Complex Analysis Qualifying Exam Solutions

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1 Complex Analysis Qualifying Exam Solutions May, 04 Part.. Let log z be the principal branch of the logarithm defined on G = {z C z (, 0]}. Show that if t > 0, then the equation log z = t has exactly one root in G. z Solution. Let z = re iθ where r > 0 and π < θ < π. Then log z = t z becomes log r + iθ = t r cos θ i t sin θ. r Since θ and t sin θ have opposite signs when θ 0, it follows that if z G, then r log z = t if and only if z = r where r > 0 satisfies log r = t. But log r is strictly z r increasing and t is strictly decreasing. Hence, the Intermediate Value Theorem r implies that log r = t has exactly one root in {r R r > 0}. Therefore, log z = t r z has exactly one root in G.. Let G = C \ [, ]. Prove that f(z) = z has an analytic square root on G but does not have an analytic logarithm on G. Solution. To see that f has a square root, consider the function g defined on G by the formula ( ) g(z) = f(z) e arg(z )+arg(z+), where the arguments are chosen in [0, π). g is a well defined continuous function on G satisfying g = f. (g is analytic since z is locally - on C \ {0}; cf. Proposition.0 pg. 39 Conway). To see that f does not have an analytic logarithm on G, recall that this is equivalent to f having a primitive on G. But if γ is a path in G that winds once f around [, ] in the counterclockwise direction, then γ f f dz = γ ( z + ) dz = 4πi 0. z +

2 Therefore, f f cannot have a primitive (cf. Corollary. pg. 66 Conway). 3. Prove that the zeros of the polynomial p(z) = z n + c n z n +... c z + c 0 all lie in the open disk with center 0 and radius R = + c n c + c 0. Solution. Note that R = if and only if p(z) = z n and that in this case the assertion is obviously true. Therefore, we may assume that R >. If z = R, then z n p(z) = c n z n +... c z + c 0 ( c n c + c 0 ) (R (n ) + R (n ) ) = ( R ) ( R n ) R = ( R n ) < R n = z n. Therefore, by Rouche s Theorem, the 0 s of p all lie in the open disk with center 0 and radius R. 4. Let G be a connected open set in C and let a G. Prove that if F H(G) is a normal family, then D = {f H(G) f(a) = 0 and f F} is a normal family. Solution. Assume that F H(G) is normal. That D is normal, will follow from Montel s Theorem if we can show that D is locally bounded. Accordingly, fix b G. Since G is assumed connected, there exists a rectifiable path γ : [0, ] G satisfying γ(0) = a and γ() = b. Since F is normal and {γ} is compact, there exists a constant M such that g F max g(γ(t)) M. t [0,]

3 Also, since F is normal, there exist r > 0 and M such that B(b, r) G and g F sup g(z) M. z B(b,r) It follows that if f D and w B(b, r), then f(w) = f (z) dz + γ γ [b,w] f (z) dz + f (z) dz [b,w] f (z) dz Thus, D is locally bounded. M γ + M r. 5. Show that if G C is a simply connected subset of C, f : G G is analytic, and f(z) is not identically equal to z, then f has at most one fixed point in G. Solution. Recall that Problem on the 0B Final Exam was to show that if g : D D is analytic and g(z) is not identically equal to z, then g can have at most one fixed point in D. The simple proof of this fact was based on an application of Schwarz s Lemma. Since G is assumed simply connected and C, it follows by the Riemann Mapping Theorem that there exists an analytic bijection φ : G D. If we define g by g = φ f φ, then g : D D is analytic and f(z) is not identically equal to z if and only if g(z) is not identically equal to z. Furthermore, a G is a fixed point for f if and only if φ(a) is a fixed point in D for g. Therefore, the assertion of this problem follows from the result in the previous paragraph. 6. Show that if G is an open subset of C, then there exist ideals in H(G) that are not finitely generated. Solution. Construct an infinite sequence of distinct points {z k } k= no limit points in G. For each n define in G that has I n = {f H(G) f(z k ) = 0 for each k n}. 3

