A variant of Cauchy's argument principle for analytic functions which applies to curves containing zeroes
AA variant of Cauchy’s argument principle for analyticfunctions which applies to curves containing zeroes
Maher Boudabra, Greg MarkowskyJune 24, 2020
Abstract
It is known that the Cauchy’s argument principle, applied to an holomorphicfunction f , requires that f has no zeros on the curve of integration. In this shortnote, we give a generalization of such a principle which covers the case when f haszeros on the curve, as well as an application. The argument principle, one of the fundamental results in complex analysis, can be for-mulated as follows (see [2] or [1]).
Theorem 1.
Suppose that (cid:13) is a smooth Jordan curve in a domain U , and a function f is analytic on U with no zeroes on (cid:13) . Then the number of zeroes of f inside (cid:13) is equal to (cid:13) f (z)f (z) dz: (1.1) This is also equal to the winding number of the curve f ((cid:13)) around . The equation (1.1) shows the importance of the condition that f not vanish on (cid:13) , sinceotherwise the integral would diverge, but it is the final statement that we will focus on. Wehave found the following variant on this statement which holds without the requirementthat f be nonzero on (cid:13) . Theorem 2.
Let f be a non zero holomorphic function defined on an open domain U and (cid:13) be a smooth Jordan curve lying inside U . Then for any line L passing throughthe origin there exist at least
2m + (cid:21) distinct points on (cid:13) mapped to L by f , where m is the number of zeros of f inside (cid:13) and (cid:21) is the number of zeros of f on (cid:13) , all countedaccording to multiplicities. We note that Theorem 2 is an immediate consequence of the Theorem 1 when thereare no zeroes on (cid:13) , since any line will be intersected at least twice by f ((cid:13)) each time that f ((cid:13)) winds around . Naturally the difficulty arises when there are zeroes on (cid:13) . A firstattempt may be to factor out the zeroes, for instance writing f (z) = g(z)h(z) , where g(z) is a polynomial with zeroes only on (cid:13) and h does not vanish on (cid:13) , and then to tryto apply Theorem 1 to h and add the zeroes from g . However this approach fails, as theintersections of f ((cid:13)) with L are in general not respected by the factorization, and in anyevent zeroes of g on (cid:13) of high order must contribute many intersections of f ((cid:13)) with L .We will prove Theorem 2 in the next section, but first we discuss the motivation forthis result, in particular the application which inspired it. We thank Mohammed Zerrakfor bringing this problem to our attention. Problem:
Let a ; : : : ; a n be a sequence of real numbers, and for simplicity as-sume a ; a n
6= 0 (this requirement can easily be removed if required). Let
P ((cid:18)) = a r X i v : . [ m a t h . C V ] J un nj=0 a j cos(j(cid:18)) and Q((cid:18)) = P nj=0 a n(cid:0)j cos(j(cid:18)) . Let Z P be the number of (cid:18) 2 [0; 2(cid:25)) such that P ((cid:18)) = 0 , and Z Q defined analogously. Show that Z P + Z Q (cid:21) 2n .A quick solution using complex analysis can be provided, as follows. Let f (z) =P nj=0 a j z j . Then P ((cid:18)) = <(f (e i(cid:18) )) , and it may be checked that Q((cid:18)) = <(g(e i(cid:18) )) , where g(z) = z n f ( ) . If there are no zeroes of f on the unit circle fjzj = 1g , then by Theorem1 the curve <(f (e i(cid:18) )) will intersect the imaginary axis at least f times, where m f isthe number of zeroes (counting multiplicities) of f inside the unit disk fjzj < 1g , and theanalogous statement holds for g . We see that Z P + Z Q (cid:21) 2(m f + m g ) . However, thenumber of zeroes of g in fjzj < 1g is the same as the number of zeroes of f in fjzj > 1g ,and since f has no zeroes on the unit circle we see that m f +m g = n . The result thereforefollows in this case.The role of Theorem 2 is to extend this solution in the case that f has zeroes on theunit circle. Let (cid:21) denote the sum of the multiplicities of the zeroes of f on fjzj = 1g , andnote that the conjugates of these zeroes must be zeroes of g with the same multiplicities.Applying Theorem 2, we see that Z P + Z Q = 2m f + (cid:21) + 2m g + (cid:21) , and as m f + m g + (cid:21) = n the result follows.As a final comment before proving Theorem 2, we point out that the number
2m + (cid:21) is the best possible, as the following examples show. Let (cid:13) be the unit circle, and let f (z) = (z + 1) n . Then, taking z = e i(cid:18) , we can check that (e i(cid:18) + 1) n = 2 n e in(cid:18)=2 cos n ((cid:18)=2) ,so the number of intersections that f ((cid:13)) has with the real axis is the same as the numberof zeroes of =(e in(cid:18)=2 ) = sin(n(cid:18)=2) in [0; 2(cid:25)) , and this is n . This shows essentially thatthe (cid:21) in the expression
2m + (cid:21) is sharp. The is even easier, as we may take (cid:13) againas the unit circle and f (z) = z n , and f (e i(cid:18) ) will intersect any line exactly times as (cid:18) ranges from to . Proof.
