Complex Analysis - Maths KU

5.5 Cauchy’s Integral Formulas and Their Consequences 283
(b) f(z) =(z − 2 − 2 √ 3 i) 2
(c) f(z) =−2iz 2 +5
[Hint: In parts (b) and (c), it may help to recall from calculus how to find the
relative extrema of a real-valued function of real variable t.]
Focus on Concepts
28. (Cauchy’s Integral Formula) Suppose f is analytic within and on a circle C
of radius r with center at z0. Use (1) to obtain
f(z0) = 1
� 2π
f(z0 + re
2π
iθ ) dθ.
0
This result is known as Gauss’ mean-value theorem and shows that the
value of f at the center z0 of the circle is the average of all the values of f on
the circumference of C.
29. (Fundamental Theorem of Algebra) Suppose
p(z) =anz n + an−1z n−1 + ···+ a1z + a0
is a polynomial of degree n>1 and that z1 is number such that p(z1) =0.
Then
(a) Show that p(z) =p(z) − p(z1) =an(z n − z n 1 )+an−1(z n−1 − z n−1
1 )+···+
a1(z − z1).
(b) Use the result in part (a) to show that p(z) =(z − z1)q(z), where q is a
polynomial of degree n − 1.
(c) Use the result in part (b) to give a sound explanation why the equation
p(z) = 0 has n roots.
30. Use Problem 29 to factor the polynomial
Do not use technology.
p(z) =z 3 +(3− 4i)z 2 − (15 + 4i)z − 1+12i.
31. (Morera’s Theorem) Sometimes in the proof of Theorem 5.15 continuity of
f ′ in D is also assumed. If this is the case, then (3) of Section 5.1 and Green’s
theorem can be used to write �
f(z) dz as
C
�
�
�
f(z) dz = udx− vdy + i vdx + udy
C
C
��
=
C
�
− ∂v
� ��
∂u
− dA + i
∂x ∂y
� �
∂u ∂v
− dA,
∂x ∂y
R
where R denotes the region bounded by C. Supply the next step(s) in the proof
and state the conclusion.
32. (Maximum Modulus Theorem) Critique the following reasoning:
Consider the function f(z) =z 2 +5z − 1 defined the closed circular region
region defined by |z| ≤1. It follows from the triangle inequality, (10) of Section
1.2, that
�
�z 2 +5z − 1 � � ≤|z| 2 +5|z| + |−1|.
R