If a e we get rotation by therefore in general we get

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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 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 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. Done. 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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