1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

3.1 • DIFFERENTIABLE AND ANALYTIC FUNCTIONS 97

We can establish Equation (3-8) from Theorem 3.1. Letting h (z) = f (z) g (z)

and using Definition 3.1, we write

h' (.zo) = Jim h(z)- h(zo) = Jim f (z)g(z)- 1 (zo)g(zo).

z-.io z - zo z-zo z - zo

If we subtract and add the term f (zo) g (z) in the numerator, we get

h' (zo) = Jim f (z) g (z) - f (zo) g (z) + f (zo) g (z) - f (zo) g (zo)

z-zo z-zo

= Jim f (z) g (z) - f (zo) g (z) + lim f (zo) g (z) - f (zo) g (zo)

z-zo Z - zo z-zo Z - Zo

Ii

f(z)-f(zo)

1

. ( )+/( ) Ii g(z)- g(zo)

= m ungz zo m.

z-zo z - zo z~zo z-zo z - zo

Using the definition of the derivative given by Equation (3-1) and the continuity
of g, we obtain h' (zo) = f' (zo) g (zo) + f (zo) g' (zo), which is what we wanted
to establish. We leave the proofs of t he other rules as exercises.
The rule for differentiating polynomials carries over to the complex case as
well. If we let P (z) be a polynomial of degree n, so

P(z) = ao + a1z + a2z^2 + · · · + a,.zn,

then mathematical induction, along with Equations (3-5) and (3-7), gives

P' (z) = a1 + 2a2z + 3a3z^2 + · · · + nanzn- i.

Again, we leave the proof of this result as an exercise.