The number of zeros of f on and inside (cid:13) is finite as the latter is a Jordan curve.Let z ; :::; z k be the roots of f on (cid:13) , where we denote by (cid:21) j the multiplicity of each z j and D j = D j (") be the disc of radius " centered at z j . The radius " is chosen small enough so D j remains inside U and contains no zeroes of f other than z j . Now consider the Jordancurve (cid:13) " constructed from (cid:13) by replacing each part (cid:13) \ D j by the arc of the circle @D j lying outside of (cid:13) ; in order to guarantee that (cid:13) " is itself a Jordan curve we may need todecrease " further. The figure illustrates (cid:13) " .Figure 2.1: The bold curve is (cid:13) " .No zeroes of f lie on (cid:13) " , and there are m + (cid:21) zeroes of f inside (cid:13) " , so by Theorem 1there are at least
2m + 2(cid:21) points on (cid:13) " which are preimages of points in L . The trick is toshow that not too many of them can be on the components of (cid:13) " which lie in the boundaryof some @D j . The intuition behind this is easy: if we shrink " sufficiently, then the pointson @D j which are preimages of points on L should be approximately equidistributed on2 D j , so that about half of them lie inside and half lie outside (cid:13) . Only the ones on theoutside of (cid:13) lie on (cid:13) " , and any other points on (cid:13) " which are preimages of points in L mustlie on (cid:13) proper. The trick is making this rigorous.Inside D j , f is of the form f (z) = (z (cid:0) z j ) (cid:21) j g j (z) where g j (z j ) 6= 0 . If we let z = z j + "e (cid:18)i then we have arg(f (z j + "e (cid:18)i )) = (cid:21) j (cid:18) + arg(g j (z)): We claim that j d arg(g j )d(cid:18) (z j + "e (cid:18)i ) j " = o(1): (2.1)Note that this implies from above that j d arg(f)d(cid:18) (z j + "e (cid:18)i ) j " = (cid:21) j + o(1): (2.2)We shall use the approximation (2.2), and will prove (2.1) later on in a separate lemma.Now, we may shrink " if necessary so that d arg(f)d(cid:18) (z j + "e (cid:18)i ) is positive on all @D j . Thisfact, combined with Theorem 1, yields that @D j contains exactly j preimages of pointsin L , since the argument of the curve may not change direction in order to create extrapreimages. In other words, j j f (z) 2 Lg = 2(cid:21) j : Denote by u = z j + re (cid:18) i ; :::; u ‘ = z j + re (cid:18) ‘ i the points of (cid:13) " \ @D j such that f (u t ) 2 L ,arranged in anti-clockwise order. Note that arg(f (u t )) (cid:0) arg(f (u t(cid:0)1 )) = (cid:25) for all t , since f is continuous and orientation preserving. Using the mean value theorem and (2.2) wehave (cid:12)(cid:12)(cid:12) arg(f(u t ))(cid:0)arg(f(u t(cid:0)1 ))(cid:18) t (cid:0)(cid:18) t(cid:0)1 (cid:12)(cid:12)(cid:12) = (cid:25)(cid:18) t (cid:0)(cid:18) t(cid:0)1 (cid:20) (cid:21) j + o(1); whence ‘ (cid:20) ‘ X t=1 ((cid:18) t (cid:0)(cid:18) t(cid:0)1 )((cid:21) j +o(1))(cid:25) = ((cid:21) j +o(1))(cid:25) ((cid:18) ‘ (cid:0) (cid:18) ) (cid:20) (cid:21) j +"(cid:14) j (cid:25) ((cid:25) + (cid:17) " ); where (cid:25) + (cid:17) " denotes the angular length of the circular arc @D j \ (cid:13) " . Note that thesmoothness of (cid:13) implies that (cid:17) " ! 