4 It is straightforward to prove that for each n, I n is an ideal and also that I, defined by I = I n, n= is an ideal. We claim that I is not finitely generated. For if f, f,..., f m generate I, then, as there exists N such that f k I N for k =,,..., m, it would follow that I = I N. On the other hand, the Weierstrass Factorization Theorem implies the existence of an f I N+ with f(z N ) 0, i.e., I I N. Part.. Let G = {z C z < and z 3 > 3 }, K = G = {z C z and z 3 3 }, and A(K) denote the space of continuous functions on K that are analytic on G equipped with the uniform norm on K. For the purposes of this problem, a Laurent polynomial ia a function of the form N n= N a nz n and a Laurent series is an infinite sum of the form n= a nz n. Determine the truth or falsity of the following assertions. Be sure to justify your answers. (a) The polynomials are dense in H(G). (b) The polynomials are dense in A(K). (c) If f is analytic on a neighborhood of K, then f can be uniformly approximated on K by a Laurent polynomial. (d) If f H(G) then f can be represented on G by a Laurent series. Solution to (a). This assertion is true. Fix a compact open exhaustion {K n } of G that has the property that C \ K n is connected for each n. For example, the sets K n defined by K n = {z C z n n + and z 3 3 n + n } have this property. To prove that the polynomials are dense in H(G), fix f H(G). For each n, since C \ K n is connected, Runge s Theorem implies that there exists a polynomial 4

5 p n such that f(z) p n (z) n. For such a sequence of polynomials, p n f in H(G). Solution to (b). This assertion is false. Note that the function f(z) = is in z A(K). If {p n } is a sequence of polynomials and p n in A(K), then there exists z n such that max p z K z n(z) <. But if we set f(z) = zp n (z), as f is analytic on a neighborhood of D, the Maximum Principle implies that = f(0) max f(z) = max p z = z = z n(z) <. This contradiction implies that the polynomials are not dense in A(K). Solution to (c). This assertion is true. C \ K has two components, C = {z z 3 < 3 } and C = {z z > } { }. Furthermore, 0 C and C. Therefore, by Runge s Theorem, if f is analytic on a neighborhood of K, f can be uniformly approximated on K by rational functions R whose only poles are at 0 and. But if R is a rational function whose only poles are at 0 and, then R is a Laurent polynomial. Solution to (d). This assertion is false. A particularly simple counterexample is obtained by considering the function f(z) = (z a) where a 3 = 3. f has two Laurent series representations, and f(z) = f(z) = n= n=0 a n, a < z zn z n a n+, z < a. If a,, then neither of these series converge on G. 3. (a) Prove that the formula π f(z) = lim 0+ 5 t z dt

6 defines an analytic function in {z C Re z > 0}. (b) Show that if f is defined as in part (a), then f has an analytic continuation to {z C Re z > } \ {0} with a simple pole at 0 with residue. Solution. (a) Let G = {z C Re z > 0}. In Conway s language, the assertion is that the integral π t z 0 dt converges uniformly in G. Fix a compact set K G. Since K is compact, if we let ρ = min Re z, z K t then ρ > 0. Since has a removable singularity at 0, there exists δ (0, ) and a constant c such that 0 < t < δ = c t. If, β (0, δ) with < β and z K, it follows that β t z dt β β t Re z dt c t tρ dt = c ρ (βρ ρ ). Noting that either Morera s Theorem or Leibniz s Rule imply that π t z dt is analytic on G when (0, π ), this estimate implies that if we choose a sequence { n } in (0, π), then { π/ t z dt } n is a Cauchy sequence in H(G). Since H(G) is complete, it follows that there exists f H(G) such that π n t z dt f in H(G). Furthermore, the function f does not depend on the choice of sequence { n }. This proves that there exists an analytic function f on G satisfying π f(z) = lim 0+ t z dt. 6

