By R.B. Burckel

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**Additional info for An Introduction to Classical Complex Analysis: Vol. 1**

**Sample text**

Zo) and h'(f(zo» are just such as to make 0 and e continuous at Zo andf(zo), respectively. '(zo) + 8(z)] and for all w sufficiently near f(zo) we get from (2) (4) hew) - h(f(zo» = (w - f(zo))[h'(f(zo)) + e(w)]. '(zo) + o(z)][h'(f(zo» + e(f(z»], § 1. Hoiomorphic and Harmonic Functions 43 that is, (5) (h 0 f)(z) - (h 0 f)(zo) = (f'(zo) Z - Zo + S(z»)[h'(f(zo» + e(f(z»]. Let z approach zoo Then S(z) andf(z) approach S(zo) = 0 andf(zo) respectively and so e(f(z» approaches e(f(zo» = 0 and the assertion (iii) follows from (5).

1 Proof: lim..... co nil.. = 1 and for any a > 0, lim..... co all.. Set x .. = nil" - 1. )" = I~ (;)XL ~ (;)x~ = n(n; ~ = 1. 2 1) x~. Therefore 05,Xll 5,Jn:'1 and so x .. _ O. If a ~ I, then I 5, al'" a < 1, apply this conclusion to Ita. o< 5, nil" for large n, so al'" _ 1. If Remarks: We are, of course, assuming known the theory of rational roots and powers on [0, 00). This is an easy consequence of the completeness of IR. ) Irrational powers on [0,00) can be dealt with at the same level of elementariness but will not be needed.

Then of course this set has a connected complement in C .. , but its complement in C is the totally disconnected set C\{(O, 0, I)} and so has infinitely many bounded components. 14 is true: any two points on any curve lie on an arc which lies on the curve. For a proof of this surprisingly non-trivial fact (due to R. L. MOORE [1916/17]) see TIETZE [1919], MAZURKIEWICZ [1920], KAMlYA [1931] or G. T. WHYBURN [1932]. 29. I9(ii) is a famous result of CARATHEoDORY [1911] (p. 2(0): if S c: IRk, then every element of co Sis a convex combination of k + 1 or fewer points of S.