Complex Analysis Slide 9: Power Series

Complex Analysis Slide 9: Power Series MA201 Mathematics III Department of Mathematics IIT Guwahati August 2015 Complex Analysis Slide 9: Power Series 1 / 37

Learning Outcome of this Lecture We learn Sequence of Complex Numbers and Series of Complex Numbers Sequences & Series of Functions: Pointwise, Absolute, Uniform Convergence Power Series Taylors Theorem / Taylor Series Laurent Theorem/ Laurent Series Complex Analysis Slide 9: Power Series 2 / 37

Sequence of Complex Numbers A sequence of complex numbers is a map from a : N C given by a(n) = a n for n N. It is written as {a n } or (a n ) or < a n >. Definition Let {a n } be a sequence of complex numbers. If there exists a complex number a such that for each ɛ > 0, there exists a natural number N 0 such that a n a < ɛ for all n N 0 then we say that {a n } converges to a. a is called the limit of the sequence {a n }. We write it as {a n } a as n or lim n a n = a. Examples: {a n = (1/n) + 2i} converges to 2i. {a n = n (1/n) + i ((n + 1)/n)} converges to 1 + i. Complex Analysis Slide 9: Power Series 3 / 37

Results If {a n } converges then the limit of {a n } is unique. If {a n } converges then the set S = {a n : n N} is bounded. If {a n } converges then { a n } converges. But converse is NOT true. {a n = x n + i y n } converges to a = x + i y if and only if {x n } x and {y n } y. That is, {a n } a if and only if {R(a n )} R(a ) and {I(a n )} I(a ). Complex Analysis Slide 9: Power Series 4 / 37

Series of Complex Numbers a n = a 0 + a 1 + a 2 + is called an (infinite) series of complex numbers. Definition Let a n be a series of complex numbers. Define the sequence of partial sums by s 0 = a 0 and s n = n a k. If there exists a complex k=0 number s such that the sequence {s n } of partial sums converges to s then we say the series a n converges to s and we write it as a n = s. Complex Analysis Slide 9: Power Series 5 / 37

If the sequence of partial sums does not converge then we say that the series a n diverges. Examples: Let {a n = (1/n 2 ) + i(1/2) n } for n N. Then a n converges. Let {a n = (1/n!) + i(1/2) n } for n = 0, 1,. Then a n converges e + 2i. Let {a n = (1/n) + i(1/2) n } for n N. Then a n diverges. We say that the series a n converges absolutely if a n converges. Results: If a n converges then {a n } 0 as n. If a n converges absolutely then a n converges. But converse is NOT true. Similarly, we can define Sequence of Complex Functions and Series of Complex Functions. Complex Analysis Slide 9: Power Series 6 / 37

Sequence of Functions: Pointwise Convergence Let f n : D C C, for n = 0, 1, be the functions defined on a set D. We say that the sequence {f n (z)} of functions converges (pointwise) to a function f(z) in D, if for each point z 0 D and for each ɛ > 0, there exists a natural number N 0 that may depend on both ɛ and the point z 0 such that f n (z 0 ) f(z 0 ) < ɛ for all n N 0. In this case, we write it as lim n f n(z) = f(z) for z D. If for some point z 0 D, the sequence {f n (z 0 )} does not converge or tends to then we say that the sequence {f n (z)} diverges at the point z = z 0. Example: Let f n (z) = z n for z D = {z C : z < 1} where n N. Let f(z) = 0 for all z D. Then, {f n (z)} converges pointwise to f(z) in D. Complex Analysis Slide 9: Power Series 7 / 37

Series of Functions: Pointwise Convergence Definition Let f n : D C C, for n = 0, 1, be the functions defined on a set D. The series f n (z) of functions converges (pointwise) to a { } n function S(z) in D if the sequence S n (z) = f k (z) of partial sums k=0 converges (pointwise) to the function S(z) in D. In this case, we write it as S(z) = f n (z) for z D. Example: Let f n (z) = z n for z D = {z C : z < 1} where n N. Let S(z) = 1/(1 z) for all z D. Then, f n (z) converges pointwise to S(z) in D. Complex Analysis Slide 9: Power Series 8 / 37

