Complex qual study guide James C. Hateley

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1 Complex qual study guide James C. Hateley General Complex Analysis Problem: Let p(z) be a polynomial. Suppose that p(z) for R(z) >. Prove that p (z) for R(z) >. Solution: Let p(z) be such a polynomial. Suppose that a i C are the zeros of p(z). Then we have n n p(z) = c (z a i ), p (z) = c (z a j ) Now the quotient is given by i= p (z) p(z) = n i= z a i i= i j If p (z) =, then the sum above must be equal to zero. Now if R(a i ) this implies that R(z a i ) >, by the assumption of the hypothesis. So there is a w such that w = z a i = x + ıy, where x >. Now we have = z a i w = x + ıy = x ıy ( ) x + y R > z a i Since each term in the sum has a positive real part we have p (z) for R(z) >. z a i Problem: Describe those polynomials a + bx + cy + dx + exy + fy with real coefficients that are the real parts of analytic functions on C. Solution: let u(x, y) = a + bx + cy + dx + exy + fy. u(x, y) being the real part of a holomorphic function implies that u(x, y) is harmonic. So u(x, y) = and xx u = d, yy u = f d + f = Problem: Prove or disprove that there exists an analytic function f(z) in the unit disc D(, ) such that ( ) ( f = f ) = n n n 3, n N Solution: Suppose there exists such a function with the above property. Since f(z) is analytic, it is uniquely determined by a cauchy sequence. Now {/n} is clearly cauchy, hence ( ) f = n n 3 f(z) = z 3 ( but clearly f(z) is an odd function and hence f ). Therefore there is no such function that n n3 satisfies the hypothesis Problem: Let p(z) be a polynomial such that all the roots of p(z) lie in D(, ). Prove that the roots of p (z) lie in D(, ). Proof: We will prove something stronger, that all the roots of p (z) lie inside the convex hull of the roots of p(z). Suppose all the roots of p(z) lie in D(, ), Let {a i } be the set of roots of p(z) including multiplicities, then we have n p(z) = c (z a i ), a i D(, ), c C i=

2 Taking the logarithmic derivative we have p (z) p(z) = n i= z a i In particular, if z is a zero of p (z) and still p(z), then n = z a i which implies This may also be written as ( n i= i= n i= z a i z a i = ) z a i z = n i= z a i a i. Taking their conjugates, we see that z is a weighted sum with positive coefficients that sum to one. If p(z) = p (z) =, then z = z + a i, and is still a convex combination of the roots of p(z). Since the unit disk is a convex hull of the roots of p(z), then all roots of p (z) are inside the unit disk. Problem: Prove or disprove that there is a sequence of analytic polynomials {p n (z)}, n N, so that p n (z) z 4 as n uniformly for z D(, ). Solution: The statement is not true. Suppose that there exists such a sequence of analytic polynomials such that p n (z) z 4. Then for all n we have d d z p n(z) = since p n (z) is analytic. However d d z z4 = 4 z 3 for all z C. Clearly 4 z 3 for all z C Problem: The Bernoulli polynomials φ n (z) are defined by the expansion Prove the following two statements: a) φ n (z + ) φ n (z) = nz n e tz e t = φ n (z) t n Proof: Let B(z) = etz e t, then B(z + ) B(z) = etz. Now by definition of B(z) we have the following expansion φ n (z + ) t n φ n (z) t n (tz) n = reindexing we have φ n (z + ) t n Hence we have φ n (z + ) φ n (z) = (n + )z n b) φ n+(n + ) n + = n k= k k Proof: By the previous part we have φ n (z) t n = φ n+ (n + ) φ n+ (n) = (n + )n n (tz) n (n )! φ n+ (n) φ n+ (n ) = (n + )(n ) n φ n+ () φ n+ () = (n + ) n.

