ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

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46 <strong>ANALYSIS</strong> QUALS γR Solution. The problem is false as stated, as some assumption of boundedness must be made. A counterexample is to let X be l ∞ , the space of bounded sequences (x1, . . . , xn, . . .), and let A : (x1, . . . , xn, . . .) ↦→ (x1, x2/2, . . . , xn/n, . . .). Then A is linear and surjective, so 0 ∈ ρ(A), but 1/n ∈ ρ(A) for any n ∈ N, so ρ(A) is not open. That said, we alter the problem and change the definition of ρ(A). Let ρ(A) = {λ ∈ C : (λ − A) −1 exists and is bounded} Let G be the set of operators on X with bounded inverse. We claim that G is open. Let S ∈ G and suppose that T − S < S −1 1 . Then S −1 T − I = S −1 (T − S) < 1, so we have that (S −1 T ) −1 = (I − (I − S −1 T )) −1 = (I − S −1 T ) i exists. Therefore T −1 = (S −1 T ) −1 S −1 , so T is invertible. Therefore, G is open. Define f : C → B(X) by α ↦→ (αI − T ). Then f is continuous, so f −1 (G) is open. But f −1 (G) = ρ(A), so ρ(A) is open. Problem 8: Determine the number of zeros of the polynomial p(z) = z 4 + z 3 + 4z 2 + 2z + 3 in the right half-plane {z : Re z > 0}. Solution. Let ΓR be the contour that is the right half of the circle of radius R centered at the origin. Let γR be the semi-circle part of the contour, and let γR ˆ be the segment from Ri to −Ri. Note that on {Re z = 0}, we have that Re p(it) = t 4 − 4t 2 + 3 = (t 2 − 3)(t 2 − 1), and Im p(it) = −t 3 + 2t = −t(t 2 − 2). Thus, there are no zeros on {Re z = 0}. Thus, we can use the argument principle in this contour, which says that p ΓR ′ (z) p(z) dz = 2πi(number of zeros in the interior of ΓR). By taking R → ∞, we can calculate the total number of zeros in {z : Re z > 0}. On γR, p(z) = z 4 + O(|z| 3 ), and p ′ (z) = 4z 3 + O(|z| 2 ), so p′ (z) p(z) = 4z−1 + O(|z| −2 ). Thus, p ′ (z) dz = p(z) π 0 Ri · 4 · e−θ R dθ + O(|R|−1 ) = 4πi + O(|R| −1 ) → 4πi as R → ∞. On ˆ γR, note that critical points (points where p crosses either the real or imaginary axes) are at t = 0, ±1, ± √ 2, ± √ 3 for z = it. Suppose R > √ 3. Then on ( √ 3i, Ri), p takes values in the 4th quadrant (by noticing that the real part is positive and the imaginary part is negative). Similarly, we
<strong>ANALYSIS</strong> QUALS 47 see that on ( √ 2i, √ 3i), p is in the 3rd quadrant; on (i, √ 2i), p is in 2nd quadrant; on (0, i), p is in 1st quadrant; on (−i, 0), p is in 4th quadrant; on (− √ 2i, −i), p is in 3rd quadrant; on (− √ 3i, − √ 2i), p is in 2nd quadrant; and on (−Ri, √ 3i), p is in the 1st quadrant. Thus, moving from Ri to −Ri gives two clockwise revolutions around the origin, meaning the winding number is 2, which means that p γR ˆ ′ (z) p(z) dz → −4πi as R → ∞. Hence, ΓR p ′ (z) p(z) dz → 4πi − 4πi = 0, so there are no zeros in {Re z > 0}. Problem 9: Let f(z) be analytic for 0 < |z| < 1. Suppose there are C > 0 and m ≥ 1 such that f (m) (z) ≤ C , 0 < |z| < 1. |z| m Show that f(z) has a removable singularity at z = 0. Solution. Suppose that f(z) has Laurent series ∞ cnz n = −1 cnz n ∞ + cnz n . Then f (m) (z) = n=−∞ −1 n=−∞ cn n=−∞ n=0 n! (n − m)! zn−m ∞ + n=0 cn n! (n − m)! zn−m . Then as f (m) (z) ≤ C |z| m for 0 < |z| < 1, we have that cn = 0 for n < 0. ∞ Thus, f(z) = cnz n , so f is analytic on D. n=0 Problem 10: Let J = {iy : 1 ≤ y < ∞} and H be the open upper half plane. Consider the domain D = H\J. Find a bounded harmonic function u : D → R such that u(x + iy) → 0 as y ↓ 0 and u(z) → 1 as z → J. It is fine to represent the solution in terms of a composition of conformal maps. Proof. Let’s map this slit half plane to something better, keeping track of where we’re sending J. Let f1(z) = z−i z+i . Then f1 takes D to the slit disc, and takes J to the slit. Let f2(z) = √ z, under the branch cut that makes sense (positive real axis). Then f2 takes the slit disc to the upper half disc, taking the image of J to R ∩ D. Let f3(z) = z−1 z+1 . Then f3 takes the upper half disc to the second quadrant of the plane and takes the image of J to the negative real axis. Let f4(z) = iz2 . This takes the second quadrant to the right half plane, taking the image of J to the positive imaginary axis. Let f5(z) = Log(z), with the negative real axis branch cut. Then f5 takes the right half plane to the horizontal strip between iπ/2 and −iπ/2, which takes J to the line Im(z) = π/2.
Page 1 and 2: ANALYSIS QUALIFYING EXAM PROBLEMS B
Page 3 and 4: ANALYSIS QUALS 3 Proof. (a) First,
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Page 15 and 16: ANALYSIS QUALS 15 Then B has the s
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Page 27 and 28: ANALYSIS QUALS 27 Solution. Picard
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Page 37 and 38: Problem 10: Let the power series f(
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Page 49: ANALYSIS QUALS 49 If z0 ∈ K, choo

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