Complex Analysis - Maths KU

Be careful when using Ln z as an
antiderivative of 1/z.
2i
y
C
Figure 5.41 Contour for Example 4
3
x
☞
5.4 Independence of Path 269
Hence, we have shown that
F (z +∆z) − F (z)
lim
= f(z) or F
∆z→0 ∆z
′ (z) =f(z). ✎
If f is an analytic function in a simply connected domain D, it is necessarily
continuous throughout D. This fact, when put together with the results
in Theorem5.6 and (7), leads to a theoremwhich states that an analytic
function possesses an analytic antiderivative.
Theorem 5.8 Existence of an Antiderivative
Suppose that a function f is analytic in a simply connected domain D.
Then f has an antiderivative in D; that is, there exists a function F such
that F ′ (z) =f(z) for all z in D.
In (21) of Section 4.1 we saw for |z| > 0, −π 0, in other words, D is the first quadrant in the z-plane. In this
case, Ln z is an antiderivative of 1/z since both these functions are analytic
in D. Hence by (4),
� 2i
From(14) of Section 4.1,
3
1
dz =Lnz|2i 3 =Ln2i−Ln 3.
z
Ln 2i = log e 2+ π
2 i and Ln 3 = log e 3