Page 44 - DMTH402_COMPLEX_ANALYSIS_AND_DIFFERENTIAL_GEOMETRY

Unit 4: Integration

In other words, F’(z) = f(z), and so, just as promised, f has an antiderivative! Let’s summarize Notes
what we have shown in this section:
Suppose f : D  C is continuous, where D is connected and every point of D is an interior point.
Then f has an antiderivative if and only if the integral between any two points of D is path
independent.

4.4 Summary



If  : D  C is simply a function on a real interval D = [, ], then the integral  (t)dt of


course, simply an ordered pair of everyday 3 grade calculus integrals:
rd
  

  (t)dt   x(t)dt i y(t)dt,

  
where g(t) = x(t) + iy(t).
A Riemann sum associated with the partition P is just what it is in the real case:

n
*

S(P)   f(z ) z ,
j
j
j 1

where z is a point on the arc between z and z, and zj = z – z .
*
j
j–1
j–1
j
j
Suppose D is a subset of the reals and  : D  C is differentiable at t. Suppose further that

g is differentiable at (t). Then let’s see about the derivative of the composition g((t). It is,
in fact, exactly what one would guess. First,
g((t)) = u(x(t), y(t)) + iv(x(t), y(t)),
where g(z) = u(x, y) + iv(x, y) and (t) = x(t) + iy(t).

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
In other words, F’(z) = f(z), and so, just as promised, f has an antiderivative! Let’s summarize
what we have shown in this section:
Suppose f : D  C is continuous, where D is connected and every point of D is an interior
point. Then f has an antiderivative if and only if the integral between any two points of D
is path independent.

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