Page 148 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 13: Schwarz's Reflection Principle

Notes
Example: Show that the circle of convergence of the power series
f(z) = 1 + z + z + z + z +
8
2
4
is a natural boundary of its sum function
Solution. We have
 n
f(z) = 1 +  z 2
n 0

Evidently, |z| = 1 is the circle of convergence of the power series. We write

q 
f(z) = 1 +  z 2 n   z 2 n = f (z) + f (z), say.
2
1
0 q 1

Let P be a point at z = r e 2ip/ q
2 lying outside the circle of convergence, where p and q are integers
and r > 1.
We examine the behaviour of f(z) as P approaches the circle of convergence through radius
vector.

P

Now, z 2 n  r e 2 ip2 /2 q  r e  ip2 n q 1 
n
n
n
2

2

q
n
2


 f (z) = 1 +  r e  ip2 n 1 q
1
n 0

which is a polynomial of degree 2 and tends to a finite limit as r1
q
 n n 1 q
Also f (z) =  r e  ip2  
2
2
q 1

Here, n > q so 2 n+1-q is an even integer and thus
e  ip2 n 1 q  = 1

 n
 f (z) =  r 2  as r1
2
q 1

Thus, f(z) = f (z) + f (z), when z = e 2 ip/2 q .

2
1
Hence, the point z = is a singularity of f(z). This point lies on the boundary of the circle |z| = 1.
But any arc of |z| = 1, however small, contains a point of the form e 2 ip /2 q , where p and q are

integers. Thus, the singularities of f(z) are everywhere dense on |z| = 1 and consequently
|z| = 1 constitutes the natural boundary for the sum function of the given power series.
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