AA Proof of Sendov’s Conjecture
G.M. Sofi
Department of Mathematics,Central University of Kashmir,Ganderbal-191201(India).
Email: gmsof i @ cukashmir.ac.in Abstract
Sendov’s conjecture asserts that if all the zeros of a polynomial p lie in the closedunit disk, then there must be a critical point of p within unit distance of each zero. Theconjecture has been proved to be true for many special cases (See [1]). Here is a proof ofthe conjecture for all polynomials. Keywords:
Zeros , crictical points.
The Sendov’s conjecture is an unproven conjecture in complex analysis. This conjecturewas included in the collection
Research P roblems in F unction T heory in 1967 by ProfessorHayman. The conjecture is due to the Bulgarian Mathematician B. Sendov. Till present theconjecture has been found to be true for many special cases(see[1]). What follows is a proofof the conjecture .
Theorem 1 . Let p ( z ) = z n + a n − z n − + a n − z n − + · · · + a z + a be a non-constant poly-nomial with all its zeros z , z , · · · , z n lying inside the closed unit disk | z | ≤
1. Then each ofthe disks | z − z i | ≤ , i = 1 , , · · · , n must contain a zero of p (cid:48) . Proof. [we infact prove that if p has all its zeros z , z , · · · , z n lying inside any closed disk | z − a | ≤ r . Then each of the disks | z − z i | ≤ r, i = 1 , , · · · , n must contain a zero of p (cid:48) andthen taking r = 1 and a = 0 we get the required conclusion.] Case1: (cid:80) ni =1 z i = 0Let ζ i , i = 1 , , · · · , n − p . Assume to the contrary that theconclusion of the theorem does not hold. Then there exists a zero of p say z such that | z − ζ i | > r for all 1 ≤ i ≤ n −
1. Therefore by Gauss-Lucas theorem all ζ i s must lie inthe crescent-shaped region outside | z − z | ≤ r but inside | z − a | ≤ r (as indicated in Figure1 a r X i v : . [ m a t h . G M ] M a r ). We can draw a line through z parallel to the line through the points of intersection ofthe circles | z − a | ≤ r and | z − z | ≤ r (see Figure 1) and it is clear that all the vectors z − ζ i , i = 1 , , · · · , n − z cannotbe the centroid of these z − ζ i s .Figure 1: z − ζ i s are exactly the antiparallel vectors of the ones indicated.But˜ z = z − (cid:80) ni =1 z i n = z − (cid:80) n − i =1 ζ i n − (cid:80) n − i =1 ( z − ζ i ) n −
1a contradiction.