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MATH 800, Spring 2020 Daily Update

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MATH 800, Spring 2020 Daily Update MATH 800, Spring 2020 Daily Update: January 22, Wednesday We began with the fundamental concepts in complex analysis - identifying the complex numbers with pairs of real numbers equipped with the algebraic operations of addition and multiplication and discussing various basic properties of complex numbers, polar representation, complex exponentials, Cauchy- Schwarz and triangle inequalities. We then spent some time working with complex @ @ polynomials and defining the partial differential operators @z and @z¯, which are linear, satisfy the Leibniz rule and play an important role in complex analysis. January 24, Friday We defined holomorphic functions and proved that their real and imaginary parts satisfy the Cauchy-Riemann equations and are harmonic functions. We then worked on the question of the existence of a holomorphic function, whose real part is a given harmonic function and proved the result in the case of an open rectangle. We also discussed the existence of holomorphic antiderivative. This concludes chapter 1 and the assigned Homework set I is due on February 5. Week 2, January 27-31 We spent this week discussing integrals in the complex plane setting and defined integral of a continuous function f over a smooth curve γ(t) = γ1(t) + iγ2(t):[a; b] ! U in an open set U as follows Z Z b 0 0 f(z)dz = f(γ(t))(γ1(t) + iγ2(t))dt: γ a In the case of a holomorphic function f : U ! C we have the following Proposition 1 Z @f(z) dz = f(γ(b)) − f(γ(a)): γ @z Note that the condition that the function f is holomorphic is very important. In that case, the particular parametrization of γ is not important. That is, for an increasing C1 function φ and forγ ~ := γ ◦ φ, we have Z Z f(z)dz = f(z)dz γ γ~ Next, we defined complex differentiability for a function f : U ! C at a point z0 2 U by the limit f(z) − f(z ) lim 0 z!z0 z − z0 0 if this limit exists. We denote the limit by f (z0). It turns out that this is equivalent with the function f being holomorphic. We proved the following theorem. Theorem 1 f 2 C1(U) is holomorphic if and only if f 0(z) exists for all z 2 U. In 0 @f this case, f = @z . 1 2 Finally, we extended the antiderivative condition as follows. Proposition 2 Let P be a point in Ω ⊂ C, a simply connected open set. Suppose that F is holomorphic in ΩnP and continuous on Ω. Then there exists H, holomorphic on Ω, such that H0(z) = F . Week 3, February 3-7 We spent this week discussing and proving Cauchy integral formula and Cauchy integral theorem. Denote the integral over a closed curve γ by H γ and assume that the positive direction is counterclockwise. We have: Cauchy Theorem Let Ω be simply connected domain and f is holomorphic on Ω. Then, for any closed C1 curve γ : [0; 1] ! Ω we have I f(z)dz = 0: γ Cauchy integral formula Let Ω be a simply connected domain in C and f a holomorphic function on it. Let γ be a C1 closed curve in Ω, which winds once around z 2 Ω. Then, 1 I f(ξ) f(z) = dξ: 2πi γ ξ − z We started the proof of this by computing the complex integral I 1 = 2πi jξ−z0j=r ξ − z whenever z : jz − z0j = r. We then proved the results for the special case when γ is the boundary of a disc, and extended it to a general closed curve. Monday, February 10 We worked on applications of Cauchy integral formula and theorem. Theorem Let U be an open subset of C and f a holomorphic function on U. Then f 2 C1(U) and for every integer k and every curve γ around z 2 U I (k) k! f(ξ) f (z) = k+1 dξ 2πi γ (ξ − z) As a corollary, we saw that if f is a holomorphic function on U, then f 0 is also holomorphic on U. The next theorem shows that the Cauchy theorem is reversible. Theorem (Morera) Let Ω be an open connected subset of C and f continuous function on U. Assume that for every closed C1 curve γ, we have I f(z)dz = 0 γ Then f is holomorphic inside Ω. Finally, we discussed "holomorphic extension". Proposition Let U be an open subset of C and f a holomorphic function on U. Let γ be a C1 closed curve in U and φ 2 C(γ). For z 2 Int(γ), define 1 I φ(ξ) f(z) = dξ 2πi γ ξ − z Then f is holomorphic inside γ. 3 Wednesday, February 12 During this and the next class, we focused on complex P1 k power series. We first proved Abel's lemma, which states that if k=0 ak(z − P ) converges at some z, then the series converges at each ! 