3 3 Rearraging and using the recursive relation above we have φ n+ (n + ) = (n + )n n + φ n+ (n) = (n + )n n + (n + )(n ) n + φ n+ (n ) φ n+(n + ) n +. = (n + ) k= = n k= k k n k= k k Complex Integration Computer the area of the image of the unit disk D = {z : z < } under the map f(z) = z + z. Solution: Denote Ω = f(d), and let dσ denote the surface measure, then an integral for the surface area is given by ; dσ = J(u, v) da Now let f(z) = f(x, y) = u(x, y) + ıv(x, y), then we have Computing J(u, v), we have Ω f(x, y) = x + ıy + (x + ıy) J(u, v) = + x y D = x + x y y + x = ( + x) + y + ı (y + xy) Converting to polar coordinates we come to the integral π dσ = + r cos(θ) + r )r drdθ = 3π Ω Problem: Evaluate the integral: sin (x) x dx Solution: Consider the following contour Γ: γ := t t [ R, /R] γ := e ıt /R t [π, π] Γ = γ 3 := t t [/R, R] γ 4 := Re ıt t [, π] Now our function has a removable singularity at x =, so consider the following f(x) = eıx x R(f(x)) = cos(x) x = sin( x) x Now for the integral around Γ we have Γ f(z) dz = πı Res(f(z)) = πı lim z Now for the integral on γ we have γ f(z) dz = /R R e ıt d dz z f(z) = πı lim z ıe ız = πı( ı) = π t dt e ıt dt as R t

4 4 For the integral on γ we have γ f(z) dz = = ı π π π Now letting R and using L Hospitals rule we have For the integral on γ 3 we have Now for γ 4 we have ı γ 3 f(z) dz = Putting this all together we have Taking real parts we have Hence we have π = sin (t) t π π R /R e ıeıt /R (ıe ıt /R) (ı) ıe ıt /R e ıt π ( ) e ıeıt /R ıe ıt /R e ıt /R e ıeıt /R e ıt /R t dt γ 4 f(z) dz = π = π + dt = π Taylor and Laurent series = ı e ıt π π = π π dt = π e ıt t Re ıt R e ıt Rıe ıt dt Re ıt Re ıt t dt + e ıt sin (t) sin (t) sin (t) t dt + t dt = t Problem: Find the largest disc centered at in which the Taylor series for will converge. Solution: The singularities of + z = (z ) k k= t dt as R occur at ±ı. First consider the taylor series centered at ; + z + z = ( ) k z k k= instead of recomputing the coeffients and taking the limsup, notice the raduis of converge for the series at is. Since the series must avoid the sigularities the radius will be the distance from the center to the closest singularity, i.e. r = inf{ ı, + ı } = Problem: Find the raduis of convergence for the series: z n and z n dt dt

5 5 Solution: For the first one we have lim sup z n n Now lim n /n =, hence z <. For the second series we have lim sup z n n /n /n Hence the series converges for z < < z < lim sup /n n ( < z < lim sup(n)/n) /(n )! = n Problem: Let f be a non-constant entire function. Prove that if a polynomial. ( Solution: Consider g(z) = f z and consider the Laurent expansion: g(z) = a k z k lim f(z) =, then f must be z ), then lim z g(z) =. Now suppose that g(z) has a pole of order k + a k+ z k + + a + a z + z k g(z) = z k Now z k g(z) c = f() as z. This implies by continuity that z k g(z) ( c + )z k for large z. Hence z k g(z) is a polynomial of at most degree k. Now we have: k ( ) k g(z)z k = a n z n a n f = g(z) = z z k n f(z) = k a n z k n n= k Problem: Show that for R >, there is N R such that when n > N R, the function P n (z) = + z + z + + zn, z R. n z k Solution: First notice that P n (z) = k! and that P n(z) e z uniformly as n on compact sets k= of C. Fix R > n z k m ɛ > N R s.t. k! z k k! = n z k k! ɛ, n > m > N R. This implies that which implies that k= ez k= n z k k! < ɛ, k= k=m n > N R e z n z k < ɛ + k!, n > N R, z D(, R) k= n z k R > N R s.t. k!, n > N R k= Problem: Let f(z) be analytic on C {} and have a simple pole at z = with residue λ. Prove that for every R >, lim n Rn f (n) () ( )n λ = a n z n

