Complex Analysis - Maths KU

1.2 Complex Plane 13
From (8) with z2 replaced by −z2, we also find
|z1 − z2| ≥ � � |z1|−|z2| � �. (10)
In conclusion, we note that the triangle inequality (6) extends to any finite
sum of complex numbers:
|z1 + z2 + z3 + ···+ zn| ≤|z1| + |z2| + |z3| + ···+ |zn| . (11)
The inequalities (6)–(10) will be important when we work with integrals of a
function of a complex variable in Chapters 5 and 6.
EXAMPLE 3 An Upper Bound
�
�
Find an upper bound for �
−1
�z
4 �
�
�
− 5z +1�
if |z| =2.
Solution By the second result in (3), the absolute value of a quotient is the
quotient of the absolute values.Thus with |−1| = 1, we want to find a positive
real number M such that
1
|z4 ≤ M.
− 5z +1|
To accomplish this task we want the denominator as small as possible.By
(10) we can write
�
� z 4 − 5z +1 � � = � �z 4 − (5z − 1) � � ≥ � � � �z 4 � � −|5z − 1| � � . (12)
But to make the difference in the last expression in (12) as small as possible, we
want to make |5z − 1| as large as possible.From (9), |5z − 1| ≤|5z| + |−1| =
5|z| + 1.Using |z| = 2, (12) becomes
�
� �� � 4
z − 5z +1� ≥ ��z4 � � �
� −|5z − 1| � �
≥ �|z| 4 �
�
− (5 |z| +1) �
�
�
= �|z| 4 �
�
− 5 |z|−1�
= |16 − 10 − 1| =5.
Hence for |z| = 2 we have
Remarks
1
| z4 1
≤
− 5z +1| 5 .
We have seen that the triangle inequality |z1 + z2| ≤|z1| + |z2| indicates
that the length of the vector z1 + z2 cannot exceed the sum of the lengths
of the individual vectors z1 and z2.But the results given in (3) are
interesting.The product z1z2 and quotient z1/z2, (z2 �= 0), are complex
numbers and so are vectors in the complex plane.The equalities |z1z2| =
|z1| |z2| and |z1/z 2 | = |z1| / |z2| indicate that the lengths of the vectors
z1z2 and z1/z2 are exactly equal to the product of the lengths and to the
quotient of the lengths, respectively, of the individual vectors z1 and z2.