2 D(P; r), where r = jz −P j. This allows to define the radius of convergence of a power series as X k r := supfj! − P j : ak(! − P ) convergesg: We proved the root test for the radius of convergence. Lemma (root test) 1 1=k a)r = 1=k if lim supk!+1jakj > 0; or lim supk!+1jakj 1=k b) + 1 if lim supk!+1jakj = 0: Next proposition states that inside the disc of convergence, the power series con- verges uniformly and absolutely. P1 k Proposition Let k=0 ak(z − P ) be a power series with radius of convergence r. Then, for any number R with 0 ≤ R < r, the series converges uniformly and absolutely on D¯(P; R). We also proved that the series obtained by term by term differentiation of the power series of a holomorphic function f is convergent and equal to the corresponding derivative. Finally, we established uniqueness of the power series, i.e. if 1 1 X j X j aj(z − P ) = bj(z − P ) j=0 j=0 and both series are convergent on D(P; r), then aj = bj for every j. Friday, February 14 A complex function is analytic if there exists a power series P1 j f(z) = j=0 aj(z − P ) for all z : jz − P j < r. We proved that holomorphic functions are analytic functions and the converse is also true. Theorem If f is a holomorphic function on the set U ⊂ C and P 2 U; D(P; r) ⊂ U. Then 1 X f (j)(P ) f(z) = (z − P )j j! j=0 for all z : jz − P j < r. The radius of convergence at each point is at lest equal to the distance between P and @U. Monday, February 17 We discussed Cauchy estimates and corollaries, including the Liouville's Theorem and the fundamental theorem of algebra. Theorem Let f be a holomorphic function on an open set U ⊂ C. Let P 2 U ¯ and r > 0 be such that D(P; r) ⊂ U. Set M = supz2D¯(P;r) jf(z)j. Then, for each k = 1; 2;::: we have the estimate Mk! jf (k)(P )j ≤ : rk We say that a function fC ! C is entire, if it is holomorphic on the whole C. Theorem (Liouville) An entire bounded function is a constant. Corollary Assume that f : C ! C is an entire function and there exist a real number C and a positive integer k such that jf(z)j ≤ jzjk for all z with jzj > 1. Then f(z) is a polynomial of degree at most k. 4 Finally, we used Liouville's theorem to give a proof of the fundamental theorem of algebra. Corollary If p(z) is a holomorphic polynomial of degree k, then there are k com- plex numbers α1; α2; : : : αk (not necessarily distinct) and a nonzero constant C such that p(z) = C:(z − α1):(z − α2) ::: (z − αk): Wednesday, February 19 We studied the uniform limits of holomorphic func- tions and proved the following theorem. Theorem Let U be an open set in C. Let ffjg be a family of holomorphic functions on U, which converges uniformly over the compact subsets to f. Then f is holomorphic on U. Thus, the set of holomorphic functions is closed under the operation uniform con- vergence over the compact subsets. Moreover, we have the following corollary. Corollary Let U be an open set in C. Let ffjg be a family of holomorphic functions on U, which converges uniformly over the compact subsets to f. Then for (k) each positive integer k we have that ffj g converges uniformly over the compact subsets to f (k). Friday, February 21 We studied the zeros of a holomorphic function f and showed that if f 6= const, then it can not have too many zeros. Theorem Let U be an open and connected subset of C and f : U ! C be holo- morphic. Then the set of zeros Z = fz 2 U : f(z) = 0g does not have accumulation 1 points inside of U unless f(z) = 0. Equivalently, if there exist a sequence fzjgj=1 and z0 2 Z such that zj ! z0, then f = 0. This does not exclude the possibility that for a non-trivial holomorphic function 1 there is an accumulation point z0 on @U. For example, the function f(z) = sin( 1−z ) 1 has zeros zn = 1 − nπ , which accumulate at 1 2 @D(0; 1). We also proved a number of useful corollaries.
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