1550251515-Classical_Complex_Analysis__Gonzalez_

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in the open neighborhood A since at its center we have

h(zo) = lf(zo) - w'I < e =^1 / 2 m

so thatμ < %m also. However, for z E oA we have

h(z) = lf(z) - w'I = lf(z) - f(zo) + f(zo) - w'I

;::: lf(z) - f(zo)l - lf(zo) - w'I

> g(z) - i/2m;::: l/2m

Chapter 6

which shows that the minimum does not occur on oA. Hence there is a '
point ( E A such that h( () = μ.

Now, letting w' = u' +iv', we have

h^2 (z) = lf(z) - w'l^2 = (u -u')^2 + (v - v')^2

and the function h^2 (z) must also have a minimum at (. But this implies
that

1 8h

2

• ~ = ( u -u ') u,,, + ( v - v ') v,,, = 0
  2 ux
  1 oh^2
  2 oy = ( u - u')uy + ( v -v')vy = 0 (6.13-2)

at ( = (e, 17). Since ( E A, it follows from assumption (2) that the

determinant of the system (6.13-2), namely,

I~: ~:I

evaluated at (e,17), does not vanish. Hence u(e,11) = u' and v(e,11) = v',

or J(() == w'. Thus w' E f(A).

Theorem 6.24 Let w = f(z) E 'D(A), A open. If lfzl = lfzl = 0 at

z 0 E A, then f may or may not be one-to-one in a neighborhood of z 0 •

If lfzl = lfzl -=/= 0 and ?Jdz - ?Jdz = 0 at zo, then f is not one-to-one in

some neighborhood of z 0 •

Proof If If z I = lfz I = 0, then f z = fz = 0 at zo, so f is monogenic, as well

as conjugate monogenic, at zo, with f'(zo) = 0. If the condition also holds

in a neighborhood of z 0 , then f^1 ( z) = 0 in that neighborhood. In this case

f is a constant function in the same neighborhood, so it is not one-to-one.

If the condition holds at z 0 only, f may or may not be one-to-one in a

neighborhood of zo, as the following examples show.