Complex Analysis - Maths KU

3.1 Differentiability and Analyticity 151
33. In this problem you are guided through the start ofthe proofofthe proposition:
If functions f and g are analytic at a point z 0 and f(z0) =0,g(z0) =0,
but g ′ (z0) �= 0,then lim
z→z0
f(z)
g(z) = f ′ (z0)
g ′ (z0) .
Proof We begin with the hypothesis that f and g are analytic at a point z0.
Analyticity at z0 implies f and g are differentiable at z0. Hence from (12) both
limits,
f ′ (z0) = lim
z→z0
f(z) − f(z0)
z − z0
and g ′ (z0) = lim
z→z0
g(z) − g(z0)
z − z0
exist. But since f(z0) =0,g(z0) = 0, the foregoing limits are the same as
f ′ (z0) = lim
z→z0
f(z)
z − z0
f(z)
Now examine lim and finish the proof.
z→z0 g(z)
and g ′ (z0) = lim
z→z0
g(z)
.
z − z0
34. In this problem you are guided through the start ofthe proofofthe product
rule.
Proof We begin with the hypothesis that f and g are differentiable at a point
z; that is, each ofthe following limits exist:
f ′ (z) = lim
∆z→0
(a) Justify the equality
f(z +∆z) − f(z)
∆z
and g ′ (z) = lim
∆z→0
g(z +∆z) − g(z)
∆z
d
f(z +∆z)g(z +∆z) − f(z)g(z)
[f(z)g(z)] = lim
dz ∆z→0
∆z
� �
f(z +∆z) − f(z)
g(z +∆z) − g(z)
= lim
g(z +∆z)+f(z) .
∆z→0 ∆z
∆z
(b) Use Definition 2.9 to justify lim g(z +∆z) =g(z).
∆z→0
(c) Use Theorems 2.2(ii) and 2.2(iii) to finish the proof.
35. In Problem 21 you were asked to prove that f(z) =¯z was nowhere differentiable.
In the event you used the polar form of the complex number ∆z you may skip
this problem. Ifyou didn’t use ∆z = |∆z| (cos θ + i sin θ), then continue.
(a) If f(z) =¯z, show that
f(z +∆z) − f(z) cos θ − i sin θ
lim
= lim
∆z→0 ∆z
∆z→0 cos θ + i sin θ .
(b) Explain succinctly why the result in part (a) shows that f is nowhere
differentiable.