Looking At Complex Power Series

With complex power series we move one more step toward Laurent series and then calculus of residues.  Terms of a complex series depend no the complex variable z. Most such series are in the form of power series, such as:

å^¥ _{n = 0}   z _n  =  1 + z + z² + z ³ + …..

We understand that in many cases such series will only converge if z is confined to a certain region. In the case of the series above, it converges (according to the ratio test), provided that ÷ z÷ < 1. 

In effect, the above power series converges for all points inside a circle of radius R = 1. This is called the radius of convergence.  This concept of radius of convergence can be applied to every power series. Thus, if a power series is convergent on a circle of some radius r then it is absolutely convergent everywhere inside this circle.  The key point is that R must be finite in order for us to assess a convergence. Else, we say the series diverges.


New Example: 

Consider the complex series:  å^¥ _{n = 1}   (z- a) ⁿ  / n 

Using the ratio test we look at:

lim _{n ® ¥}      [(z- a) ^{n +1}   / n + 1 /   (z- a) ⁿ  / n ]

= (z – a) lim _{n ® ¥}   [n/  n + 1]  =  ÷ z – a ÷

We thereby have convergence for all z such that: 
 ÷ z – a ÷ < 1

And we have divergence for all z such that:  ÷ z – a ÷ >  1

 We have ambiguous or no conclusive test result for: ÷ z – a ÷ = 1

Example (3):    å^¥ _{n = 1}   a ⁿ  (z- a) ⁿ 

Apply the ratio test to write: lim _{n ® ¥}      [a ⁿ⁺¹ (z- a) ^{n +1}    /  a ⁿ  (z- a) ⁿ ]

=  ÷ z – a ÷  lim _{n ® ¥}   [a ⁿ⁺¹ /  a ⁿ ] =   ÷ z – a ÷ × L

We have convergence for all z such that:

 ÷ z – a ÷ × L < 1 or ÷ z – a ÷  < 1/ L

We have divergence for all z such that:

÷ z – a ÷ × L >  1 or ÷ z – a ÷  >  1/ L

No test result for:  ÷ z – a ÷ =   1/ L

Using Taylor’s Theorem in the complex context:

Recall Taylor’s Theorem:

 Let f be a function that is continuous together with its first n + 1 derivatives on an interval containing a and x. Then the value of the function at x is given by:

f(x) = f(a) + f’(a) × (x – a) + f” (a)/ 2! (x – a)² + f”” (a) / 3! (x – a)³ +.......

R_n (x,a)

Where  R_n (x,a)  denotes the remainder.

This can also be applied to the case of complex functions. If f(z) is analytic inside a circle C with center at a (see diagram) then for all z inside C:

f(z) = f(a) + f’(a) × (z – a) + f” (a)/ 2! (z – a)² + f”” (a) / 3! (z – a)³ +......

R_n (z,a)

By reference consider the diagram shown.  Here, we let z be any point inside the circle C. We next construct a circle C1 with center at a, and which encloses z. Then by Cauchy’s integral formula:

f(z) =  1/2 pi  ò_C1  f(w) dw /(w – z)

 One can also show, using appropriate substitutions, that the remainder:

R_n (z,a) =  1/ 2 pi ò_C1  (z- a / w – a)ⁿ f(w) dw /(w – z)

And: lim _{n ® ¥}   R_n = 0. Interested readers are invited to show this by taking:

1 / w – z  = 1/ {w – a) – (z – a) = 1/ (w – a)[ 1/ 1 –(z –a)/(w – a)]

then expanding into terms of a complex series and substituting into Cauchy’s integral formula.

Problems for Math Mavens:

1) Confirm that for the power series: å^¥ _{n = 0}   z _n 

The radius of convergence is R= 1. (I.e. apply the ratio test to show this)

2) Consider the complex power series:

å^¥ _{n = 0}   (z) ⁿ  / n! 

Show that the radius of convergence R = ¥

3) In the Maclaurin series we use the Taylor series but with a = 0. Use the Maclaurin series to expand the function f(z) = exp(-z) to show that the radius of convergence R = ¥.