Mathematical Methods for Physics and Engineering - Matematica.NET

24.11 TAYLOR AND LAURENT SERIESFurther, it may be proved by induction that the nth derivative of f(z) is alsogiven by a Cauchy integral,f (n) (z 0 )= n! ∮f(z) dz. (24.48)2πi C (z − z 0 )n+1Thus, if the value of the analytic function is known on C then not only may thevalue of the function at any interior point be calculated, but also the values ofall its derivatives.The observant reader will notice that (24.48) may also be obtained by theformal device of differentiating under the integral sign with respect to z 0 inCauchy’s integral formula (24.46):f (n) (z 0 )= 1 ∮∂ n [ ] f(z)dz2πi (z − z 0 )= n!2πi∮C ∂z0nCf(z) dz(z − z 0 ) n+1 .◮Suppose that f(z) is analytic inside and on a circle C of radius R centred on the pointz = z 0 .If|f(z)| ≤M on the circle, where M is some constant, show that|f (n) (z 0 )|≤ Mn!R . (24.49)nFrom (24.48) we have|f (n) (z 0 )| = n!∮2π ∣and on using (24.39) this becomes|f (n) (z 0 )|≤ n! M2π RThis result is known as Cauchy’s inequality. ◭Cf(z) dz(z − z 0 ) n+1 ∣ ∣∣∣,Mn!2πR =n+1R . nWe may use Cauchy’s inequality to prove Liouville’s theorem, which states thatif f(z) is analytic and bounded for all z then f is a constant. Setting n =1in(24.49) and letting R →∞, we find |f ′ (z 0 )| =0andhencef ′ (z 0 ) = 0. Since f(z) isanalytic for all z, wemaytakez 0 as any point in the z-plane and thus f ′ (z) =0for all z; this implies f(z) = constant. Liouville’s theorem may be used in turn toprove the fundamental theorem of algebra (see exercise 24.9).24.11 Taylor and Laurent seriesFollowing on from (24.48), we may establish Taylor’s theorem forfunctionsofacomplex variable. If f(z) is analytic inside and on a circle C of radius R centredon the point z = z 0 ,andz is a point inside C, then∞∑f(z) = a n (z − z 0 ) n , (24.50)n=0853