NPTEL web course on Complex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 1 / 17
Complex Analysis Module: 5: Consequences of Complex Integration Lecture: 2: Taylor series A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 2 / 17
Consequences of complex integration Taylor Series A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 3 / 17
In this part, we find certain series expansions that are consequences of Cauchy Integral Formula. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 4 / 17
Taylor theorem Theorem Let f (z) be analytic at all points within a circle C 0 with center z 0 and radius ρ 0. Then for every point z within C 0, we have f (z) = f (z 0 ) + f (z 0 )(z z 0 ) + f (z 0 ) (z z 0 ) 2 +... + f n 2! n! (z z 0) n +... (z z 0 ) n = f (z 0 ) + f (n) (z z 0 ) n (z 0 ) = f (n) (z 0 ), n! n! n=1 where the power series converges to f (z) on the disc z z 0 < ρ 0. n=0 A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 5 / 17
The representation of f (z) as an infinite series (z z 0 ) n f (z) = f (n) (z 0 ) is known as Taylor Series. n! n=0 Using the Taylor series expansion, we can write f (n) (z 0 ) = 1 f (z) dz. 2πi (z z 0 ) n+1 We prove this result using Cauchy Integral formula. When z 0 = 0, this reduces to f (z) = f (0) + c n=1 which is known as Maclaurin s series. z n n! f (n) (0) Taylor s series at origin is called Maclaurin s series. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 6 / 17
For the validity of the expansion as a Taylor series, it is essential that f (z) be analytic at all points inside the circle C 0 for then the convergence of Taylor series for f (z) is assured. Hence the greatest radius of C 0 is the distance from the point z 0 to the singularity of f (z) which is nearest to z 0, since we require the function to be analytic at all points within C 0. If f (z) is analytic in a region Ω, containing the point a such that f (a) and all its derivatives f n (a) vanishes, then f(z)=0 in Ω. If f is analytic at z 0, the Taylor series for f around z 0 can be obtained by termwise differentiation of the Taylor series for f around z 0 and converges in the same disk as the series for f. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 7 / 17
Let f and g be analytic functions with Taylor series f (z) = a j (z z 0 ) j and g(z) = b j (z z 0 ) j around the point z 0, then (i) the Taylor series for cf(z), c a constant, is (ii) the Taylor series for f (z) ± g(z) is ca j (z z 0 ) j ; (a j ± b j )(z z 0 ) j. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 8 / 17
The Cauchy product of two Taylor series f (z) = and g(z) = h(z) = a j (z z 0 ) j b j (z z 0 ) j is defined to be the series c j (z z 0 ) j, where c j = j a j i b i. Let f and g be analytic functions with Taylor series f (z) = a j (z z 0 ) j and g(z) = b j (z z 0 ) j around the point z 0. Then the Taylor series for the product fg around z 0 is given by the Cauchy product of these two series. i=0 A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 9 / 17
Comparison with real functions An analytic function is infinitely differentiable and always has a Taylor series expansion with a non-zero radius of convergence. A real function f can also be infinitely differentiable. But it is sometimes impossible to expand it in a power series. An example of such a function is { e 1/x 2, x 0 f (x) = 0, x = 0. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 10 / 17
Comparison with real functions If a real valued function f (x) defined on the whole of real line is extended as complex valued function f (z) analytically, it may not be analytic in the entire complex plane. For example, e x is real analytic function for all real x and its complex extension e z is also an analytic (entire) function. Whereas, 1/(1 + x 2 ) is a real analytic function for all real x. But its complex extension 1/(1 + z 2 ) ceases to be analytic at z = ±i A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 11 / 17
Example The function f (z) = exp(z) is entire. Hence Macluaurin series expansion is valid for all z. Since f (n) (0) = 1 we get f (z) = exp(z) = n=0 z n n! A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 12 / 17
Example The function f (z) = sin z is entire. Hence Macluaurin series expansion is valid for all z. Since f (2n) (0) = 0 and f (2n 1) (0) = ( 1) n, we get f (z) = sin z = ( 1) n z 2n+1 (2n + 1)! n=0 Differentiating both sides of the function gives cos z = ( 1) n z2n (2n)! n=0 Since sinh z = i sin(iz), the Maclaurin series expansion for sinh z and cosh z are also possible. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 13 / 17
Example Expand log(1 + z) in a Taylor Series about z=0 and determine the region of convergence for the resulting series. Solution: Here f (z) = log(1 + z). We find f (z), f (z), f (z)... f n (z) and their values at the point 0. Using the above values, we write the series as f (z) = z z2 2 + z3 3 + z4 zn +... + ( 1)n 1 4 n +... The point z = 1 is the singularity of log(1 + z) nearest the point z=0. Thus the series converges for all values of z within the circle z = 1. A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 14 / 17
Example Expand about the point z 0 = 2i. f (z) = 1 1 z A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 15 / 17
We illustrate the use of geometric series in deriving the Taylor series. Solution Write 1 1 z = 1 1 z + 2i 2i = 1 1 1 2i 1 z 2i 1 2i = 1 1 2i + 1 (1 2i) 2 (z 2i) + 1 (1 2i) 3 (z 2i)2 + The distance from the center z 0 = 2i to the nearest singularity z = 1, is 5. Hence the circle of convergence is z 2i < 5 A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 16 / 17
Exercises Expand in a Maclaurin series and also give the radius of convergence of the function f (z) = ze z2 Use Maclaurin series for e z to expand about z 0 = 1. f (z) = (z 1)e z A.Swaminathan and V.K.Katiyar (NPTEL) Complex Analysis 17 / 17