0 as " ! 0 .Figure 2.2: The dotted line is the tangent of (cid:13) at z j .Thus, by again shrinking " if necessary, we obtain ‘ (cid:20) (cid:21) j : There are ‘ + 1 points on @D j \ (cid:13) " which are preimages of points in L , and applyingthis around every z j and combining these estimates yields at least3 m + 2(cid:21) (cid:0) k X j=1 ((cid:21) j + 1) points on (cid:13) \ (cid:13) " which are preimages of points in L . Recalling that (cid:21) = P kj=1 (cid:21) j givesat least
2m + k X j=1 ((cid:21) j (cid:0) 1) points on (cid:13) \ (cid:13) " which are preimages of points in L . Finally, the points z ; : : : ; z k areall mapped to L by f , since L passes through the origin, and taking these into accountcompletes the proof of the theorem.It remains only to prove the lemma used earlier on the derivative of the argument of g . Lemma 3.
We have j d arg(g j )d(cid:18) (z j + "e (cid:18)i ) j " = o(1): Proof.
The Taylor expansion of g j (z = z j + "e (cid:18)i ) at z j is given by g j (z j + "e (cid:18)i ) = g j (z j ) + X r=1 g (r)j (z j )r! " r e r(cid:18)i = g j (z j ) + "’ j ((cid:18)): Notice that the function ’ j is differentiable (with respect to (cid:18) ) with bounded derivative.Set g j (z j ) = a + bi and ’ j ((cid:18)) = "(cid:11)((cid:18)) + "(cid:12)((cid:18))i . Without loss of generality we may assumethat a; b > 0 and hence for small " we get a + "(cid:11)((cid:18)); b + "(cid:12)((cid:18)) > 0 . In particular, we canexpress g j (z j + "e (cid:18)i ) as g j (z j + "e (cid:18)i ) = arctan( b+(cid:12)((cid:18))a+(cid:11)((cid:18)) ): Therefore d arg(g j )d(cid:18) (z) = "(cid:12) ((cid:18))(a+"(cid:11)((cid:18)))(cid:0)"(cid:11) ((cid:18))(b+"(cid:12)((cid:18)))(a+"(cid:11)((cid:18))) (cid:2) = O(")= o(1): The condition that (cid:13) be smooth can be weakened to piecewise smooth if required. In thatcase we can associate an interior angle (cid:12) j and an interior angle (cid:11) j to each of the zeroes z j ; see the figure below. The same reasoning applies, except that an upper bound on thenumber of preimages of points on L on each @D j is given by b (cid:12)(cid:25) (cid:21) j c + 1 where b(cid:1)c is thefloor function. A bit of algebra yields the following generalization of Theorem 2. Theorem 4.
Let f be a non zero holomorphic function defined on an open domain U and (cid:13) be a piecewise smooth Jordan curve lying inside U . Suppose the zeroes of f lyingon (cid:13) are z ; : : : ; z k . Let (cid:21) j be the multiplicity of the zero at z j , and let (cid:11) j be the interiorangle of (cid:13) at z j . Then for any line L passing through the origin there exist at least
2m + k X j=1 d(cid:21) j (cid:11) j (cid:25) e distinct points on (cid:13) mapped to L by f , where m is the number of zeros of f inside (cid:13) counted according to multiplicities and d(cid:1)e denotes the ceiling function. (cid:11) j at z j . References [1] R. Remmert.
Theory of complex functions , volume 122. Springer Science & BusinessMedia, 2012.[2] W. Rudin.