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Course: MATH 703, Fall 2011
School: South Carolina
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Variables Complex Notes for Math 703. Updated Fall 2011 Anton R. Schep CHAPTER 1 Holomorphic (or Analytic) Functions 1. Definitions and elementary properties In complex analysis we study functions f : S C, where S C. When referring to open sets in C and continuity of functions f we will always consider C (and its subsets) as a metric space with respect to the metric d(z1 , z2 ) = |z1 - z2 |, where | | denotes...

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South Carolina - MATH - 703
Homework 1, Additional Problem. 1 (1) Let 1 &lt; p &lt; a real number and let q be defined by 1 = p + 1 . q 1 a. Let f (t) = p tp + 1 - t. Show (by means of calculus), that f (t) 0 for all t 0. q p q a b. Show that ab ap + bq for all a, b &gt; 0. (Hint: Take t = b
South Carolina - MATH - 703
Homework 2, Additional Problem. (1) Let (X, d) be a metric space and let A X be a non-empty subset. Define d(x, A) = infcfw_d(x, y) : y A. a. Prove d(x, A) = 0 if and only if x A. b. Show that |d(x, A) - d(y, A)| d(x, y), for all x, y X.1
South Carolina - MATH - 703
Homework 4, Additional Problem. (1) Let (X, d) be a compact metric space and f : X X a mapping such that d(f (x), f (y) &lt; d(x, y) for all x = y. a. Show that there exists x0 X such that f (x0 ) = x0 . (Hint: Consider infcfw_d(x, f (x) : x X, show that it
South Carolina - MATH - 703
Homework 7 1. Express in the form a + bi. a. (1 + i)20 . b.1-2i . 2+i2. Solve z 2 - 4z + (4 + 2i) = 0. 3. Describe the sets whose points satisfy the following relations. Which of these sets are regions (i.e., open and connected sets)? a. |z + i| 1. b. d
South Carolina - MATH - 703
(1)(2)(3)(4)(5)Homework 8. Prove that if z = x + iy and f (z) = (|xy|), then the real part and imaginary part of f satisfy the Cauchy-Riemann equations at z = 0, but f is not differentiable at z = 0. Let G C be an open and connected set and let f : G
South Carolina - MATH - 703
Homework 9. (1) Let cn &gt; 0 in R. Prove that cn+1 cn+1 lim n cn lim n cn lim . lim cn cn In particular, if limn cn+1 exists, then limn n cn = limn cn+1 . cn cn (2) Let an 0 and bn 0. Assume that both (an ) and (bn ) are bounded sequences. (a) Prove that li
South Carolina - MATH - 703
Homework 10 dz, using a branch of log z, where is the join of the line segments (1) Evaluate [1 - i, 1 + i], [1 + i, -1 + i],and [-1 + i, -1 - i], starting at 1 - i and traversing the curve once (see figure 1).1 zFigure 1. (2) Compute2ecos t [cos(sin
South Carolina - MATH - 703
Homework 11 (1) Evaluate (without parametrizing, but using Cauchy's Integral Theorem) for a. (t) = 1 + eit (0 t 2). b. (t) = -i + eit (0 t 2). c. (t) = 2eit (0 t 2). d. (t) = 3i + 3eit (0 t 2). (2) Let C with | = 1. Compute2 01 1+z 2dzdt 1 - 2 cos t +
South Carolina - MATH - 703
Homework 13. (1) Let G be open and connected and f, g analytic on G such that f (z)g(z) = 0 for all z G. Prove that either f (z) = 0 for all z G or g(z) = 0 for all z G. (2) (Quals '02) Let f, g : cfw_z : |z| &lt; 1 C be analytic functions such that |f (z)|
South Carolina - MATH - 703
Homework 14. (1) (Schwarz's lemma) Let f be a holomorphic function on B(0, 1) with |f (z)| 1 for all |z| &lt; 1 and f (0) = 0. a. Define f1 (z) = f (z) for z = 0 in B(0, 1). Prove that z = 0 is a removable z singularity of f1 . b. Prove that |f1 (z)| 1 on B(
South Carolina - MATH - 703
Math 703 Course Outline Fall 2011 TTH 2:00 -3:15A Second Course in Mathematical Analysis by: J. and H. Burkill Professor: Anton R. Schep Office: LeConte 300C Webpage: http:/www.math.sc.edu/~schep/math703-2011.html Phone: 7-6190 Email: schep@math.sc.edu O
