Download e-book for iPad: Complex Analysis by Serge Lang (auth.)

By Serge Lang (auth.)

ISBN-10: 1475718713

ISBN-13: 9781475718713

ISBN-10: 147571873X

ISBN-13: 9781475718737

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By definition, there exist positive integers m, k with k > m such that IZk - vi < r/2 and IZm - vi < r/2. Then IZk - Zm I < r and this contradicts the property of our sequence {zn} because Zk lies in the disc Dm. This proves the theorem. Let S be a set of complex numbers, and let I be some set. Suppose that for each i E I we are given an open set Vi' We denote this association by {Vhel> and call it a family of open sets. The union of the family is the set V consisting of all Z such that Z E Vi for some i E I.

We say that V is open if for every point (X in V there is a disc D«(X, r) centered at (x, and of some radius r > 0 such that this disc D«(X, r) is contained in V. We have illustrated an open set in Fig. 8. Note that the radius r of the disc depends on the point (x. As (X comes closer to the boundary of V, the radius of the disc will be smaller. Examples of Open Sets. The first quadrant, consisting of all numbers + iy with x > 0 and y > 0 is open, and drawn on Fig. 9. z= x Figure 8 18 COMPLEX NUMBERS AND FUNCTIONS (a) [I, §4] (b) Figure 9 On the other hand, the set consisting of the first quadrant and the vertical and horizontal axes as on Fig.

Is taken over all elements z E A and wEB. If B consists of one point, we also write d(A, w) instead of d(A, B). We shall leave the next two results as easy exercises. 6. Let S be a closed set of complex numbers, and let v be a complex number. There exists a point W E S such that d(S, v) = Iw - vi. 7. Let K be a compact set of complex numbers, and let S be a closed set. There exist elements d(K, S) [Hint: Consider the function ZH Zo E = Izo - K and Wo E S such that wol. 8. Let S be compact. Let r be a real number > O.

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Complex Analysis by Serge Lang (auth.)


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