Prove: If f(z) is analytic inside and on the boundary C of a simply-connected region R and a is any point inside C then
Proof. The function f(z)/(z-a) is analytic inside and on C except at the point z = a.. Construct a circle Γ of radius ρ at point a as shown in Fig. 1. Then by the principle of the deformation of contours
The right member of 1) can be written as
We compute the second term first. On the circle Γ
3) z = a + ρ(cos θ + i sinθ) .
Noting that a and ρ are constant, we get
4) dz = ρ(-sin θ + i cos θ)dθ = iρ(cos θ + i sin θ)dθ
From 3) we get
z - a = ρ(cos θ + i sinθ)
and the second term of 2) becomes
Equation 2) then becomes
In the above equation let us take limits as ρ 0. The left side and the first term of the right side will remain unchanged. We will show that the limit of the second term is equal to zero. In order to show that it approaches zero as ρ 0 note that
This means that the quantity [f(z)- f(a)]/(z-a) is bounded i.e.
Applying property 5 of the integral which states that
where |f(z)| M ( i.e. M is an upper bound of |f(z)| on C) and L is the length of C, we get
which approaches zero as ρ 0.
Equation 6) then becomes
Mathematics, Its Content, Methods and Meaning. Vol II
Spiegel. Complex Variables (Schaum)
Wylie. Advanced Engineering Mathematics