Page 622 - Handbook Of Integral Equations

When considering the point at infinity, we must replace the difference z – z 0 by 1/z.If z 0 ∈ L, then
1
we define the order of the function to be equal to n.
2
Let N Ω and P Ω (N L and P L ) be the numbers of zeros and poles on the domain (on the contour,
respectively), where each zero and pole is taken according to its multiplicity. Let [δ] L denote the
increment of the variable δ when going around the contour in the positive direction. As usual, by
the positive direction we mean the direction the domain under consideration remains to the left of
the contour.
THE PRINCIPLE OF ARGUMENT. Let f(z) be a single-valued analytic function in a multiply
connected domain Ω bounded by a smooth contour L = L 0 + L 1 + ··· + L m except for finitely many
points at which f(z) may have poles, and let f(z) be continuous in the closed domain Ω ∪ L (except
for these poles) and have at most finitely many zeros of integer order on the contour. In this case,
the following formula holds:
1 1
N Ω – P Ω + (N L – P L )= [arg f(z)] L .
2 2π
THE GENERALIZED LIOUVILLE THEOREM. Assume that a function f(z) is analytic on the entire
complex plane except for points a 0 = ∞, a k (k =1, ... , n), where it has poles, and that the principal
parts of the Laurent series expansions of f(z) at the poles have the form
0 2
0
Q 0 (z)= c z + c z + ··· + c 0 z m 0 at the point a 0 ,
1 2 m 0
k k k
1 c 1 c 2 c
Q k = + 2 + ··· + m k at the points a k .
z – a k z – a k (z – a k ) (z – a k ) m k
Then f(z) is a rational function, and can be represented by the formula
n
1
f(z)= C + Q 0 (z)+ Q k ,
z – a k
k=1
where C is a constant. In particular, if the only singularity of f(z) is a pole of order m at infinity,
then f(z) is a polynomial of degree m,
m
f(z)= c 0 + c 1 z + ··· + c m z .
The following notation is customary:
(a) f(z) is the function conjugate to a given function f(z);
z
z
(b) f(¯) is the function obtained from f(z) by replacing z by ¯, i.e., y by –y in f(z);
¯
¯
(c) f(z) is the function defined by the condition f(z)= f(¯z).
If z = x + iy and f(z)= u(x, y)+ iv(x, y), then
¯
z
f(z)= u(x, y) – iv(x, y), f(¯)= u(x, –y)+ iv(x, –y), f(z)= u(x, –y) – iv(x, –y).
k
In particular, if f(z) is given by a series f(z)= n
c k z , then
k=0
n n n
¯
k k k
z
z
z
f(z)= ¯ c k ¯ , f(¯)= c k ¯ , f(z)= ¯ c k z .
k=0 k=0 k=0
For a function represented by a Cauchy type integral
1 ϕ(τ)
f(z)= dτ,
2πi L τ – z
we have
1 ϕ(τ) 1 ϕ(τ) 1 ϕ(τ)
¯
f(z)= – dτ, f(¯)= dτ, f(z)= – dτ.
z
z
2πi L ¯ τ –¯ 2πi L τ –¯ z 2πi L ¯ τ – z
¯
Note that if a function satisfies the condition f(z)= f(z), then it takes real values for all real values
of z. The converse assertion also holds.
© 1998 by CRC Press LLC

© 1998 by CRC Press LLC
Page 605