Absolute Convergence Let f n : D C C, for n = 0, 1, be the functions defined on a set D. Definition We say that the sequence {f n (z)} of functions converges absolutely to a function g(z) in D, if for each point z 0 D, the sequence { f n (z 0 ) } converges (pointwise) to g(z 0 ). Definition The series f n (z) converges absolutely to a function T (z) in D if the { } n sequence S n (z) = f k (z) converges (pointwise) to the function T (z) in D. k=0 Complex Analysis Slide 9: Power Series 9 / 37

Uniform Convergence Let f n : D C C, for n = 0, 1, be the functions defined on a set D. Definition We say that the sequence {f n (z)} of functions converges uniformly to a function f(z) in the set D, if for each ɛ > 0, there exists a natural number N (that may depend only on ɛ) such that f n (z) f(z) < ɛ for all n N and for all z D. Definition The series f n (z) converges uniformly to a function S(z) in D if the { } n sequence S n (z) = f k (z) of partial sums converges uniformly to the function S(z) in D. k=0 Complex Analysis Slide 9: Power Series 10 / 37

We now introduce a special type of series of functions, namely, power series. Power Series Complex Analysis Slide 9: Power Series 11 / 37

Power Series Definition A power series about a point z 0 is an infinite series of the form a n (z z 0 ) n. Example-1: The geometric series z n is one of the easiest n=1 examples of a power series. z n Example-2: is another example of a power series. n n=1 (z 3) n Example-3: is another example of a power series. 4n Complex Analysis Slide 9: Power Series 12 / 37

Convergence of Power Series For which values of z does the geometric series z n converge? It is easily seen that 1 z n+1 = (1 z)(1 + z + z 2 + + z n ) so that 1 + z + + z n = 1 zn+1 1 z. If z < 1 then lim z n = 0 and so the geometric series is convergent with z n = 1 1 z. If z > 1 then lim z n = and the series diverges. Complex Analysis Slide 9: Power Series 13 / 37

Recall: Limit Superior of Real Sequences Let {a n } be a sequence of real numbers. lim sup a n = lim (sup{a n, a n+1, }). n n lim inf n a n = lim n (inf{a n, a n+1, }). Other Notation: lim sup is also denoted by lim. Further these concepts lim sup and lim inf are defined only for real sequences and NOT for complex sequences. Results: For a real sequence, lim sup a n and lim inf a n always exist and it may be + or also. Always lim inf a n lim sup a n. If {a n } converges then lim inf a n = lim a n = lim sup a n. Complex Analysis Slide 9: Power Series 14 / 37

Basic Result on Convergence of Power Series Theorem For a given power series 0 R, by then: 1 R a n (z z 0 ) n define the number R, = lim sup a n 1 n (Cauchy-Hadamard Formula) n 1 if z z 0 < R, the series converges absolutely; 2 if z z 0 > R, the series diverges; 3 if 0 < r < R, the series converges uniformly on {z : z r}. Moreover, the number R is the only number having the above said three properties. Complex Analysis Slide 9: Power Series 15 / 37

Continuation of Previous Slide In the previous theorem: The number R is called the radius of convergence of the power series. The circle z z 0 = R is called the circle of the convergence of the series. The open disk z z 0 < R is called the domain of convergence or disk of convergence of the series. Examples: The power series The power series The power series k n z n has radius of convergence R = 1/ k. z n has radius of convergence R =. n! 5 n2 z n has radius of convergence R = 0. Complex Analysis Slide 9: Power Series 16 / 37

