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Unformatted text preview: p(z)
bounded, since |f(z)| Æ 0 as |z|ÆÏ. Thus, f(z)—and hence p(z)—are constant; a
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
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 x-axis 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.
Question What happens when f'(z) = 0?
Answer Looking at the above argument, we find that the tangent vector at the im...
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