Page 43 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Complex Analysis and Differential Geometry

Notes 260 728
=   i
27 2187
Now, instead of assuming f has an antiderivative, let us suppose that the integral of f between
any two points in the domain is independent of path and that f is continuous. Assume also that
every point in the domain D is an interior point of D and that D is connected. We shall see that
in this case, f has an antiderivative. To do so, let z be any point in D, and define the function F
0
by
F(z)   f(z)dz,
C z
where C is any path in D from z to z. Here is important that the integral is path independent,
z
0
otherwise F(z) would not be well-defined.

Notes Also we need the assumption that D is connected in order to be sure there
always is at least one such path.

Now, for the computation of the derivative of F:

F(z + z) – F(z) =  f(s)ds
L  z
where L is the line segment from z to z + z.
z
Figure 4.2

1
Next, observe that  ds   Thus, f(z) = z  (f(s) – f(z))ds.
z.
L  z  L  z
Now then,

1 1 z max{ f(s) f(z) : s L }
 z  (f(s)– f(z))ds   z     z
L  z
 max{|f(s) – f(z)| : s  L }.
z


We know f is continuous at z, and so lim max{ f(s) f(z) : s L }  0. Hence,

z
 z 0 

lim F(z   z) F(z)  f(z) = lim   1  (f(s) f(z))ds 

0
 z 0  z  z   z 
 L  z 
= 0
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