Radius of Convergence as the limit of Ratios of Coefficients The radius of the convergence of a power series can be calculated sometimes from the ratio of the coefficients as follows. Theorem If a n (z z 0 ) n is a given power series with radius of convergence R, then R = lim n a n a n+1 if this limit exists (including the limit tending to + in the extended real number system). Example: The power series z n n! has radius of convergence R =. Complex Analysis Slide 9: Power Series 17 / 37

On Circle of Convergence - What happens? On the circle of convergence C : z z 0 = R, power series may converge on C, diverge on C, or converge on some part of C and diverge on the remaining part. The power series z n diverges at all points on the circle of convergence z = 1, since z n does not tend to 0 as n. z n The power series series diverges at the point z = 1 and n n=1 converges at the point z = 1. One can show that this power series converges at all points on the circle z = 1 except at the point z = 1. z n The power series converges at all points on the circle of n2 n=1 convergence z = 1, since z n n 2 1 n 2 <. Complex Analysis Slide 9: Power Series 18 / 37

Theorem Picard s Theorem: Consider the power series a n z n and suppose that: 1 The coefficients a n are real nonnegative numbers. 2 a n a n+1 for n = 1, 2, 3,. 3 {a n } 0 as n. Then the power series a n z n converges at all points of the circle z = 1, except possibly at z = 1, so its radius of convergence is at least 1. Using the above theorem and using the fact 1 n diverges, one can z n conclude that converges at all points on the circle z = 1 except n n=1 at the point z = 1. Complex Analysis Slide 9: Power Series 19 / 37

Properties Let a n (z z 0 ) n and b n (z z 0 ) n be power series with radius of convergence R 1 and R 2 respectively. Then, Sum: (a n + b n )(z z 0 ) n has the radius of convergence R min(r 1, R 2 ). Scalar Multiplication: λa n (z z 0 ) n where λ 0 has the radius of convergence R = R 1. Product: c n (z z 0 ) n where c n = n a k b n k (Cauchy Product) k=0 has the radius of convergence R min(r 1, R 2 ). Complex Analysis Slide 9: Power Series 20 / 37

Properties (Continuation of Previous Slide) Product Coordinatewise: convergence R R 1 R 2. a n b n (z z 0 ) n has the radius of Division Coordinatewise: If b n 0 for all n then has the radius of convergence R R 1 /R 2. a n b n (z z 0 ) n Division of Two Series: If r is the largest real number such that bn (z z 0 ) n 0 for all z {z : z z 0 < r} then an (z z 0 ) n bn (z z 0 ) n has the radius of convergence R min(r, R 1, R 2 ). Complex Analysis Slide 9: Power Series 21 / 37

What can we say about the sum function of a Power Series? Let a n (z z 0 ) n have the radius of convergence R > 0. Let us denote the sum function of this series by f(z). That is, an (z z 0 ) n =: f(z) for z B R (z 0 ). Questions: Is the sum function f(z) differentiable/ analytic in B R (z 0 )? Is the series formed by termwise differentiation na n (z z 0 ) n 1 convergent? If so, what is its radius of convergence R? Set g(z) := na n (z z 0 ) n 1 for z B R (z 0 ). Is g(z) = f (z) for n=1 z B R (z 0 ). What about k-times differentiated series n(n 1) (n k + 1)a n (z z 0 ) n k? n=k Is there any relation between the coefficients a n s and f(z)? Complex Analysis Slide 9: Power Series 22 / 37

Sum function of a Power Series is analytic Theorem: Let Then, a n (z z 0 ) n have radius of convergence R > 0. The function defined by f(z) := a n (z z 0 ) n is analytic in B R (z 0 ) = {z C : z z 0 < R}. For each k 1, n(n 1) (n k + 1)a n (z z 0 ) n k has the n=k radius of convergence R. f (k) (z) = n(n 1) (n k + 1)a n (z z 0 ) n k for z B R (z 0 ). n=k For n = 0, 1,, the coefficient a n = f (n) (z 0 ). n! Complex Analysis Slide 9: Power Series 23 / 37

