Terms of a complex series depend on 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.

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 again 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:

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}**+***nd 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 here,:

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.

__Practice Problems__

1) Consider the power series:

å

^{¥}_{n = 0}(z)^{n}_{ }**/**n!
Show that the radius of convergence R = ¥

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

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