Solutions to Complex Series Problems

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

 Has a radius of convergence R = 1

 

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

for which the series converges provided z < 1.

(I.e. if  ÷ z ^{n +1}   /   z ⁿ  ÷  <  k, for all k sufficiently large and k < 1 then:

å  z _n   converges absolutely.

For example, expanding the series terms:

 

å^¥ _{n = 0}   z _n    = 1 + z  + z² + z³ +…..

Let z = ½ :

 

å^¥ _{n = 0}   z _n    =  1 + ½  + (½) ²  + (½) ³

 

 

=   1 + ½ + 1/4  + 1/8 +      ……

 

 

So  since the sum is limiting as terms are added, not expanding, we see the series converges absolutely.   BUT this only obtains for all points INSIDE a circle of radius R =1, so the radius of convergence is R =1.

2)Consider the power series:

 

å^¥ _{n = 0}   (z) ⁿ  / n! 

Show that the radius of convergence R = ¥

Apply the ratio test:

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

 

  lim _{n ® ¥}     a ⁿ  _/  a ^{n +1}       =  n + 1 ® ¥ 

as n tends to infinity, hence R = ¥

3) The Maclaurin series for the normal exponential function f(z) = exp(z) is:

 

å^¥ _{n = 0}   (z) ⁿ  / n!  = 1 + z + z² / 2!  + z³ / 3!  +…..

If we now replace z by (-z) we get:

 

å^¥ _{n = 0}   (-z) ⁿ  / n!  = 1 -  z + z² / 2!  -   z³ / 3!  +…..

Which is the Maclaurin series for exp(-z)

 

We have for all z for this series, that as n  ® ¥ ,  the series converges in the limit L = 0 (Check a number of terms to satisfy yourself!)

 

Then for radius of convergence:  R = 1/L   =  1/0 = ¥