Example for Previous Theorem The power series R =. The function f(z) = sin z = Observe that ( 1) n z 2n+1 (2n + 1)! d dz (sin z) = cos z = has the radius of convergence ( 1) n z 2n+1 (2n + 1)! has the radius of convergence R =. ( 1) n z 2n (2n)! is analytic in C. and this series also Note: In previous theorem, like termwise differentiation of power series, termwise integration (indefinite integral) is also valid for power series. For example, by doing termwise integration of power series of cos z, we can get the power series of sin z. Complex Analysis Slide 9: Power Series 24 / 37

To think Last theorem says, Given power series, its sum function is analytic (and hence infinitely many times differentiable) in the disk/domain of convergence of the power series. Now, think about converse of above statement. Is it true statement? Question: Let D be an open set and let z 0 D. Given that f(z) is analytic in D. Whether f can have power series representation about z 0? That is, whether f(z) = a n (z z 0 ) n for z B R (z 0 ) D for some R > 0? If the answer is YES, then is there more than one such power series possible? Complex Analysis Slide 9: Power Series 25 / 37

Analytic function has a Power Series Representation Taylor Theorem: Let f(z) be analytic in B R (z 0 ) = {z C : z z 0 < R}. Then, f(z) has a power series expansion around z 0 given by f(z) = a n (z z 0 ) n for z B R (z 0 ) where a n = f (n) (z 0 ) = 1 f(w) dw for n = 0, 1, 2, where n! 2πi C r (w z 0 ) n+1 C r = {z C : z z 0 = r} for any r with 0 < r < R. This series is called the Taylor series of f about the point z 0 and has radius of convergence R. Further, the Taylor series of f about that point z 0 is unique. Proof: Worked out on the board. Complex Analysis Slide 9: Power Series 26 / 37

Example for Taylor Theorem Example: Find the power series of f(z) = e z about the point z 0 = i. Observe that for each n N, f (n) (z) = e z for z C. This gives that f (n) (i) = e i for n = 1, 2,. For each n = 0, 1, 2,, a n = f (n) (i) n! = ei n!. Therefore, the Taylor series of e z about the point z 0 = i is given by e z = e i (z i)n n! and it has radius of convergence R =. Complex Analysis Slide 9: Power Series 27 / 37

Remarks The Taylor series of f(z) about the point z 0 = 0 is called the Maclaurin series of f. If f(z) is analytic in z z 0 < R for some R > 0, then by Taylor theorem, f(z) can be approximated with arbitrarily high precision by a polynomial P n (z) of sufficiently high degree. Alternative Way to find Radius of Convergence of Taylor Series: If f(z) is analytic at z 0, then the radius of convergence R of the Taylor series of f(z) about z = z 0 is the distance from z 0 to the point (singularity) nearest to z 0 at which f(z) fails to be analytic. That is, R = z 0 z where z is the singularity of f nearest to z 0. Complex Analysis Slide 9: Power Series 28 / 37

Comparison between Real Functions and Complex Functions By Taylor theorem, if a function f(z) is infinitely differentiable in an open set D, then f(z) can be expanded in power series in D. This result is not true in case of real valued functions of a real variable. For example, the function f(x) = exp( 1/x 2 ) for x R \ {0} and f(0) = 0 is infinitely many times differentiable in the neighborhood of x 0 = 0, but f(x) can not be represented by a power series about the point x 0 = 0. A real Taylor series of a real valued function f of a real variable converges if and only if the Taylor remainder term goes to zero. In a complex Taylor series, the remainder term is irrelevant; the Taylor series will converge to in the largest disk that one can fit inside the domain of analyticity of f. Complex Analysis Slide 9: Power Series 29 / 37

