SERIES AND LIMITS
r=−expiθ. Therefore, on the the circle of convergence we have
P(z)=
1
1+expiθ
.
Unlessθ=πthis is a finite complex number, and soP(z) converges at all points on the
circle|z|=2exceptatθ=π(i.e.z=−2), where it diverges. Note thatP(z)isjustthe
binomial expansion of (1 +z/2)−^1 , for which it is obvious thatz=−2 is a singular point.
In general, for power series expansions of complex functions about a given point in the
complex plane, the circle of convergence extends as far as the nearest singular point. This
is discussed further in chapter 24.
Note that the centre of the circle of convergence does not necessarily lie at the
origin. For example, applying the ratio test to the complex power series
P(z)=1+
z− 1
2
+
(z−1)^2
4
+
(z−1)^3
8
+···,
we find that for it to converge we require|(z−1)/ 2 |<1. Thus the series converges
forzlying within a circle of radius 2 centred on the point (1, 0) in the Argand
diagram.
4.5.2 Operations with power series
The following rules are useful when manipulating power series; they apply to
power series in a real or complex variable.
(i) If two power seriesP(x)andQ(x) have regions of convergence that overlap
to some extent then the series produced by taking the sum, the difference or the
product ofP(x)andQ(x) converges in the common region.
(ii) If two power seriesP(x)andQ(x) converge for all values ofxthen one
series may be substituted into the other to give a third series, which also converges
for all values ofx. For example, consider the power series expansions of sinxand
exgiven below in subsection 4.6.3,
sinx=x−
x^3
3!
+
x^5
5!
−
x^7
7!
+···
ex=1+x+
x^2
2!
+
x^3
3!
+
x^4
4!
+···,
both of which converge for all values ofx. Substituting the series for sinxinto
that forexwe obtain
esinx=1+x+
x^2
2!
−
3 x^4
4!
−
8 x^5
5!
+···,
which also converges for all values ofx.
If, however, either of the power seriesP(x)andQ(x) has only a limited region
of convergence, or if they both do so, then further care must be taken when
substituting one series into the other. For example, supposeQ(x) converges for
allx, butP(x)onlyconvergesforxwithin a finite range. We may substitute