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
÷ z – a ÷ < 1
And we have divergence
for all z such that: ÷ 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
Using Taylor ’s Theorem in the complex context:
Recall Taylor ’s
Theorem:
f(x) = f(a) + f’(a) × (x
– a) + f” (a)/ 2! (x – a)2 + f”” (a) / 3! (x – a)3 +.......
Rn (x,a)
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)
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 = ¥.
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