Functions. Limits and Continuity. Arcs and Curves 145lim Argf(z) = Arg(lim f(z)) = Argf(a)
z~a z~aprovided that J(a) + \J(a)\ f. 0. In particular, 'for J(z) = z we have that
Arg z is continuous on the principal region z + \z\ f. 0.
For a = 0, Arg z is discontinuous since it cannot be suitable defined at
this point. For any a= -r = reirr (r > 0), Arg z is also discontinuous, sincethen limz->a Arg z does not exist. In fact, for any given E > 0 first choose
8 < r, and let a be the absolute value of the angle from the negative realaxis to either tangent from the origin 0 to the circle \z -a\ = 8 (Fig. 3.10).
Then if z E N5(a) and Imz > 0, we have
8
\Argz-7r\<a<tana= (r 2 ·.....:o 2 ) 1 / 2 <e.whenever 8 < tr/(l + E^2 )^112 • Hence limz__.a,Imz>O Argz = 7!".
On the other hand, if z E N5(a) and Imz < 0, we have
8 '
\Argz-(-7r)\ <a< (r2 -82)1/2 < Ewhenever 8 < ff /(1 + E^2 )^1 l^2. Hence limz->a,Im z<O Arg z
limz_,a Arg z does not exist in the sense of Definition 3.1.
-7!", soRemark If f and g fail to be continuous at a, some of the functions f + g,
J - g, f g, f / g, Re J, Im f may have a removable discontinuity at a. For
example, both f(z) = z + 1 - l/z and g(z) = z + l/z are discontinuous
at z = 0 [in the ordinary sense, i.e., in (<C, d)], but f + g has a removable
discontinuity at z = 0. In fact, for z f. 0 we have (f + g)(z) = 2z + 1,
and limz__, 0 (1+g)(z)=1. Hence if we define (f + g)(O) = 1, the function
so defined becomes continuous at z = 0. Note that (f + g)(O) cannot be
defined directly [as f(O) + g(O)] since both f and'g are undefined at z = 0.
yxFig. 3.10