6 6 Proof: Since f(z) has simple pole at one we have the Laurent expansion. Define g(z) = f(z) expansion Also we have the geometric series for f(z) = λ z + a n (z ) n λ, then g(z) is an entire function. Now for z < f(z) has the taylor z λ z = f(z) = λ z f (n) () (z ) λ + (z )) = λ ( ) n (z ) n This implies that the series for g(z) about is f (n) () ( f g(z) = (z ) n λ ( ) n (z ) n = (z ) n (n) ) () λ( ) n Now for z < we have that (z ) n f (n) () λ( ) n. But since g(z) is entire we this holds for any z C Hence for any R > we have So the result is shown. R n f (n) () λ( ) n as n Problem: Find the radius of convergence R of the series z n n and show the series converges uniformly on D(, R ). What is the radius of convergence R of the derivative of this series? Does it converge uniformly on D(, R )? Solution: Denote the above series by f(z). By taking limsup we find that the raduis of convergence is. Let z D(, ), then we have z n n n < hence the series converges uniformly on D(, ). Now the derivative f (z) of the series will converge on the unit disc by Abel s theorem. But will not converge uniformly on D(, ), since if z =, the series diverges. Problem: Let f(z) be analytic in the punctured unit disk U = {z : < z < } such that thre is a positive interger n with f n (z) z n for all z U. Show that z = is a removable singularity for f(z) Solution: Let g(z) = z n f (n) (z), then g(z) for all z D(, ). This implies that z = is a removable singularity of g(z). Now consider the Laurent series expansion for g(z) inside D(, ). g(z) = a n z n f (n) (z) = a n z k n n= k Now f(z) has the laurent expansion f(z) = b n z k f (n) ( ) n (n + k )! (z) = b n z n k (k!)

7 7 but since the Laurent expansion is unique, we must have b k = for all k <. Which implies that f(z) has a Taylor expansion about z =. Therefore f(z) has a removeable singularity at z =. Problem: Let f(z) be analytic in the disk U = { z < }, with f() = f () =. Show that ( z g(z) = f defines an analytic function on U. Moreover, show that the above function g(z) n) satisfies if and only if f(z) = cz. g(z) = f(z) Solution: Consider the Taylor expansion for f(z) with the conditions f() = f () =, this implies that f(z) = z n k= k= n z k = z h(z) for some holomorphic function h(z). Plugging in z/n we have ( z ( f = n) z z k n k ) = z n h ( z n Since f(z/n) is analytic for all n it suffices to show that the series ) ( z f converges normally on n) U. Let K be a compact set of U Since h(z) is analytic on K it is continuous. Hence h(z) attains it s maximum on K, denote this value as M. Now we have ( z z f = n) ( z ) n h z n n M ( z Hence if z K then the seris f convergies absolutely on K. Hences by the Weierstrass M- n) test ( z f converges uniformly on K, which implies the series converges normally since K was an n) arbitrary closed set in U. Therefore g(z) is analytic in U, since it s the normal limit of analytic function on U. For the second part, if f(z) = cz, then we have ( z f = c n) z n = cz On the other hand, suppose k= g(z) = f(z) n = f(z) n. consider the talyor expansion for f(z), plugging this in we have ( ) ( ) ( g(z) = z k ) n = n k z k but k= k= ( ( z z g(z) = f = n) k n k = k= n ) n z k Since the power series for an analytic function is unique, this implies that = for all k. Therefore f(z) = a z

8 8 Applications of Cauchy s Interal formula Problem: Let U C be a connected open set, and γ be a closed curve in U. Suppose that for any function f(z) holomorphic on U we have f(z)dz =. Does it imply that γ is homotopic to a constant curve? Solution: No γ does not have to a constant curve, consider the function f(z) = z, on the punctured disk U = D(, ) {}. Then f(z) is holomorphic on U, now fix r (, ) and let γ = re ıt + for t [, π], then by Cauchy s theorem we have f(z)dz =. but re ıt + is clearly not a constant curve. γ Problem: Let f(z) be entire holomorphic function on C such that f(z) cos(z). Prove f(z) = c cos(z) for some constant c. Solution: Consider g(z) = follows: f(z), then g(z), hence g(z) is a bounded function. Define ĝ(z) as cos(z) { g(z) if cos(z) ĝ(z) = lim g(z) z w if cos(w) = Then ĝ(z) is a bounded entire function. Hence by Lioville s theorem it must, i.e. ĝ(z) = c for some c C. It follows from the definition of g(z) that f(z) = c cos(z) Problem: Prove that there is no entire analytic function such that {z C : f (n) (z) = } = R Solution: First there exists an N such that S = {z C : f (N) (z) = } is dense in R, if not then R is a countable union of nowhere dense sets, which is a contradiction to the Baire Cateogry theorem. Now let z S, then for every ɛ >, the disc D(z, ɛ) contains infintely many points in S. Now let ζ D(z, ɛ) such that ζ / S, now by Cauchy estimates we have f (N) (ζ) N! π z+z =ɛ Now consider the change of variables w = z ɛ, then N+ N! f(z) dz = N! π z ζ π z+z =ɛ D(w,ɛ) N+ f(z) dz z ζ f(w + ɛ ) w N+ as ɛ. This implies that f (N) (ζ) =, hence ζ S which is a contradiction to ζ / S. Therefore there cannot exist such a function Problem: Find all entire functions f(z) on C satisfying f(z) z e x, z = x + ıy C Solution: First notice the following: z e x = ze x = ze x e ıy = ze z

