A first course in complex analysis - download pdf or read online

By Beck M., Marchesi G., Pixton G.

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The complex number zk (0), considered as a vector, is the tangent vector to γk at the point a. Then f transforms the curve γk to the curve f (γk ), parameterized as f (zk (t)). If we differentiate f (zk (t)) at t = 0 and use the chain rule we see that the tangent vector to the transformed curve at the point f (a) is f (a)zk (0). Since f (a) = 0 the transformation from z1 (0) and z2 (0) to f (a)z1 (0) and f (a)z2 (0) is a dilation. A dilation is the composition of a scale change and a rotation and both of these preserve the angles between vectors.

If we remember that ∞ corresponds to being arbitrarily far away from the origin we can visualize a line plus infinity as a circle passing through ∞. 3 can be expressed very simply: any M¨obius transformation of C transforms circles to circles. For example, the transformation f (z) = z+i z−i transforms −i to 0, i to ∞, and 1 to i. The three points −i, i and 1 determine a circle—the unit circle |z| = 1—and the three image points 0, ∞ and i also determine a circle—the imaginary axis plus the point at infinity.

Proof 2 . Suppose (by way of contradiction) that p does not have any roots, that is, p(z) = 0 for all z ∈ C. Then Cauchy’s formula gives us 1 1 = p(0) 2πi γR 1/p(z) dz z where γR is the circle of radius R around the origin. Notice that the value of the integral does not depend on R, so we have 1 dz 1 = lim . 3 we have |z p(z)| ≥ 12 |ad | |z|d+1 for all large z, where d is the degree of p(z) and ad is the leading coefficient of p(z). 1(d) and the formula for the circumference of a circle we see that the integral can be bounded as dz 1 1 2 2 ≤ · · (2πR) = 2πi γR zp(z) 2π |ad | Rd+1 |ad | Rd and this has limit 0 as R → ∞.

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