1550251515-Classical_Complex_Analysis__Gonzalez_

Elementary Functions 245 d(z1,z2) = d(z2,z1). d(z1,z3) ~ d(z1,z2) + d(z2,z3), where equality holds only if z1, z2, z 3 are on t ...

246 Chapter 5 From (5.11-5) it follows that ldzl/2y (y > 0) is a differential invari- ant in Poincare's model. This result ca ...

Elementary Functions 247 Next, letting dy = 0, then dx = 0, in (5.11-7) we get dsx = dx/y and dsy = dy / y. This leads to the fo ...

248 .,,,.....-----.................. "' \ \ / \/ -~----------~r------~- Fig. 5.19 Chapter 5 asymptotic triangle (i.e., a triangl ...

Elementary Functions 249 2. Let w = l/z with z = x + iy, w = u +iv. Show that x = u/(u^2 + v^2 ), y = -v /( u^2 + v^2 ), and pro ...

250 Chapter 5 In general, only six of the twenty-four ratios are different. In fact, if the value of one of them is r, the value ...

Elementary Functions 25. If a is an arbitrary point such that !al '/' R, then the function F(z) = R 2 lal z -a a az -R^2 251 has ...

252 Chapter 5 the four points z 1 , z 2 , z 3 , z 4 lie on a circle (as well as their images). A conjugate transformation carrie ...

Elementary Functions 253 For e -:j:. 0, lal = ldl, the second inequality implies that be - be -:j:. 0. However, the point Zo = a ...

254 with lal f: 0, lbl f: 0, lei f: 0. Then be= lbl ieieih'-,8) and this is real if 'Y-/3=mr for some integer n. Also, we must h ...

Elementary Functions which gives a+ 8 = 2k7r and ial^2 ± lbllci = 1. Thus we find that /3 = k1r - (2q + 1 )'ff = (2k - 2q - 1 )' ...

256 Chapter5 We recall (Exercises 1.2, problem 24) that an equation of the form Lz + Mz + N = 0 has no solution if ILi = IMI and ...

Elementary Functions 257 Finally, suppose that IAI = IBI-::/:- 0 and AC= BC. Then Az+Bz+C = 0 defines a straight line that may e ...

258 Chapter 5 By introducing, if necessary, an appropriate factor, we may assume the conditions above to be and Hence general bi ...

Elementary Functions 259 where A = a1 + ia2, B = a1 + ia 2 , C = 2(c 1 + ic 2 ), N = 2c 3. Then we find that 1 a1 + a1 a1 -a1 C1 ...

260 Chapter 5 Show that w = Az + Bz + C, with IAI i= IBI transforms circles into ellipses. By examining the conditions D 1 = 0, ...

Elementary Functions 261 The factorization (5.14-3) shows that a polynomial of degree n has ex- actly n zeros or roots (counting ...

262 Chapter^5 so there exists R 1 > 0 such that for lzl > Ri we have Now, given M > 0, the inequality IP(z)I > %1anl ...

Elementary Functions 263 inverse image, namely, z = a and z = oo, respectively. It is said, in the case of those exceptional poi ...

264 Chapter^5 where 0 < B 1 - Bo :::; 27r /n, i.e., the interior of the angle formed by the rays Lo and L 1 (Fig. 5.20). Then ...