South Carolina - MATH - 703
Solutions for HW 1 Problem 10;5. (1) y f (AC A) y = f (x) for some x AC A y f (A) for some A C y AC f (A). (2) Let y f (AC A). This implies y = f (x) for some x AC A. Hence y f (A) for all A C, i.e., y AC f (A) Assume now f is injective. Let y AC f (A). T
South Carolina - MATH - 703
Solutions for HW 2 Problem 25:8. Let xn x, yn y in the normed space V and n , n in the scalars. Then we have n xn - x = n xn - n x + n x - x |n | xn - x + |n - | x . Now (n ) is bounded, so |n | xn - x 0, and similarly |n - | x 0. Hence we have n xn - x 0
South Carolina - MATH - 703
Solutions for HW 3 Problem 47: 4. Sets consisting of one point are obviously connected. Let E R \ Q contain at least two points a &lt; b. Then there exist a rational number r with a &lt; r &lt; b. Let G1 = (-, r) and G2 = (r, ). Then G1 , G2 are disjoint open sets
South Carolina - MATH - 703
Solutions for HW 4 Problem 55: 7. For each n pick xn Fn . We first show that (xn ) is a Cauchy sequence. Let &gt; 0. Then there exists N such that (Fn ) &lt; for all n N . Let n, m N and assume n &lt; m. Then (xn , xm ) (Fn ) &lt; . Hence (xn ) is a Cauchy sequence.
South Carolina - MATH - 703
Solutions for HW 5 Problem 66: 3. Define: 1 fn (x) = 2nx - 1 01 for n &lt; x 1 2 1 for n &lt; x n 2 for 0 x n .Then fn B(, 1), but (fn ) has no uniformly convergent subsequence, as (fn , fm ) = 1 for all n = m, where denotes the uniform metric. Problem 66: 5.
South Carolina - MATH - 703
Solutions for HW 5 Problem 107: 2. Let M = | + | and let &gt; 0. Then there exists N such that supx |fn (x) - f (x)| &lt; 2M for all n N and supx |gn (x) - g(x)| &lt; 2M for all n N . Now sup |fn (x)+gn (x)-f (x)-g(x)| | sup |fn (x)-f (x)|+| sup |gn (x)-g(x)| &lt; +
South Carolina - MATH - 703
Homework 7 1. Express in the form a + bi. a. (1 + i)20 . i Solution: 1 + i = 2e 4 . Hence (1 + i)20 = 210 e5i = -210 . b.1-2i . 2+iSolution:1-2i 2+i=1-2i 2-i 2+i 2-i=-5i 5= -i.2. Solve z 2 - 4z + (4 + 2i) = 0. Solution: z 2 - 4z + (4 + 2i) = (z -
South Carolina - MATH - 703
Solutions Homework 8. (1) Prove that if z = x + iy and f (z) = (|xy|), then the real part and imaginary part of f satisfy the Cauchy-Riemann equations at z = 0, but f is not differentiable at z = 0. v v Solution: u(x, y) = |xy| and v(x, y) = 0. Hence x (x
South Carolina - MATH - 703
Solutions Homework 9. (1) Let cn &gt; 0 in R. Prove that cn+1 cn+1 lim n cn lim n cn lim . lim cn cn Solution: Let A = lim cn+1 and let &gt; 0. If A = , then there is nothing to prove, cn so assume A &lt; . Then there exists N such that cn+1 &lt; A + for all n N . Th
South Carolina - MATH - 703
Homework 10 dz, using a branch of log z, where is the join of the line segments (1) Evaluate [1 - i, 1 + i], [1 + i, -1 + i],and [-1 + i, -1 - i], starting at 1 - i and traversing the curve once (see figure 1).1 zFigure 1. Solution: Let F (z) = log |z|
South Carolina - MATH - 703
Solutions homework 11 1 (1) Evaluate (without parametrizing, but using Cauchy's Integral Theorem) 1+z2 dz for a. (t) = 1 + eit (0 t 2). 1 1 1 1 Solution: f (z) = 1+z2 = 2i ( z-i - z+i ). Hence both z = i and z = -i are outside the curve and thus f (z) dz
South Carolina - MATH - 703
Solutions Homework 12 (1) (Quals 1995) Let f be an entire function on C and assume that |f (z)| A|z|k + B for some constants A, B, integer k and all z C. Prove that f is a polynomial. Solution: Let R &gt; 0 and R = Reit , 0 t 2. From Cauchy's integral formul
South Carolina - MATH - 703
