Power series in a complex variable

COMPLEX VARIABLES
where i and j are the unit vectors along the x- and y-axes, respectively. A similar
expression exists for ∇v, the normal to the curve v(x, y) = constant. Taking the
scalar product of these two normal vectors, we obtain
∂u ∂v
∂u ∂v
+
∂x ∂x ∂y ∂y
∂u ∂u ∂u ∂u
+
= 0,
=−
∂x ∂y ∂y ∂x
∇u · ∇v =
where in the last line we have used the Cauchy–Riemann relations to rewrite
the partial derivatives of v as partial derivatives of u. Since the scalar product
of the normal vectors is zero, they must be orthogonal, and the curves u(x, y) =
constant and v(x, y) = constant must therefore intersect at right angles.
Use the Cauchy–Riemann relations to show that, for any analytic function f = u + iv, the
relation |∇u| = |∇v| must hold.
From (24.9) we have
|∇u|2 = ∇u · ∇u =
∂u
∂x
2
+
∂u
∂y
2
.
Using the Cauchy–Riemann relations to write the partial derivatives of u in terms of those
of v, we obtain
2 2
∂v
∂v
|∇u|2 =
+
= |∇v|2 ,
∂y
∂x
from which the result |∇u| = |∇v| follows immediately. 24.3 Power series in a complex variable
The theory of power series in a real variable was considered in chapter 4, which
also contained a brief discussion of the natural extension of this theory to a series
such as
f(z) =
∞
an z n ,
(24.10)
n=0
where z is a complex variable and the an are, in general, complex. We now
consider complex power series in more detail.
Expression (24.10) is a power series about the origin and may be used for
general discussion, since a power series about any other point z0 can be obtained
by a change of variable from z to z − z0 . If z were written in its modulus and
argument form, z = r exp iθ, expression (24.10) would become
f(z) =
∞
an r n exp(inθ).
n=0
830
(24.11)
24.3 POWER SERIES IN A COMPLEX VARIABLE
This series is absolutely convergent if
∞
|an |r n ,
(24.12)
n=0
which is a series of positive real terms, is convergent. Thus tests for the absolute
convergence of real series can be used in the present context, and of these the
most appropriate form is based on the Cauchy root test. With the radius of
convergence R defined by
1
= lim |an |1/n ,
n→∞
R
(24.13)
the series (24.10) is absolutely convergent if |z| < R and divergent if |z| > R.
If |z| = R then no particular conclusion may be drawn, and this case must be
considered separately, as discussed in subsection 4.5.1.
A circle of radius R centred on the origin is called the circle of convergence
of the series
an z n . The cases R = 0 and R = ∞ correspond, respectively, to
convergence at the origin only and convergence everywhere. For R finite the
convergence occurs in a restricted part of the z-plane (the Argand diagram). For
a power series about a general point z0 , the circle of convergence is, of course,
centred on that point.
Find the parts of the z-plane for which the following series are convergent:
(i)
∞
zn
,
n!
n=0
(ii)
∞
n=0
n!z n ,
(iii)
∞
zn
.
n
n=1
(i) Since (n!)1/n behaves like n as n → ∞ we find lim(1/n!)1/n = 0. Hence R = ∞ and the
series is convergent for all z. (ii) Correspondingly, lim(n!)1/n = ∞. Thus R = 0 and the
series converges only at z = 0. (iii) As n → ∞, (n)1/n has a lower limit of 1 and hence
lim(1/n)1/n = 1/1 = 1. Thus the series is absolutely convergent if the condition |z| < 1 is
satisfied. Case (iii) in the above example provides a good illustration of the fact that
on its circle of convergence a power series may or may not converge. For this
particular series, the circle of convergence is |z| = 1, so let us consider the
convergence of the series at two different points on this circle. Taking z = 1, the
series becomes
∞
1 1 1
1
= 1 + + + + ··· ,
n
2 3 4
n=1
which is easily shown to diverge (by, for example, grouping terms, as discussed in
subsection 4.3.2). Taking z = −1, however, the series is given by
∞
(−1)n
n=1
n
= −1 +
1 1 1
− + − ··· ,
2 3 4
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