Complex Analysis - Maths KU

310 Chapter 6 Series and Residues
Remarks
(i) Ifzn = an + ibn then the nth term of the sequence of partial
sums for �∞ k=1 zk can be written Sn = �n k=1 (ak + ibk) =
�n k=1 ak + i �n k=1 bk.Analogous to Theorem 6.2, �∞ k=1 zk converges
to a number L = a+ib if and only if Re(Sn) = �n k=1 ak converges to
a and Im(Sn) = �n k=1 bk converges to b.See Problem 35 in Exercise
6.1.
(ii) When written in terms of summation notation, a geometric series
may not be immediately recognizable as equivalent to (2).In summation
notation a geometric series need not start at k = 1 nor
does the general term have to appear precisely as azk−1 .At first
glance �∞ ik+2
k=3 40 does not appear to match the general form
2k−1 �∞ k=1 azk−1of a geometric series.However by writing out three
terms,
∞�
k=3
40 ik+2
=
2k−1 a
� �� �
40 i5
2
2 +
az
� �� �
40 i6
+
23 az 2
� �� �
40 i7
+ ···
24 we are able make the identifications a =40 � i5�22� and z = i/2 on
the right-hand side of the equality.Since |z| = 1
2 < 1 the sum of the
series is given by (5):
∞�
k=3
40 ik+2
=
2k−1 40 i5
2 2
1 − i
2
= −4+8i.
(iii) Although
�
we have not proved it, it bears repeating: A power series
∞
k=0 ak(z − z0) k , z �= z0, always possesses a radius of convergence
R that is either positive or ∞.We have seen in the discussion prior
to Example 6 that the ratio and root tests lead to
�
1 �
= lim �
R n→∞�
an+1
an
�
�
�
� and
1 �
n
= lim |an|
R n→∞
assuming the appropriate limit exists.Since these formulas depend
only on the coefficients, it is easy to make up examples where neither
limn→∞|an+1/an| nor limn→∞ |an| 1/n exist.What is R if neither of
these limits exist? See Problems 45 and 46 in Exercises 6.1.
EXERCISES 6.1 Answers to selected odd-numbered problems begin on page ANS-18.
In Problems 1–4, write out the first five terms of the given sequence.
1. {5i n } 2. {2+(−i) n }
3. � 1+e nπi�
4. {(1 + i) n } [Hint: Write in polar form.]