Solutions homework 13. (1) Let G be open and connected and f, g analytic on G such that f (z)g(z) = 0 for all z G. Prove that either f (z) = 0 for all z G or g(z) = 0 for all z G. Solution: Assume neither f (z) = 0 for all z G or g(z) = 0 for all z G. The
South Carolina - MATH - 703
Solutions Homework 14. (1) (Schwarz's lemma) Let f be a holomorphic function on B(0, 1) with |f (z)| 1 for all |z| &lt; 1 and f (0) = 0. a. Define f1 (z) = f (z) for z = 0 in B(0, 1). Prove that z = 0 is a removable z singularity of f1 . Solution: Since f is
South Carolina - MATH - 703
Math 703 Course Outline Fall 2011 TTH 2:00 -3:15A Second Course in Mathematical Analysis by: J. and H. Burkill Professor: Anton R. Schep Office: LeConte 300C Webpage: http:/www.math.sc.edu/~schep/math703-2011.html Phone: 7-6190 Email: schep@math.sc.edu O
South Carolina - MATH - 703
Math 703 Course Outline Fall 2011 TTH 2:00 -3:15A Second Course in Mathematical Analysis by: J. and H. Burkill Professor: Anton R. Schep Office: LeConte 300C Webpage: http:/www.math.sc.edu/~schep/math703-2011.html Phone: 7-6190 Email: schep@math.sc.edu O
South Carolina - MATH - 555
Math 555/704I Course OutlineSpring 2011 Text : A First Course in Real Analysis Springer, Undergraduate Texts in Mathematics by: Sterling K. Berberian Supplemented with notes.Professor : Anton R. Schep Office : LeConte 300C Email : schep@math.sc.edu Web
South Carolina - MATH - 555
Homework 1. (1) Prove that [-1, 1) is not compact by using the definition of a compact set (to get credit for the problem, use the definition and not any theorems about compact sets). (2) What is an interior point? Prove that1 4is an interior point of (
South Carolina - MATH - 555
Homework 3, Additional Problems. (1) Let (X, d) be a metric space. a. Let Ei X (i cfw_1, , n) be a finite collection of subsets of X. Prove that n Ei = n Ei . i=1 i=1 b. Let Ei (i I) now be an arbitrary collection of subsets of X. Prove that iI Ei iI Ei a
South Carolina - MATH - 555
Extra problems Homework 4. (1) Let f, g : R R be uniformly continuous functions. Assume that both f and g are bounded. Prove that the product f g is uniformly continuous. (2) A function f : R R is periodic, if there exists a c R such that f (x + c) = f (x
South Carolina - MATH - 555
Extra problems Homework 7. (1) Let f : [0, 1 R be a continuous function. Prove that 1 lim n nn 1f (k/n) =k=1 0f (x) dx.(Hint: Use uniform continuity to show that for &gt; 0 there exists a &gt; 0 such that S() - s() &lt; for all subdivisions with norm N () &lt; .
South Carolina - MATH - 555
Homework 9. Prove that f (x) = limn fn (x) exists for all x R. Does (fn ) (1) Let fn (x) = converge uniformly to f ?x2n . 1+x2n(2) Define fn : [0, 1] [0, 1] by fn (x) = xn (1 - x). Prove that fn converges uniformly to 0. (3) Prove that nx + sin(nx2 ) n
South Carolina - MATH - 555
Solutions homework 1. (1) Prove that [-1; 1) is not compact by using the definition of a compact set (to get credit for the problem, use the definition and not any theorems about compact sets).1 Proof: Let On = (-1, 1 - n ). Then [-1; 1) n On , but [-1;
South Carolina - MATH - 555
Solutions homework 2. Page 32 Problem 6: Clearly d(x, y) 0 and d(x, y) = 0 if and only if x = y. Hence property (i) holds. It is also clear that d(x, y) = d(y, x), so it remains to show that the triangle inequality holds. Let x, y, z X. If x = y, then d(x
South Carolina - MATH - 555
Solutions homework 3. Page 69 Problem 10: a. Assume A and B are neigborhoods of x. Then there exist r1 &gt; 0 such that Ur1 (x) A and r2 &gt; 0 such that Ur2 (x) B. let r = mincfw_r1 , r2 . Then r &gt; 0 and Ur A B. Hence A B is a neighborhood of x. b. Let Br (c)
