## Saturday, December 7, 2013

### 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 = ¥