7 (b) We adapt the trick used in class to analytically continue the Riemann ζ function to the critical strip {z 0 < Re z < }. Let and f (z) = π 0 f (z) = ( ) t z dt t π 0 t tz dt. Noting that has a removable singularity at 0, it follows by a straightforward sin z z modification of the analysis in part (a) that f is a well defined analytic function on {z Re z > }. Also, if Re z > 0, then f (z) = ( π )z z by straightforward calculation. It follows that if Re z > 0, then Hence, f(z) = f (z) + f (z) = f (z) + ( π )z z. f (z) + ( π )z z, which is analytic in {z C Re z > } \ {0} with a simple pole at 0 gives the desired continuation of f. Res(f, 0) = lim z (f (z) + ( π )z z 0 z ) = ( π )0 =. 3. Let G be an open set in C and let (S (G), ρ) denote the sheaf of germs of analytic functions on G. (a) Prove that the sheaf topology on S (G) is Hausdorff. (b) Let Γ(t) = (γ(t), [f t ] γ(t) ), 0 t be a function from [0, ] into S (G) and for each t [0, ] choose an open set D t such that γ(t) D t G and f t is analytic on D t. Prove that Γ is continuous if and only if γ is a path in G and {(f t, D t ) 0 t } is an analytic continuation along γ. 7

8 Solution to (a). Assume that (a, [f] a ), (b, [f] b ) S (G) with (a, [f] a ) (b, [f] b ). Either a b, or a = b and [f] a [f] b. If a b, choose disjoint neighborhoods D a, D b G of a and b respectively such that f is analytic on D a and g is analytic on D b. Then N(f, D a ) = {(z, [f] z ) z D a } and N(g, D b ) = {(z, [g] z ) z D b } are disjoint neighborhoods of (a, [f] a ) and (b, [f] b ) respectively. Solution to (b). First assume that Γ(t) = (γ(t), [f t ] γ(t) ), 0 t is a continuous function from [0, ] into S (G) and that D t is an open set such that γ(t) D t G and f t is analytic on D t. We wish to show γ is a path in G and {(f t, D t ) t [0, ]} is an analytic continuation along γ. Since γ = ρ Γ, that γ is a path in G follows immediately follows from the continuity of Γ and ρ. To prove that {(f t, D t ) t [0, ]} is an analytic continuation along γ, fix t [0, ]. Since Γ is continuous and N(f t, D t ) is a neighborhood of Γ(t), there exists δ > 0 such that s [0, ] and s t < δ = Γ(s) N(f t, D t ). But Γ(s) = (γ(s), [f s ] γ(s) ) and N(f t, D t ) = {(z, [f t ] z ) z D t }. Therefore, s [0, ] and s t < δ = γ(s) D t and [f s ] γ(s) = [f t ] γ(s). This proves that {(f t, D t ) t [0, ]} is an analytic continuation along γ. Conversely, assume that γ is a path in G, Γ(t) = (γ(t), [f t ] γ(t) ), 0 t is a function from [0, ] into S (G), and that {(f t, D t ) t [0, ]} is an analytic continuation along γ. We wish to show that Γ is continuous. Accordingly, fix t [0, ] and an open set Ω in S (G) such that Γ(t) Ω. By the definition of the sheaf topology, there exists an open set U in G such that Γ(t) N(f t, U) Ω. Since {(f t, D t ) t [0, ]} is an analytic continuation along γ, there exists δ > 0 such that s [0, ] and s t < δ = γ(s) D t and [f s ] γ(s) = [f t ] γ(s). Since γ is a path, there exists δ > 0 such that Therefore, if δ = min{δ, δ }, then s [0, ] and s t < δ = γ(s) U. s [0, ] and s t < δ = γ(s) D t U and [f s ] γ(s) = [f t ] γ(s). But γ(s) D t U and [f s ] γ(s) = [f t ] γ(s) imply that Γ(s) = (γ(s), [f s ] γ(s) ) {(z, [f t ] z ) z D t U} = N(f t, U) Ω. Summarizing, we have shown that if t [0, ] and Ω is a neighborhood of Γ(t) in S (G), then there exists δ > 0 such that Γ(s) Ω whenever s [0, ] and s t < δ. Therefore, Γ is continuous. 8