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 + z^{2}+ 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

Apply the ratio test to write:
lim^{¥}_{n = 1}a^{n}(z- a)^{n}_{ }_{}_{ 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)

R

**×**(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 (

**) then for all z inside C:***see diagram*
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**_{C}**1**^{ }f(w) dw /(w – z)
One can also show, using appropriate
substitutions, that the remainder:

R

_{n}(z,a) = 1/ 2 pi**ò**_{C}1^{ }(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 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 = ¥.

*the Maclaurin series*we use the
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