South Carolina - MATH - 555
Solutions homework 4. Page 108 Problem 4: Let (xn ) be a Cauchy sequence and let &gt; 0. Then there exists &gt; 0 such that |x - y| &lt; implies |f (x) - f (y)| &lt; . For this there exist N such that |xn - xm | &lt; for all n, m N . Hence |f (xn ) - f (xm )| &lt; for all
South Carolina - MATH - 555
Solutions homework 5. Page 128 Problem 3: Using the substitution y = -x we get g(y) - g(-c) f (-y) - f (c) f (x) - f (c) = =- . y - (-c) y - (-c) x-c Now letting y -c- is the same as letting x c+ , from which the problem follows. Page 128 Problem 4. As g(
South Carolina - MATH - 555
Solutions homework 6. Page 141 Problem 2. If f g, then mk (f ) = infcfw_f (x) : xk-1 x xk mk (g) = infcfw_g(x) : xk-1 x xk and thus sf () sg () for any subdivision . By definition this implies thatb bfa ag.The corresponding inequalities for the upp
South Carolina - MATH - 555
Solutions homework 7. Page 182 Problem 1. The answer is 2 times the sum of the geometric series with c = 1 , 3 1 so the answer is 2 1- 1 = 3. 3 Page 182 Problem 4. For n 1 we have s2n+2 = s2n + a2n+1 + a2n+2 s2n (since a2n+1 +a2n+2 0), so (s2n ) is an inc
South Carolina - MATH - 555
Page 188 Problem 6. follows now fromn 1 1 xSolutions homework 8. dx = 2 n - 2 as n . The divergence of the seriesn+11 A simpler proof follows from the fact that diverges. n Page 188 Problem 7. From ln x x - 1 it follows that ln(1 + n) (1 + n) - 1 = n.
South Carolina - MATH - 555
Solutions homework 9. Prove that f (x) = limn fn (x) exists for all x R. Does (fn ) (1) Let fn (x) = converge uniformly to f ? Solution: There are 3 cases: |x| &lt; 1, |x| = 1, and |x| &gt; 1. In case |x| &lt; 1, then x2n 0 as n , which implies that fn (x) 0 as n
South Carolina - MATH - 555
Math 555/704I Course OutlineSpring 2011 Text : A First Course in Real Analysis Springer, Undergraduate Texts in Mathematics by: Sterling K. Berberian Supplemented with notes.Professor : Anton R. Schep Office : LeConte 300C Email : schep@math.sc.edu Web
South Carolina - MATH - 554
Extra Credit problems, MATH 554/703I 1. Let (an ) be a sequence of real numbers such that for some = 0 we have an - = 0. n an + lim What can you say about (an ). 2. Let , &gt; 0. Prove thatnlimnn + n = maxcfw_, .3. Assume an a in R. Put sn = a1 +an . Pr
South Carolina - MATH - 554
Math 554/703I Course OutlineFall 2010 Text : A First Course in Real Analysis Springer, Undergraduate Texts in Mathematics by: Sterling K. BerberianProfessor : Anton R. Schep Office : LeConte 300C Email : schep@math.sc.edu Web page : /www.math.sc.edu/~sc
South Carolina - MATH - 554
Homework 1, Additional Problems. (1) Prove that F = cfw_0, 1 with the usual multiplication and usual addition, except that 1 + 1 := 0, is a field. (2) Suppose x &gt; 0 and y &gt; 0 are rational numbers. Determine with proof the validity of the following stateme
South Carolina - MATH - 554
Solutions homework 1. Extra Problem 1: First observe that F is closed under the addition and multiplication as defined. To check (A1) (a + b) + c = a + (b + c), note that this holds if two of the three elements equal 0, as then the sums on both sides equa
South Carolina - MATH - 554
Solutions homework 2. Page 8 Problem 14: Note first that the function y = 2x + x2 has a minimum at x = -1, so 2x + x2 -2 + 1 = -1 for all x. Hence by Problem 13 we have (1 + x)2n = (1 + 2x + x2 )n 1 + n(2x + x2 ) 1 + 2nx. Page 8 Problem 17. Assume n = r2
South Carolina - MATH - 554