9 Let g(z) = f(z) f(z), then g(z) is a bounded function since g(z) = zez ze z <. Hence the discontinuity at z = is removable. Define f(z) ze z ĝ(z) = z f(z) lim z ze z z = Now since f(z) and ze z are entire we have ĝ(z) as a bounded entire function. Hence by louvilles theorem ĝ(z) = k C. So we have f(z) kze z, where k. Problem: Complete the following problems: a) State the Lioville s theorem Lioville s theorem states that a bounded entire function is constant. b) Prove the Lioville s theorem by calculating the following integral f(z) (z a)(z b) dz and taking the limit R. z =R Solution: Suppose that f(z) is a bounded entire function, that is, there exists M R such that f(z) < M for all z C. Fix R >, now for a, b D(, R) the integral is bounded by; f(z) (z a)(z b) dz πrm (R a )(R b ) as R z =R Now by direct computation we have f(z) dz = πı (Res f(a) + Res f(b)) (z a)(z b) z =R f(b) f(a) = πı = b a This implies that f(b) = f(a) for all a, b C, hence f(z) is constant. 9 Problem: Find the number of zeros of the function f(z) = z 5 + 8z in the annulus < z <. Solution: Let D = {z C : < z < }, then D = {z C : z = or z = }. Now consider the function g(z) = z 5 + 8z. Then f(z) g(z) = on D. Now g(z) = z 5 + 8z = z z 4 + 8, on D and = min z g(z) z D So we have f(z) g(z) = < + f(z) on D. Also both f(z) and g(z) are holomorphic on D. Hence By Rouche s theorem f(z) and g(z) have the same number of zeros in D. Now g(z) = z(z 4 + 8), which implies z = and z 4 = 4. So the set of zeros that lie in D are So g(z) has 4 roots in D f(z) has 4 roots in D z = 4 /4 e ıπk/4, k =,,, 3 Problem: Find all roots of the equation z + sin(z) = in the unit disc. Solution: Clearly z = is a root, to show that this is the only root consider f(z) = z, g(z) = sin(z). Let z D(, ), now by convexity of e z we have g(z) = eız e ız e z + e z = e + e <

10 This implies that g(z) < = z = f(z) z D(, ) Hence by Rouche s theorem f(z) and f(g) + g(z) have the same number of roots in D(, ). z has root in D(, ), therefore z + sin(z) has root in D(, ), which is z = Problem: If f(z) is an entire function satisfying the estimate f(z) + z z C Show that f(z) is a polynomial and determine the best upperbound for the degree of f(z). Solution: First observe that 44 < < 45. Let R > and consider the Cauchy estimate for f (n) () on D(, R). f (n) () ( + R ) Now if we consider the Taylor expansion for f(z) about z =, we have f(z) = a n z n z C where a n = f (n) () If n >, let α = n > then we have the estimates let a n = f (n) () R n ( + R ) R n = R n + R α as R Hence we have a n = for all n >, which implies that a n = for all n 45. Therefore f(z) is a polynomial of degree at most 44. Problem: Show that f(z) = αe z z has only one zero in U = { z < } if α < /3 and no zeros if α > 3. Solution: For α < /3, let g(z) = z then we have f(z) g(z) = αe z α e x < e < = g(z) on U 3 Thus by Rouche s thoerem f(z) and g(z) have the same number of zeros in U. Therefore f(z) has exactly one zero in U since g(z) = z has one zero. For α > 3, let h(z) = αe z, then we have f(z) g(z) = z = 3 e < α e αez = h(z) on U Thus by Rouche s thoerem f(z) and h(z) have the same number of zeros in U. Therefore f(z) has no zeros in U, since h(z) no roots. Problem: Show that there is a holomorphic function defined in the set Ω = {z C : z > 4} Whose derivative is z (z )(z )(z 3). Is there a holomophic function on Ω whose derivative is z (z )(z )(z 3)? Solution: Let γ be a closed curve lying outside of Ω. Now if there exists such a function F (z), such that F z (z) = f(z) = (z )(z )(z 3)

