros in .
Theorem 3.3.6 (Rouché).
Let be open and be two holomorphic functions over . Let be a simple, closed, rectifiable curve, and for all
Then and have the same number of zeros enclosed by and do not vanish on .
Proof.
It is obvious that has no zeros on . Otherwise, such that , implying that which is impossible. Similarly, has no zeros on , since is an impossibility.
Let and denote the number of zeros of and enclosed by , respectively. By the Argument Principle (Theorem 3.3.4),
Let with . Then,
From (3.3.4), by dividing both sides by , we obtain . Then lies in the open disk , which will never intersect or enclose . Then by Lemma 3.1.2,
as desired.
By the Fundamental Theorem of Algebra (Theorem 3.3.1), any polynomial in the form (, , where ) has at least one complex zero. Consider the function , with a zero at with multiplicity . By Rouché’s Theorem (Theorem 3.3.6), since such that over , and have the same number of zeros, counting multiplicity.
Theorem 3.3.7.
Let be open and connected, and be holomorphic and non-constant on .
If and , and the multiplicity of the zero at of is , then for all such that is non-vanishing on , such that , has zeros in , counting multiplicity.
Proof.
The zero at is isolated by Theorem 3.3.2. Furthermore, is continuous on and attains a positive infimum . In other words, on this set, . Hence, , we have for any .
By Rouché’s Theorem, since , it follows that and have the same number of zeros in .
We also have the following generalization of Theorem 3.3.5, which is a heuristic restatement of Theorem 3.3.7:
Theorem 3.3.8 (Hurwitz).
Let be an open and connected set, and suppose is a holomorphic function sequence that uniformly converges to a non-constant function on all compact sets of .
If and , and the multiplicity of the zero at of is , then for all such that is non-vanishing on , such that , has zeros in , counting multiplicity.
Proof.
The zero at is isolated by Theorem 3.3.2. Furthermore, is continuous on and attains a positive infimum . In other words, on this set, . By uniform convergence, such that , we have for any .
By Rouché’s Theorem (Theorem 3.3.6), since
it follows that and have the same number of zeros in .