Download Final Exam Solutions for Math 412: Complex Analysis - Prof. Scott Annin and more Exams Mathematics in PDF only on Docsity!
May 2009 Sample Final Exam Name:
Math 412
SOLUTIONS
Problem 1. (16 points) State any 8 of the following: (a): Euler’s Formula (b): Cauchy’s Theorem (c): Higher-order version of Cauchy’s Integral Formula (d): Morera’s Theorem (e): Gauss’ Mean Value Theorem (f ): Maximum Modulus Principle (g): Liouville’s Theorem (h): Fundamental Theorem of Algebra (i): Cauchy’s Residue Theorem (j): Taylor’s Theorem SOLUTION: Look these up in the textbook or class notes! Problem 2. (6 points) Identify all points in the complex plane where |z + 1| ≤ 4 − |z − 1 |. SOLUTION: Begin by looking for where equality holds: |z + 1| + |z − 1 | = 4. This is the set of points z such that the sum of the distance to 1 and to −1 is equal to 4. This is an ellipse in the complex plane with foci ±1. The major axis joins the points ±2 and the semimajor axis joins the points ±√ 3 i. Due to the inequality, we see that the collection of points in this case lie in the interior of the ellipse described above. Problem 3. (a): (6 points) Find all values of 81 /^6 SOLUTION: We are looking for all six solutions to the equation z^6 = 8 = 8ei(2πn). Taking the 1/6th power of each side, we obtain z = √^68 eiπn/^3 , which we can evaluate for n = 0, 1 , 2 , 3 , 4 , 5 to obtain the six distinct solutions: Case 1: n = 0. Here we obtain z = √^6 8. Case 2: n = 1. Here we obtain z = √^68 eiπ/^3 = √^68
2 +^ i^
√ 3 2
Case 3: n = 2. Here we obtain z = √^68 e^2 iπ/^3 = √^68
− 12 + i √ 23
Case 4: n = 3. Here we obtain z = √^68 eiπ^ = − √^6 8.
Case 5: n = 4. Here we obtain z = √^68 e^4 iπ/^3 = √^68
− 12 − i
√ 3 2
Case 6: n = 5. Here we obtain z = √^68 e^5 iπ/^3 = √^68
2 −^ i^
√ 3 2
(b): (6 points) Find all values of z such that Log(z) = 1 − i π 4. SOLUTION: Given z ∈ C, recall that Log(z) = ln |z| + iArg(z). Therefore, we are seeking values of z such that ln |z| = 1 and Arg(z) = −π/4. Hence, the polar form of z must be z = e · ei(−π/4+2πn)^ = e1+i(−π/4+2πn), where n ∈ Z. Hence, the solution set is {z ∈ C : z = e1+i(−π/4+2πn)^ for some n ∈ Z }. Problem 4. (10 points) Show that g(z) = |z|^2 is a function that is continuous everywhere but analytic nowhere. SOLUTION: Writing z = x+iy, we have |z|^2 = x^2 +y^2. Then we can write g(z) = u(x, y)+iv(x, y) where u(x, y) = x^2 + y^2 and v(x, y) = 0. Since both u(x, y) and v(x, y) are continuous real-valued functions throughout R^2 , we can apply the theorem that states that a complex-valued function is continuous if and only if its corresponding real and imaginary component functions are continuous to conclude that g(z) is continuous everywhere. However, since ux = 2x and uy = 2y and vx = vy = 0, the Cauchy-Riemann equations are only satisfied at the origin z = 0. Hence, there is no disk about any point in C throughout which g is differentiable. By definition, g is therefore nowhere analytic. Problem 5. (a): (5 points) Let f have a pole of order k at α. Show that f ′^ has a pole of order k + 1 at α. SOLUTION: Since f has a pole of order k at α, we may write
f (z) = (^) (z g−(z α))k ,
where g is analytic at α and g(α) 6 = 0 (otherwise we could factor z − α from g(z) and cancel with the denominator to reduce the order of the pole α for f. Applying the quotient rule, we see that
f ′(z) = (z^ −^ α)
kg′(z) − kg(z)(z − α)k− 1 (z − α)^2 k^ =
(z − α)g′(z) − kg(z) (z − α)k+^. Since the numerator of this latter fraction does not contain α as a root (since g(α) 6 = 0), the order of the pole α here is the power in the denominator, k + 1, as requested. (b): (5 points) Let f be entire containing k factors of α (we say that α is a zero of order k). Show that g(z) = f^
′(z) f (z)
SOLUTION: We can use the Maclaurin series for cos z:
cos z = 1 − z
2 2! +^
z^4 4! −^
z^6 6! +^... , so that
f (z) =
1 − (2 2!z) 2 + (2 4!z) 4 − (2 6!z) 6 +...
z^2 = − (^) z^12 − 2
2 2! +
4! z
6! z
(b): (4 points) Classify all singularities of f. SOLUTION: z = 0 is the only singularity–a pole of order 2. This is evident from the Laurent series in part (a). Problem 9. Suppose f is analytic on a simply connected domain D containing C 4 (0). If |f (z)| ≤ 7 for all z ∈ C 4 (0), find a sharp upper bound for (a): (4 points) |f (3)(0)|. SOLUTION: We can directly apply the Cauchy Inequalities (Theorem 6.17) with R = 4, z 0 = 0, M = 7, and n = 3: |f (3)(z 0 )| ≤ 3! 4 · 3 7 =^4264 =^2132. (b): (4 points) |f (3)(2)|. SOLUTION: We can only apply the Cauchy Inequalities (Theorem 6.17) in a disk of radius 2 in this case (larger disks centered at z 0 = 2 could protrude outside of C 4 (0), the region of guaranteed analyticity of f ). So we have R = 2, z 0 = 2, M = 7, and n = 3. Thus,
|f (3)(2)| ≤ 3! 2 · 3 7 =^428 =^214. (c): (4 points) Is f (z) = e^2 z^3 −sin^ z^ bounded? Explain. SOLUTION: NO. Assume to the contrary that f is bounded. Since f is comprised on com- positions, products, and subtractions of entire functions, f is entire. By Liouville’s Theorem, we conclude that f is constant. However, f is clearly not constant (just plug in two values of z that yield different outputs), a contradiction. Problem 10. (a): (6 points) Find all z such that ∑^ ∞ n=
n! (z − i)n
converges. SOLUTION: Let us apply the ratio test:
nlim→∞
(z^ (n−+1)!i)n+ (z−^ n!i)n
∣∣ = (^) nlim→∞
∣∣^ n z^ + 1− i
for all z ∈ C. Hence, the series is nowhere convergent. (b): (4 points) Evaluate (^) ∞ ∑ n=
n + i + 1 −^
n + i
or state that it does not converge. SOLUTION: This is a telescoping series: ∑ n=
n + i + 1 −^
n + i
i + 2 −^
i + 1
i + 3 −^
i + 2
i + 4 −^
i + 3
The nth partial sum is Sn = (^) n + 1 i + 1 − (^) i + 1^1.
We have nlim→∞ Sn^ =^ −^ i + 1^1 , so we conclude that this series converges: ∑^ ∞ n=
n + i + 1 −^
n + i
= − (^) i + 1^1.
Problem 11. (24 points) Practice two of the theorems required for the exam from the proof list. SOLUTION: Consult the list of theorems and their proofs from the textbook or class notes.