Analytic at We say that the function f(z) is analytic at z = if the function g(w) = f(1/w) is analytic at w = 0. Thus, we make the change of variable w = 1/z, and we study the behaviour of f(z) at z = by studying the behaviour f(1/w) at w = 0. Examples: f(z) = 1/z 2 is analytic at. f(z) = e 1/z is analytic at. Complex Analysis Slide 9: Power Series 30 / 37

Power series of f(z) about z = If f(z) is analytic at z = then the function g(w) = f(1/w) is analytic at w = 0 and hence g(w) = c n w n for w < r for some r > 0. Thus, f(z) can be represented by a power series as f(z) = cn z for z > R = 1 n r and it is the power series expansion of f(z) about the point z =. Example: e 1/z = 1 n! z n for z > 0. Complex Analysis Slide 9: Power Series 31 / 37

Is power series of f possible at singular points? We now wish to investigate the possibility of representing a function by a power series near singular points. A singular point z 0 is said to be an isolated singular point of f(z) if f(z) is analytic in the punctured disk 0 < z z 0 < r for some r > 0. Then, the function f(z) can be represented by a power series about an isolated singular points. But, in this case the power series of f(z) contains the negative powers of (z z 0 ) also. Similarly, if f(z) is analytic in the annular region r 1 < z z 0 < r 2 and f(z) need not be analytic in the region z z 0 < r 1 then also f(z) can be represented by a power series in the annular region 0 < r 1 < z z 0 < r 2. Complex Analysis Slide 9: Power Series 32 / 37

Laurent Theorem Laurent Theorem: Let f be analytic in the annular region r 1 < z z 0 < r 2. Then f has a series representation given by f(z) = a n (z z 0 ) n + a n (z z 0 ) n for r 1 < z z 0 < r 2 n=1 where the coefficients a n = 1 f(w) dw for any r with 2πi z z 0 =r (w z 0 ) n+1 r 1 < r < r 2. The above series is called the Laurent Series and converges absolutely in r 1 < z z 0 < r 2. Further, it converges uniformly in R 1 z z 0 R 2 where r 1 < R 1 < R 2 < r 2. Moreover this series is unique. Complex Analysis Slide 9: Power Series 33 / 37

The Laurent series for f in the annular region is usually abbreviated In the Laurent series n= n= negative powers of (z z 0 ), namely, 1 n= a n (z z 0 ) n. a n (z z 0 ) n, the series containing the a n (z z 0 ) n = a n (z z 0 ) n is called the principal part of the Laurent series. The series containing the non-negative powers of (z z 0 ), namely, n=1 a n (z z 0 ) n is called the regular part of the Laurent series. Complex Analysis Slide 9: Power Series 34 / 37

Taylor Theorem is a special case of Laurent Theorem If f(z) is analytic at the point z 0 then the Laurent series of f(z) about the point z = z 0 does not contain any negative powers of (z z 0 ). That is, the Laurent series of f(z) has no principal part. Hence the Laurent series reduces to the Taylor series of f(z) about the point z = z 0 in this case. Complex Analysis Slide 9: Power Series 35 / 37

Worked out Example 1 Let f(z) = 1 (z 1)(z 2). 1 Find the Laurent series of f about the point z = 1 (OR) Find the power series expansion of f in the region 0 < z 1 < 1. 2 Find the Laurent series of f about the point z = 2 (OR) Find the power series expansion of f in the region 0 < z 2 < 1. 3 Find the Laurent series of f about the point z = 0 (OR) Find the power series expansion of f in the region z < 1. 4 Find the Laurent series of f about the point z = (OR) Find the power series expansion of f in the region z > 2. 5 Find the Laurent series of f in the annular region 1 < z < 2. Details are Worked Out on the Board. Complex Analysis Slide 9: Power Series 36 / 37

Worked out Example 2 Expand e 1/z in the Laurent series about the point z = 0. Details are Worked Out on the Board. Complex Analysis Slide 9: Power Series 37 / 37