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Unit 5: Cauchys Theorem
Now, lets look at the derivative of I(s). We assume everything is nice enough to allow us to Notes
differentiate under the integral:
d 1 H(t,s)
I(s) = f(H(t,s)) dt
ds 0 t
1 H(t,s) H(t,s) 2 H(t,s)
= f'(H(t,s)) f(H(t,s)) dt
0 t t s t
1 H(t,s) H(t,s) 2 H(t,s)
= f'(H(t,s)) f(H(t,s)) dt
0 t t t s
1
= f(H(t,s)) H(t,s) dt
0 t s
= f(H(1, s)) H(1,s) f(H(0, s)) H(0,s) .
s s
But we know each H(t, s) describes a closed curve, and so H(0, s) = H(1, s) for all s. Thus,
I'(s) f(H(1,s)) H(1,s) f(H(0,s)) H(0,s) 0.
s s
which means I(s) is constant! In particular, I(0) = I(1), or
f(z)dz f(z)dz
C 1 C 2
This is a big deal. We have shown that if C and C are closed curves in a region D that are
2
1
z
z
d
z
homotopic in D, and f is analytic on D, then f ( )dz f ( ) .
C 1 C 2
An easy corollary of this result is the celebrated Cauchys Theorem, which says that if f is
analytic on a simply connected region D, then for any closed curve C in D,
f(z)dz 0.
C
In court testimony, one is admonished to tell the truth, the whole truth, and nothing but the
truth. Well, so far in this chapter, we have told the truth, but we have not quite told the whole
truth. We assumed all sorts of continuous derivatives in the preceding discussion. These are not
always necessaryspecifically, the results can be proved true without all our smoothness
assumptionsthink about approximation.
Example 2:
Look at the picture below and convince your self that the path C is homotopic to the closed path
consisting of the two curves C and C together with the line L. We traverse the line twice, once
1
2
from C to C and once from C to C .
2
1
1
2
Observe then that an integral over this closed path is simply the sum of the integrals over C and
1
C , since the two integrals along L, being in opposite directions, would sum to zero. Thus, if f is
2
analytic in the region bounded by these curves (the region with two holes in it), then we know
that
f(z)dz f(z)dz f(z)dz.
C C 1 C 2
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