By R.B. Burckel

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**Extra resources for An Introduction to Classical Complex Analysis: Vol. 1**

**Sample text**

Show that rp maps D(O, 1) onto ( - 00, 0) x IR and maps qD(O, I) onto (0,00) x 1R\{I}. Hints: If w = 2 Iwl (iii) (iv) Hints: rp(z), then compute z + 1 z+ 1 1 = z _ l' z _ 1- 1= + z) 4 Iz _ 112 = Iz _ 112 Re z. 2(z Show that rp maps [-1, I) onto (-00,0] and (consequently) maps q[ -1,1] onto q( -00, O]\{I}. p(z) = z + lIz maps qD(O, 1) conformally onto q[ -2, 2J. Set s(z) = Z2 and note that in qD(O, 1) ~ (0, 00) x 1R\{I} ~ q( -00, O]\{I} ~ q[ -I, IJ all maps are one-to-one and onto. p(z) = 2rf>(rp2(Z».

2 k=O = 2 [1 + (_1)n] -=2:inxn co i = L, _ - 1 + E( - E(ix) [X2 + 4! - 1] 2! x 4-2 (2k)! + (-1)2"X4n ] (4n)! ]X4n - 2. 1 < 1 ~--=-: (4n - 2)! co [ x 2 + n~2 1] (4n)! - (4n _ 2)! x 4n-2 Therefore C(V3) < -t. Since C(O) = Re E(O) = 1, we have that C[O, V3] is a connected subset of IR which contains positives and negatives and hence zero; that is, the compact set [0, V3] r. C-l(O) is not void. T/2 = mineO, V3] r. C-1(0). T/2) = O. From (8) and (6) we get (9) S(17/2) = ± 1. However by (5) and the definition of 17 (10) S' = C > 0 in [0,17/2), § 3.

V are each compact, the finite intersection property of compacta implies that not all of them can meet the e10sed set q V. So X,. () Vc V for some n. Then the set W = Xn () V = X,. , contains X () V => K and so is non-void. Since X" is connected, it follows that X,. = W. But then X" is compact, a contradiction. 5. In this case n c (D"o) U X"o' a union of compact sets, so n is bounded. Finally notice that §5 8n = n Xn (exercisette). Connectivity of a Set The theorems of this section are technical results which will be needed later (in Chapter X).