July 16, 2017

# Download An Introduction to Classical Complex Analysis: 1 by Robert B. Burckel PDF

By Robert B. Burckel

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Extra info for An Introduction to Classical Complex Analysis: 1

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3 it follows that U u Cis connected if U is. 39 Let R be a region, C,, C2 two distinct bounded components of C\R. Then there is an unbounded, connected, closed subset of C\C, which contains Ca. Proof: Since C,, Ca are disjoint and compact, there exists an open neighborhood V of Ca diijoint from C,. Now Ca meets fi, Pick any point c2 E C2n fi and take a convex neighborhood N2 of C a which is disjoint from C, and pick ba E Na n R. Consider then two cases: Case I : R is bounded. Then C\Q has (exactly) one unbounded component C,.

Thus E(x + iy) = E(x)E(iy) = w and we have C\{O} c E(@),completing the proof of (vi). Power Series and the Exponential Function 64 Remarks: If n is a positive integer, then E(l/n) is a positive real number whose nth power is E(l), that is, E(l/n) is the unique positive nth root [E(l)]”* of E(1). If m is any integer, then E(m/n) = [E(1In)]” = [E(l)]”’”. Consequently, if we follow the universal custom and, in honor of Euler, let e denote the number E(1), then for every rational r E(r) = e‘. This equality and the homomorphism property of E (the so-called “law of exponents”) explain another universal notation, namely ea, as a name for E(z).

For arbitrary S, however, the two definitions do not agree. , a point x such that C\{x} is totally disconnected (meaning that each component is a single point). Such exotica indeed exist; see page 145 of STEENand SEEBACH [1970]. We may situate Cin C, so that x = (0,0,l). Then let S = Q:,\C, a subset of C,\{(O,O, 1)) = C. 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.