Complex Analysis - Maths KU

–1
y
z = e iθ
Figure 2.54 Figure for Example 6
x
2.6 Limits and Continuity 121
Furthermore, for z0 =1− i we have:
f(z0) =f(1 − i) =(1− i) 2 − i (1 − i)+2=1− 3i.
Since lim f(z) =f(z0), we conclude that f(z) =z
z→z0
2 − iz + 2 is continuous at
the point z0 =1− i.
As Example 5 indicates, the continuityof complex polynomial and rational
functions is easilydetermined using Theorem 2.2 and the limits in (15) and
(16). More complicated functions, however, often require other techniques.
EXAMPLE 6 Discontinuity of Principal Square Root Function
Show that the principal square root function f(z) =z1/2 defined by(7) of
Section 2.4 is discontinuous at the point z0 = −1.
Solution We show that f(z) =z 1/2 is discontinuous at z0 = −1 bydemon-
strating that the limit lim
z→z0
f(z) = lim
z→−1 z1/2 does not exist. In order to do so,
we present two ways of letting z approach −1 that yield different values of this
limit. Before we begin, recall from (7) of Section 2.4 that the principal square
root function is defined by z 1/2 = � |z|e iArg(z)/2 . Now consider z approaching
−1 along the quarter of the unit circle lying in the second quadrant. See
Figure 2.54. That is, consider the points |z| =1,π/2 < arg(z)