This preview shows page 1. Sign up to view the full content.
Unformatted text preview: p(z)
bounded, since f(z) Æ 0 as zÆÏ. Thus, f(z)—and hence p(z)—are constant; a
contradiction.
Proof S’pose p(z) is a polynomial with no zeros. Then f(z) = Exercise Set 5
p. 848 #1, 3, 7, 11, 15, 23
We now skip to Chapter 20
6. Conformal Mappings
Definition 6.1 A mapping f: DÆC is called conformal if it preserves angles between
I
curves.
19 Theorem 6.2 If f is analytic, then f is conformal at all points where f'(z) ≠ 0.
Proof. If C is any curve in D through z0, we show that f rotates its tangent vector at z0
through a fixed angle. First think of C as being represented by z = z(t). The derivative,
z'(t), in vector form, evaluated at z0= z(t0) is tangent there, and the angle it makes with
the xaxis is given by its argument. The image curve f*C , is given by z = f(z(t)). The
tangent vector to any path z = z(t) is its derivative with respect to t, thought of as a
vector, rather than a complex number. Therefore, the tangent to f*C at z0 is given by
f'(z(t0))z'(t0), and its angle is its argument, given by
argf'(z0) + argz'(t0)
= Angle independent of the path through z0 + Angle of original tangent.
Done.
Question What happens when f'(z) = 0?
Answer Looking at the above argument, we find that the tangent vector at the im...
View Full
Document
 Fall '03
 StefanWaner
 Math, Algebra, Geometry, Complex Numbers

Click to edit the document details