§4. The topology of the functions Arg(z) and Cog z
417
of the argument differ from one another by a constant multiple of27r and two
branches of .cog by a constant multiple of 27ri.
It goes without saying that the connectedness of G is essential: else take
for G the union of two disjoint open discs.
(iv) The answer to the above existence question is negative if G = C*.
Suppose that indeed we have found on C* a continuous real valued function
A(z) such that zllzl = exp[iA(z)] for all z and restrict to the z E 1[', the unit
circle in C. The map
t~exp(it) = cost + i.sin't
of I = [0,27r] into 1[' is continuous and, as we saw in nO 14, surjective; the
composite function f(t) = A[exp(it)] is therefore continuous on I. But the
function g(t) = t is also continuous and also satisfies g(t) E Arg[exp(it)] for
any t. Since f(t) - t is, for all t E I, an integer multiple of 27r, we must have
f(t) = t + 2k7r with an integer k independent of t. Now exp(it) = 1 for t = 0
or 27r and so f(O) = f(l) since f(t) = A[exp(it)] depends, by definition, only
on the point exp(it); absurd since f(t) = t + 2k7r. It is therefore impossible
to choose an argument for every nonzero complex number z so that it is a
continuous function of z on all of C* or even only on the unit circle 1['. Same
result for .cog z.
(v) Let us show that, on the contrary, it is possible to find a uniform
branch of the argument (or of the logarithm) on the open set G obtained
by excising any half-line of origin 0 from C*. Modulo a rotation z 1--+ e ia z
about the origin, we may restrict to showing this for the open set G = C - lR_,
where lR_ is the set of real numbers :S O. Since the real negative numbers
are characterised by the fact that their arguments are of the form (2k + 1)7r,
it is natural, on G, to choose for A(z) the value of the argument of z which
satisfies
(21.6)
IA(z)1 < 7r,
determined unambiguously, and yielding, for .cog, the function
(21.7)
L(z) = log Izl + i.A(z)
the image J(G) is an interval. Consider two points J(u) and J(v) of J(G), with
u, v E G and, for example, J(u) < J(v), and consider a number C Elf(u), J(v)[
not belonging to J(G). The subsets U and V of G defined by the inequalities
J(z) < c and J(z) > C are then disjoint, satisfy G = U u V, u E U, v E V,
and, finally, are open in G since J is continuous. But, by definition, one cannot
decompose a connected space as two open disjoint nonempty sets: contradiction.
The image J(G) thus satisfies Theorem 5 of Chap. III, nO 4 which characterises
the intervals. Still better: the image of a connected space under a continuous
map is again a connected space.
417
of the argument differ from one another by a constant multiple of27r and two
branches of .cog by a constant multiple of 27ri.
It goes without saying that the connectedness of G is essential: else take
for G the union of two disjoint open discs.
(iv) The answer to the above existence question is negative if G = C*.
Suppose that indeed we have found on C* a continuous real valued function
A(z) such that zllzl = exp[iA(z)] for all z and restrict to the z E 1[', the unit
circle in C. The map
t~exp(it) = cost + i.sin't
of I = [0,27r] into 1[' is continuous and, as we saw in nO 14, surjective; the
composite function f(t) = A[exp(it)] is therefore continuous on I. But the
function g(t) = t is also continuous and also satisfies g(t) E Arg[exp(it)] for
any t. Since f(t) - t is, for all t E I, an integer multiple of 27r, we must have
f(t) = t + 2k7r with an integer k independent of t. Now exp(it) = 1 for t = 0
or 27r and so f(O) = f(l) since f(t) = A[exp(it)] depends, by definition, only
on the point exp(it); absurd since f(t) = t + 2k7r. It is therefore impossible
to choose an argument for every nonzero complex number z so that it is a
continuous function of z on all of C* or even only on the unit circle 1['. Same
result for .cog z.
(v) Let us show that, on the contrary, it is possible to find a uniform
branch of the argument (or of the logarithm) on the open set G obtained
by excising any half-line of origin 0 from C*. Modulo a rotation z 1--+ e ia z
about the origin, we may restrict to showing this for the open set G = C - lR_,
where lR_ is the set of real numbers :S O. Since the real negative numbers
are characterised by the fact that their arguments are of the form (2k + 1)7r,
it is natural, on G, to choose for A(z) the value of the argument of z which
satisfies
(21.6)
IA(z)1 < 7r,
determined unambiguously, and yielding, for .cog, the function
(21.7)
L(z) = log Izl + i.A(z)
the image J(G) is an interval. Consider two points J(u) and J(v) of J(G), with
u, v E G and, for example, J(u) < J(v), and consider a number C Elf(u), J(v)[
not belonging to J(G). The subsets U and V of G defined by the inequalities
J(z) < c and J(z) > C are then disjoint, satisfy G = U u V, u E U, v E V,
and, finally, are open in G since J is continuous. But, by definition, one cannot
decompose a connected space as two open disjoint nonempty sets: contradiction.
The image J(G) thus satisfies Theorem 5 of Chap. III, nO 4 which characterises
the intervals. Still better: the image of a connected space under a continuous
map is again a connected space.