Solutions homework 3. Page 16 Problem 1: Any b B is an upper bound for A, so A is bounded above and sup A b for all b B. Now sup A is a lower bound for B and thus sup A inf B. 1 Page 16 Problem 5. For n even the numbers look like 1 + n , which are bounded
South Carolina - MATH - 554
Solutions homework 4. Page 22 Problem 1.: By 1.3.8 -an -a and -bn -b, so by Theorem 2.5.6 we have -an + (-bn ) -a + (-b), which again by 1.3.8 implies that an + bn a + b. Page 22 Problem 2. From 0 an an+1 and 0 bn bn+1 it follows that an bn an+1 bn+1 , so
South Carolina - MATH - 554
Solutions homework 5. Page 25 Problem 3.: 1 a. Take (an , bn ) = (0, n ). Then n (an , bn ) = , as there is no x &gt; 0 which also satisfies 1 x &lt; n for all n. b. Let [a, b] = [an , bn ]. We claim A = [a, b]. As (an , bn ) [an , bn ], it is obvious that A [a
South Carolina - MATH - 554
Solutions homework 6. Page 32 Problem 3.: |a + b| = |a| + |b| if and only if |a + b|2 = (|a| + |b|)2 if and only if (a + b)2 = a2 + 2|a|b| + b2 if and only if ab = |a|b| = |ab|. Now by definition x = |x| if and only if x 0 and thus ab = |ab| if and only i
South Carolina - MATH - 554
Solutions homework 7. Page 39 Problem 3.: False. Take (xn ) = (1, 0, 1, 0, ) and (yn ) = (0, 1, 0, 1, ). Then neither (xn ) or (yn ) is a null sequence, but their product is identical zero. Page 39 Problem 4. True. Note n+1+ n 1 xn = ( n + 1 - n) = 0. n+1
South Carolina - MATH - 554
Solutions homework 8. Page 47 Problem 1.: &quot; &quot; If |ank | k, then is clear that there can't be an M such that |an | M for all n (take k &gt; M to get a contradiction). &quot; Assume (an ) is unbounded. Then for all M there exists n such that |an | &gt; M . Let M = 1.
South Carolina - MATH - 554
Solutions homework 9. Page 60 Problem 2.: (i) Let x y with x, y I J. Then I s an interval implies that [x, y] I and similarly [x, y] J. Hence [x, y] I J. It follows therefore from Theorem 4.1.4 that I J is an interval. (ii) False, take e.g. I = [0, 1) and
South Carolina - MATH - 554
Solutions homework 10. 1 Page 64 Problem 3. Let a = sup A. Then for all n 1 the number a - n is not an 1 1 upperbound of A, so there exists an A such that a - n &lt; an a. Then |a - an | &lt; n , so an a. Thus a A. The proof for inf A A is similar. Page 64 Prob
South Carolina - MATH - 554
Solutions homework 12. Page 78 Problem 5. (i) Let B = C. Then obvious that B A, as each interval (r, s) A. Conversely let x A. Then there exists &gt; 0 such that (x - , x + ) A (as A is open). Now take r Q such that x - &lt; r &lt; x and s Q such that x &lt; s &lt; x +
South Carolina - MATH - 554
Solutions homework 13. Page 86 Problem 7. From 0 f (x) x it follows that f is continuous at 0. If x &gt; 0 and x rational, let xn x with xn irrational. Then f (xn ) = xn x = 0 = f (x). Hence f is discontinuous at rational x &gt; 0. Similarly, If x &gt; 0 and x irr
South Carolina - MATH - 554
Solutions homework 14. Page 99 Problem 1. Let p(x) be a polynomial of degree 2k + 1. By multiplying by a constant we can assume that the coefficient of x2k+1 is 1. Then p(x) = x2k+1 + a2k x2k + a0 2k + a1 x + a0 . Then p(n) = n2k+1 (1 + ak + + n2k+1 ) &gt; 0
A.T. Still University - ECON - 101
Universidad de Puerto Rico Rio Piedras Campus Facultad de Administracin de Empresas Departamento de Finanzas Alvin Y. Ros Arzola 801-07-6764 Asignacin Cap. 1 y 2 Captulo 1:11 What is finance? Explain how this field affects all of the activities in which
A.T. Still University - ECON - 101
13 Current Liabilities and ContingenciesPowerPoint Authors: Susan Coomer Galbreath, Ph.D., CPA Charles W. Caldwell, D.B.A., CMA Jon A. Booker, Ph.D., CPA, CIA Cynthia J. Rooney, Ph.D., CPAMcGraw-Hill/IrwinCopyright 2011 by the McGraw-Hill Companies, In