## Friday, December 6, 2013

### 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 convergenceThis 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 +.......

Rn (x,a)

Where  Rn (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 +......

Rn (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  òC1  f(w) dw /(w – z)

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

Rn (z,a) =  1/ 2 pi òC1  (z- a / w – a)n f(w) dw /(w – z)

And: lim n ® ¥   Rn = 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 = ¥.