Mathematical Methods for Physics and Engineering - Matematica.NET

COMPLEX VARIABLESwhere a n is given by f (n) (z 0 )/n!. The Taylor expansion is valid inside the regionof analyticity and, for any particular z 0 , can be shown to be unique.To prove Taylor’s theorem (24.50), we note that, since f(z) is analytic insideand on C, we may use Cauchy’s formula to write f(z) asf(z) = 12πi∮Cn=0f(ξ)dξ, (24.51)ξ − zwhere ξ lies on C. Now we may expand the factor (ξ − z) −1 as a geometric seriesin (z − z 0 )/(ξ − z 0 ),1ξ − z = 1 ∑∞ ( ) n z − z0,ξ − z 0 ξ − z 0so (24.51) becomesf(z) = 1 ∮f(ξ)2πi C ξ − z 0= 1 ∞∑2πi= 12πin=0∞ ∑n=0(z − z 0 ) n ∮( z − z0ξ − z 0) nCdξf(ξ)dξ(ξ − z 0 )n+1∞∑(z − z 0 ) n 2πif(n) (z 0 ), (24.52)n!n=0where we have used Cauchy’s integral formula (24.48) for the derivatives off(z). Cancelling the factors of 2πi, we thus establish the result (24.50) witha n = f (n) (z 0 )/n!.◮Show that if f(z) and g(z) are analytic in some region R, and f(z) =g(z) within somesubregion S of R, thenf(z) =g(z) throughout R.It is simpler to consider the (analytic) function h(z) =f(z)−g(z), and to show that becauseh(z) =0inS it follows that h(z) = 0 throughout R.If we choose a point z = z 0 in S, then we can expand h(z) in a Taylor series about z 0 ,h(z) =h(z 0 )+h ′ (z 0 )(z − z 0 )+ 1 2 h′′ (z 0 )(z − z 0 ) 2 + ··· ,which will converge inside some circle C that extends at least as far as the nearest part ofthe boundary of R, sinceh(z) isanalyticinR. But since z 0 lies in S, we haveh(z 0 )=h ′ (z 0 )=h ′′ (z 0 )=···=0,and so h(z) =0insideC. We may now expand about a new point, which can lie anywherewithin C, and repeat the process. By continuing this procedure we may show that h(z) =0throughout R.This result is called the identity theorem and, in fact, the equality of f(z) andg(z)throughout R follows from their equality along any curve of non-zero length in R, orevenat a countably infinite number of points in R. ◭So far we have assumed that f(z) is analytic inside and on the (circular)contour C. If,however,f(z) has a singularity inside C at the point z = z 0 ,thenitcannot be expanded in a Taylor series. Nevertheless, suppose that f(z) has a pole854