Mathematical Methods for Physics and Engineering - Matematica.NET

25.3 LOCATION OF ZEROSpolynomials, only the behaviour of a single term in the function need be consideredif the contour is chosen appropriately. For example, for a polynomial,f(z)+g(z) = ∑ N0 b iz i , only the properties of its largest power, taken as f(z), needbe investigated if a circular contour is chosen with radius R sufficiently large that,on the contour, the magnitude of the largest power term, |b N R N |, is greater thanthe sum of the magnitudes of all other terms. It is obvious that f(z) =b N z N hasN zeros inside |z| = R (all at the origin); consequently, f + g also has N zerosinside the same circle.The corresponding situation, in which only the properties of the polynomial’ssmallest power, again taken as f(z), need be investigated is a circular contourwith a radius R chosen sufficiently small that, on the contour, the magnitude ofthe smallest power term (usually the constant term in a polynomial) is greaterthan the sum of the magnitudes of all other terms. Then, a similar argument tothat given above shows that, since f(z) =b 0 has no zeros inside |z| = R, neitherdoes f + g.Aweakformofthemaximum-modulus theorem may also be deduced. Thisstates that if f(z) is analytic within and on a simple closed contour C then |f(z)|attains its maximum value on the boundary of C. The proof is as follows.Let |f(z)| ≤M on C with equality at at least one point of C. Now supposethat there is a point z = a inside C such that |f(a)| >M. Then the functionh(z) ≡ f(a) is such that |h(z)| > |−f(z)| on C, and thus, by Rouché’s theorem,h(z) andh(z) − f(z) have the same number of zeros inside C. Buth(z) (≡ f(a))has no zeros inside C, and, again by Rouché’s theorem, this would imply thatf(a) − f(z) has no zeros in C. However, f(a) − f(z) clearly has a zero at z = a,and so we have a contradiction; the assumption of the existence of a point z = ainside C such that |f(a)| >Mmust be invalid. This establishes the theorem.The stronger form of the maximum-modulus theorem, which we do not prove,states, in addition, that the maximum value of f(z) is not attained at any interiorpoint except for the case where f(z) is a constant.◮ Show that the four zeros of h(z) =z 4 + z +1 occur one in each quadrant of the Arganddiagram and that all four lie between the circles |z| =2/3 and |z| =3/2.Putting z = x and z = iy shows that no zeros occur on the real or imaginary axes. Theymust therefore occur in conjugate pairs, as can be shown by taking the complex conjugateof h(z) =0.Now take C as the contour OXY O shown in figure 25.5 and consider the changes∆[arg h] in the argument of h(z) asz traverses C.(i) OX: argh is everywhere zero, since h is real, and thus ∆ OX [arg h] =0.(ii) XY : z = R exp iθ andsoargh changes by an amount∆ XY [arg h] =∆ XY [arg z 4 ]+∆ XY [arg(1 + z −3 + z −4 )]{=∆ XY [arg R 4 e 4iθ ]+∆ XY arg[1+O(R −3 )] }=2π +O(R −3 ). (25.22)881