Differentiation 361one-to-one in some neighborhood of every point of A'. Hence f: A -t f(A)
is not one-to-one either.
The condition lfzl = lfzl # 0 implies that J = UxVy - UyVx = 0 and
If z 12 = i-( u; + u; + v; + v;) # 0 in A', so that at least one of the partial
derivatives in the preceding parentheses must be different from zero at each
point of A' (not necessarily the same partial derivative at different points).
Suppose that Uy ( x 0 , Yo) # 0 and that all the partials are continuous inN 0 ((x 0 ,yo)) CA' for some 8 > 0. Then, by a well-known theorem ([131],
p. 58) there exists a function 'ljJ and a positive number 81 :::; 8 such that
v = 'l/;(u) for all (x,y) in N 0 1((xo,Yo)). Hence N 0 1((x 0 ,y 0 )) is mapped by
f into the set {(u,v): v = 'lj;(u)}.
Remark If we assume that v = 'lj;(u), 'ljJ differentiable, in some region
A" C A, we have f(z) = u + i'lj;(u), andso thatfz = %(ux - iuy)[l + i'l/;'(u)]
fz = %(ux + iuy)[l + i'lj;'(u)]lfzl^2 = lfzl^2 =^1 /iu~ + u~)[l + 1/J'(u)^2 ]Hence lfzl = lfzl # 0 at every point of A" except at points such that
Ux =Uy = O, in which case lfzl = lfzl = 0. If the condition Ux = Uy = 0
holds in some subregion of A", then f(z) = c in that subregion.
Examples
1. Let w = Az + Bz + C for all z EC. In this example we have fz =A,
fz =Band 'T/l = "12 = 0 at every point. Hence, if IAI = IBI # 0 the
mapping defined by this function is not one-to-one. In fact, w = Az +
Bz+C maps the z-plane into the straight line Aw-Bw+BC-AC = 0.
If IAI = IBI = 0 the function reduces to a constant.
2. Let w = f(z) = (az+bz)/(cz+dz) for all z E C-{0}, with ad-be# O,
lei # ldl. In this case we have
(ad-bc)z
f z = (CZ+ dz)^2 '
so that lfzl = lf.zl f; 0 in C - {O}.
complex plane into the circlefz = (be - ad)z
(cz+dz)^2
This function maps the deleted(cc - dd)ww + (bd - ac)w + (bd - ac)w + aa - bb = O
(see Exercises 6.2, problem 10).- Let w = f(z) = z/(1 + lzl). We have
1+%1zl %1zl