144 Chapter^3Example The function f(z) = l/z defined from (<C,d) - {O} into (<C,d)
is not continuous at z = 0, since f(z) is not defined at z = 0. However,
by setting f(O) = oo, the extended function, as defined from (<C, d) into
(<C*, x) now becomes continuous. In fact, for any given E > 0 we have
x(f(z),f(O))=x(~,oo) = (lzl2~11)1/2 <Ewhenever lzl < o, with o = E, since that choice of o yields
lzl <I I
(lzl2+1)1/2 - z < EDefinition 3.10 A function that is not continuous at a point is said to
be discontinuous at that point. The preceding example shows that this
notion depends of the metric of the space.
Discontinuity of a function at a point may occur because either (1 )f (a)
is not defined, (2) limz_,a J(z) does not exist, or (3) limz->a f(z) ::j:. f(a). If
limz->a f(z) exists (as a finite number) but does not equal f(a), or if f(a)
is not defined, a continuous function F( z) may be obtained by defining
F(z) = f(z) for all z in Dt - {a}, and F(a) = limz_,a F(z) = limz->a f(z).
Such a discontinuity is called a removable discontinuity.
Example Let f(z) = (z^2 - 4)/(z - 2) for z ::j:. 2, f(2) undefined.
Since for z ::j:. 2 we have f( z) = z + 2, it follows that limz->2 f ( z) = 4.
Hence if we set F(z) = f(z) for z ::j:. 2, and F(2) = 4, the function F thus
defined is continuous at 2. In practice, no distinction is made between F
and f. We simply say that the definition off has been suitably completed,
so as to "restore" continuity at 2.
3.12 Properties of Continuous Functions
Theorem 3.6 Suppose that the functions f and g have a common domain
of definition D, and that a E D is an accumulation point of D. If both
f and g are continuous at a, then f ± g, Jg, f /g, ], Ref, Imf, lfl, and
Arg f, are continuous at a provided that g( a) ::j:. 0 in the case of f / g and
f(a) + lf(a)I ::j:. 0 in the case of Argf.Proof All these properties follow at once from Definition 3.8 and The-
orein 3.2. For instance, since by assumption limz,a f( z) = f (a) and
limz,ag(z) = g(a), we have, by Theorem 3.2,
lim(fg)(z) = lim f(z) · lim g(z) = f(a)g(a) = (fg)(a)
z->a z-a z->aSimilarly,