Complex Analysis - Maths KU

118 Chapter 2 Complex Functions and Mappings
EXAMPLE 4 Computing Limits with Theorem 2.2
Use Theorem 2.2 and the basic limits (15) and (16) to compute the limits
(3 + i)z
(a) lim
z→i
4 − z2 +2z
z +1
(b) lim
z→1+ √ 3i
Solution
z 2 − 2z +4
z − 1 − √ 3i
(a) ByTheorem 2.2(iii) and (16), we have:
lim
z→i z2 � � � �
= lim z · z =
z→i
lim z
z→i
· lim z
z→i
= i · i = −1.
Similarly, lim
z→i z 4 = i 4 = 1. Using these limits, Theorems 2.2(i), 2.2(ii),
and the limit in (16), we obtain:
lim
z→i
� (3 + i)z 4 − z 2 +2z � =(3+i) lim
z→i z 4 − lim
z→i z 2 + 2 lim
z→i z
=(3+i)(1) − (−1)+2(i)
=4+3i,
and lim
z→i (z +1)=1+i. Therefore, byTheorem 2.2(iv), we have:
(3 + i)z
lim
z→i
4 − z2 +2z
=
z +1
lim
z→i
� (3 + i)z 4 − z 2 +2z �
lim (z +1)
z→i
After carrying out the division, we obtain lim
7 1
−
2 2 i.
(b) In order to find lim
z→1+ √ 3i
lim
z→1+ √ � � 2
z − 2z +4 =
3i
z→i
= 4+3i
1+i .
(3 + i)z 4 − z 2 +2z
z +1
z2 − 2z +4
z − 1 − √ , we proceed as in (a):
3i
�
1+ √ �2 �
3i − 2 1+ √ �
3i +4
= −2+2 √ 3i − 2 − 2 √ 3i +4=0,
and lim
z→1+ √ � √ � √ √
z − 1 − 3i =1+ 3i − 1 − 3i = 0. It appears that
3i
we cannot applyTheorem 2.2(iv) since the limit of the denominator is 0.
However, in the previous calculation we found that 1 + √ 3i isarootof
the quadratic polynomial z2 −2z +4. From Section 1.6, recall that if z1 is
a root of a quadratic polynomial, then z −z1 is a factor of the polynomial.
Using long division, we find that
z 2 �
− 2z +4= z − 1+ √ ��
3i z − 1 − √ �
3i .
=