Complex Analysis - Maths KU

2.6 Limits and Continuity 117
Theorem 2.2 Properties of Complex Limits
Suppose that f and g are complex functions. If lim f(z) = L and
z→z0
lim g(z) =M, then
z→z0
(i) lim cf(z) =cL, c a complex constant,
z→z0
(ii) lim (f(z) ± g(z)) = L ± M,
z→z0
(iii) lim f(z) · g(z) =L · M, and
z→z0
f(z) L
(iv) lim = , provided M �= 0.
z→z0 g(z) M
Proof of (i) Each part of Theorem 2.2 follows from Theorem 2.1 and the
analogous property(9)–(12). We will prove part (i) and leave the remaining
parts as exercises.
Let f(z) =u(x, y)+iv(x, y), z0 = x0 + iy0, L = u0 + iv0, and c = a + ib.
Since lim f(z) =L, it follows from Theorem 2.1 that lim u(x, y) =
z→z0
(x,y)→(x0,y0)
u0 and lim
(x,y)→(x0,y0) v(x, y) =v0. By(9) and (10), we have
and
lim
(x,y)→(x0,y0) (au(x, y) − bv(x, y)) = au0 − bv0
lim
(x,y)→(x0,y0) (bu(x, y)+av(x, y)) = bu0 + av0.
However, Re (cf(z)) = au(x, y) − bv(x, y) and Im (cf(z)) = bu(x, y) +
av(x, y). Therefore, byTheorem 2.1,
lim cf(z) =au0 − bv0 + i (bu0 + av0) =cL.
z→z0
Of course the results in Theorems 2.2(ii) and 2.2(iii) hold for anyfinite
sum of functions or finite product of functions, respectively. After establishing
a couple of basic complex limits, we can use Theorem 2.2 to compute a large
number of limits in a verydirect manner. The two basic limits that we need
are those of the complex constant function f(z) =c, where c is a complex
constant, and the complex identity function f(z) =z. In Problem 45 in
Exercises 2.6 you will be asked to show that:
and
✎
lim c = c, c a complex constant,
z→z0
(15)
lim z = z0.
z→z0
(16)
The following example illustrates how these basic limits can be combined with
the Theorem 2.2 to compute limits of complex rational functions.