14—Complex Variables 349Necessarily iffis analytic atz 0 it will also be analytic at every point within the disk ε
|z−z 0 |< ε. This follows because at any pointz 1 within the original disk you have a disk
centered atz 1 and of radius(ε−|z 1 −z 0 |)/ 2 on which the function is differentiable.
The common formulas for differentiation are exactly the same for complex variables as
they are for real variables, and their proofs are exactly the same. For example, the product
formula:
f(z+ ∆z)g(z+ ∆z)−f(z)g(z)
∆z
=
f(z+ ∆z)g(z+ ∆z)−f(z)g(z+ ∆z) +f(z)g(z+ ∆z)−f(z)g(z)
∆z
=
f(z+ ∆z)−f(z)
∆z
g(z+ ∆z) +f(z)
g(z+ ∆z)−g(z)
∆z
As∆z→ 0 , this becomes the familiarf′g+fg′. That the numbers are complex made no difference.
For integer powers you can use induction, just as in the real case: dz/dz= 1and
Ifdzn
dz
=nzn−^1 , then use the product rule
dzn+1
dz
=
d(zn.z)
dz
=nzn−^1 .z+zn.1 = (n+ 1)zn
The other differentiation techniques are in the same spirit. They follow very closely from thedefinition. For example, how do you handle negative powers? Simply note thatznz−n= 1and use
the product formula. The chain rule, the derivative of the inverse of a function, all the rest, are close
to the surface.
14.2 Integration
The standard Riemann integral of section1.6is
∫baf(x)dx= lim
∆xk→ 0∑N
k=1f(ξk)∆xk
The extension of this to complex functions is direct. Instead of partitioning the intervala < x < binto
Npieces, you have to specify a curve in the complex plane and partitionitintoNpieces. The interval
is the complex number∆zk=zk−zk− 1.
∫
Cf(z)dz= lim
∆zk→ 0∑N
k=1f(ζk)∆zk
z 0 z^1
z 2 z 3 z (^4) z
5
z 6
ζ 1
ζ 2 ζ^3 ζ 4 ζ 5 ζ^6
Just asξkis a point in thekthinterval, so isζka point in thekthinterval along the curveC.
How do you evaluate these integrals? Pretty much the same way that you evaluate line integrals
in vector calculus. You can write this as
∫
C