Midterm Exam 2 Solutions - Complex Analysis with Applications | MATH 417, Exams of Mathematics

Download Midterm Exam 2 Solutions - Complex Analysis with Applications | MATH 417 and more Exams Mathematics in PDF only on Docsity! Math 417 – I. Nenciu Midterm 2 Solutions, March 30, 2012 Name: Instructions: Write your name at the top; answer all questions; show work to earn partial credit; answers without justification will not earn any credit. Problems: 1. Let f be an entire function such that |f(z)| ≤ |z| 12 , for all z ∈ C . (a) Show that f is a constant function. (b) Show that f ≡ 0. 2. Let f be a function which is analytic in the open unit disk D and such that |f(z)| ≤ 1 for all z ∈ D , and f(−1/2) = 0 . (a) Let α be a complex number, with |α| < 1. What is the definition of the function Bα, and what are its properties? (b) Define a function g : D → C by setting g(z) = f(z) B−1/2(z) for all z ∈ D , z 6= −1 2 . Find the value that g must take at z = −1 2 which guarantees that g becomes analytic in all of D. (JUSTIFY why this particular value works!) (c) Extend g to the whole unit disk D using the value you found in part (b). Show that |g(z)| ≤ 1 for all z ∈ D . (NOTE: g is in general NOT defined on the unit circle!) (d) Find the best possible upper bound for |f(1/2)|. 3. Let C be the unit circle, positively (i.e., counter-clockwise) oriented, and define a function g(z) = 1 2πi ∫ C e3s(s2 − 3s+ 4) (s− z)3 ds , for all z ∈ C, z /∈ C. Find g(0). 4. Let f be a function which is analytic in the unit disk D, and continuous in the closed disk D̄. Further assume that |f(z)| ≤ 2 for |z| = 1, Im(z) ≥ 0 and |f(z)| ≤ 3 for |z| = 1, Im(z) ≤ 0. (a) What can you say about the function h defined by h(z) = f(z) f(−z)? (b) Show that |f(0)| ≤ √ 6. 2 (d) If we apply the conclusion from part (c) to z = 1 2 , we find that ∣ ∣f(1/2) ∣ ∣ ∣ ∣B−1/2(1/2) ∣ ∣ = ∣ ∣g(1/2) ∣ ∣ ≤ 1 , and hence |f (1/2)| ≤ ∣ ∣B−1/2 (1/2) ∣ ∣ = 1 2 + 1 2 1 + 1 2 · 1 2 = 1 5 4 = 4 5 . Since B−1/2 itself is one of the functions which satisfy the hypothe- ses imposed for f , it follows that the bound above is indeed the best possible upper bound for |f(1/2)|. 3. Let us start by defining f : C → C f(z) = e3z(z2 − 3z + 4) , and noting that f is an entire function. By plugging in z = 0, we see that we are required to compute g(0) = 1 2πi ∫ C f(s) s3 ds . We know from the general theory that, since f is entire and C is the positively-oriented unit circle, 1 2πi ∫ C f(s) sk+1 ds = f (k)(0) k! , as this is the coefficient of zk in the power series expansion of f around 0. In other words, we find that g(0) = f ′′(0) 2 . Direct calculations show that f ′(z) = e3z ( 3(z2 − 3z + 4) + (2z − 3) ) = e3z ( 3z2 − 7z + 9 ) , and f ′′(z) = e3z ( 3(3z2 − 7z + 9) + (6z − 7) ) = e3z ( 9z2 − 15z + 20 ) . Thus g(0) = 1 · 20 2 = 10 . 5 4. (a) Since we know by hypothesis that f is analytic on D̄, and since z ∈ D̄ iff −z ∈ D̄, it follows that h is also analytic on the closed unit disk D̄. Now consider a point z ∈ ∂D. Note that Im(z) ≥ 0 ⇐⇒ Im(−z) ≤ 0 and Im(z) ≤ 0 ⇐⇒ Im(−z) ≥ 0 . This implies that, for any z ∈ ∂D, ∣ ∣h(z) ∣ ∣ = ∣ ∣f(z) ∣ ∣ · ∣ ∣f(−z) ∣ ∣ ≤ 2 · 3 = 6 . By the MaximumModulus Theorem, we then can conclude that |h(z)| ≤ 6 for all z ∈ D̄. (b) In particular, the conclusion from part (a) implies that: ∣ ∣f(0) ∣ ∣ 2 = ∣ ∣h(0) ∣ ∣ ≤ 6 =⇒ ∣ ∣f(0) ∣ ∣ ≤ √ 6 . 6