First Midterm Exam Questions with Solutions - Complex Analysis | MATH 6321, Exams of Mathematics

Download First Midterm Exam Questions with Solutions - Complex Analysis | MATH 6321 and more Exams Mathematics in PDF only on Docsity! MATH 6321 First Midterm February 18, 2009 You can use your book and notes. No laptop or wireless devices allowed. Write clearly and try to make your arguments as linear and simple as possible. The complete solution of one exercise will be considered more that two half solutions. All numbers appearing in the test are complex numbers and all functions are from C to C. You should solve at least one among ex. 1 and 2, one among 3 and 4, and one among 5 and 6. A total of 50 pts will give you an A. Name: Question: 1 2 3 4 5 6 7 Total Points: 10 10 10 10 10 10 20 80 Score: MATH 6321 First Midterm February 18, 2009 1. (10 points) Find a one-to-one analytic function f mapping the open unit disk D = {z | |z| < 1} (1) to the disk C = {z | |z − 1− 2i| < 2} (2) such that f(0) = 1 + i and f ′(0) is real and positive. (Hint: you may use the result of exercise III.3.9) Solution: We first note that the function g(z) = 2z +1+2i maps D into C. If h(z) is a Möbius transformation that maps D into itself, f(z) = g ◦ h(z) maps D into C. From III.3.9 we know that h(z) = az + b b̄z + ā (3) We thus have that f(z) = 2az + 2b b̄z + ā + 1 + 2i (4) We get f(0) = 2b ā + 1 + 2i ⇒ 2b ā = −i f ′(0) = 2|a|2 − 2|b|2 ā2 ⇒ |a| 2 − |b|2 ā2 > 0 (5) The second line implies that a is either real or pure immaginary. We can thus reduce the choice to a = 1 or a = i. But the first line tells that 2|b| = |a| thus a = 1 and b = −i/2. Page 1 of 7 MATH 6321 First Midterm February 18, 2009 4. (10 points) Let fn be analytic functions in the open unit disk D = {z | |z| < 1} that converge to a function f . Assume that fn converge uniformly to f in any closed disk Dr = {z | |z| ≤ r} of radius r < 1. Prove that f is analytic on the open unit disk. (Hint: given two analytic functions f and g, estimate |f ′(z)− g′(z)| in terms of |f(w)− g(w)|) Solution: First solution Let r < 1 and γ = {re2isπ|0 ≤ s ≤ 1}. Since fn is analytic we have f ′n(z) = 1 2πi ∫ γ fn(w) (w − z)2dw (9) for every |z| < r. This implies that |f ′n(z)−f ′m(z)| ≤ sup|z|=r |fn(z)−fm(z)|(r−|z|)−2. Since fn converge uniformly and r is arbitrary, we get that f ′ n(z) is a Cauchy sequence on every Dr with r < 1. Thus it converge uniformly to a function g on every Dr. It is now enough to observe that if fn → f uniformly and f ′n → g uniformely, than f ′ = g. In our case it is enough to observe that for every z, z′ ∈ Dr we have: |fn(z)− fn(z′)− f ′n(z)(z − z′)| ≤ C|z − z′|2 (10) where C = supn supDr f ′′ n(z) exists due to an argument similar to the above one. Taking the limit we get: |f(z)− f(z′)− g(z)(z − z′)| ≤ C|z − z′|2 (11) that means that g is the derivative of f . Second solution Let r < 1 and γ = {re2isπ|0 ≤ s ≤ 1}. Since fn is analytic we have fn(z) = 1 2πi ∫ γ fn(w) w − z dw (12) for every |z| < r. Taking the limit on n on both side we get f(z) = ∫ γ f(w) w − zdw (13) From Proposition IV.2.1 we get that f(z) is differentiable with respect to both <(z) and =(z). Since f(w)/(w−z) is an analytic function of z for every w we also get that the derivatives of f satisfies the Cauchy-Riemann equations and thus f is analytic for |z| < r. Since r was arbitrary we obtain the thesis. Page 4 of 7 MATH 6321 First Midterm February 18, 2009 5. (10 points) Let fn be analytic functions in the closed unit disk D = {z | |z| ≤ 1}. Suppose that fn converge uniformely to an analityc function f on D. Let, moreover, a (n) k be the coefficients of the Taylor series of fn around 0, i.e. fn(z) = ∞∑ k=0 a (n) k x k (14) Similarly let ak be the coefficients of the Taylor series of f around 0, i.e. f(z) = ∞∑ k=0 akx k (15) Define ²n = supk |a(n)k − ak|. (a) Show that limn→∞ ²n = 0. Solution: Let γ = {e2πis|0 ≤ s ≤ 1}. Since we have f (k)n (z) = k! 2πi ∫ γ fn(w) (w − z)k+1dw (16) we get |f (k)n (0)−f (k)(0)| ≤ k!² if n is so large that |fn(z)−f(z)| ≤ ². The thesis follows immediately from the fact that a (n) k = f (k) n (0)/k!. (b) Is the converse true? More precisely, is it true that if limn→∞ ²n = 0 then fn converges uniformly to f . Solution: Let a (n) k = 1/n if k < n and a (n) k = 0 if k > n. Clearly ²n = 1/n and limn→∞ ²n = 0. On the other hand, fn(1) = 1 so that fn does not converge uniformely to 0. Page 5 of 7 MATH 6321 First Midterm February 18, 2009 6. (10 points) Show that there exists a function f with f(0) = 0 that satisfies the differ- ential equation f ′(z) = z exp (f(z)) (17) in a region containing the origin. Show that the function f is analitic in a disc centered in z = 0 and compute the power series representation of f around z = 0. Find the radius of convergence of this series. Solution: Observe that the equation can be rewritten as d dz exp (−f(z)) = −z (18) or exp (−f(z)) = 1− z 2 2 (19) where we have used that f(0) = 1. Since 1− z2 2 6= 0 for |z| < √2 we have that f(z) = − ln ( 1− z 2 2 ) (20) This implies that f(z) as a power series expansion around z = 0 that can be imme- diately obtained from the expansion of ln(1− z). We get f(z) = ∞∑ n=0 anz n (21) where an = 0 if n is odd and a2m = 1 m zn 2m (22) if n = 2m is even. Moreover, since there is no singularity in |z| < √2 we clearly have that the convergence radius is √ 2. Page 6 of 7