11 then the following condition should be satisfied γ F (z) = It suffices to show this is true for γ = re ıt, where r >. Now F (z) has 3 simple poles {,, 3} lying inside of γ. Hence we have γ f(z) = πı (Res f() Res f() + Res f(3)) ( = πı + 3 ) = Hence by Morera s there does exists such a function F (z) such that F (z) = f(z) Now consider the function z g(z) = (z )(z )(z 3). Using the same ideas above, we have g(z) = πı (Res f() Res f() + Res f(3)) γ ( = πı ) Hence there cannot exist a homomorphic function G(z) such that G (z) = g(z) Find the integral z = 4z 7 z 8 z + dz Solution: Let f(z) = z 8 z + and g(z) = z 8, then we have Also we have f(z) g(z) = z + z + = 5 on D(, ) 5 < 8 = z 8 = g(z) on D(, ) Hence by Rouche s theorem f(z) and g(z) have the same number of zeros including multiplicties in D(, ). Since g(z) has 8 zeros, f(z) has 8 zeros in D(, ). Now observe 4z 7 z 8 z + dz = f (z) f(z) dz z = and f(z) is entire and non-vanishing on D(, ), since all of the roots of f(z) lie inside D(, ). Hence by the argument principle we have 4z 7 z 8 z + dz = (8) = 4 Normal Families z = Problem: Let F be a family of holomorphic functions on the unit disk D for which there exists M > such that f(z) dxdy M, f F. Show that F is a normal family. D Solution: We want to show f F is bounded. Consider the following construction, fix r (, ) D(, ) D(z, r) z D(,) z =

12 Since D(, ) is compact there exists a finite number of {z k } such that n D(, ) D(, ) D(z k, r) Let ɛ = inf{r, z k D(, )}, now for any f F we have f(z k ) = πɛ f(z k ) πɛ πɛ k= D(z k,ɛ) D(z k,ɛ) D(,) f(z) da f(z) da f(z) da M πɛ = M ɛ Since D(z k, ɛ) D(, ). Now this is for all z k and for all f F. Let M = max M ɛ. where the sup is taken over all possible finite covers of D(z, r). Then we have f(z) M. Hence F is a z D(,) bounded family. Therefore by Montel s theorem {f k } F f kj s.t. f kj u f Where f is holomorphic in D(, ), i.e. F is a normal family. Problem: Consider the family of functions {f α } α A that is holomorphic on a domain U. Suppose that for all z U, and for all α A we have R(f(z) (I(z)(f(z)). Prove that {f α } is a normal family. Solution: Consider the following two domains U := {z : x < y } and U := {z : x > y }. By the Riemann open mapping theorem, there exists maps φ such that φ : U D(, ) and φ such that φ : U D(, ). Now consider the following function: { φ f α, f α U h(z) α = φ f α, f α U then h α is holomorphic in U and bounded hence by Montels theorem h α is a normal family. Therefore {f α } is a normal family. Problem: Let F = {f α } be a family of holomorhiphc functions on D(, ) and for all z D(, ) Prove that F is a normal family. f (z) ( z ) + f(). Proof: Let ɛ > and consider D(, r) where r = ɛ. Now for z D(, r) we have f (z) f() z Using the triangle inequality and integrating we have f(z) = f (z) dz f (z) dz f() z dz < this is valid for all z D(, r) and for any f(z) F. Hence by Montel s theorem F is a normal family. Problem: Let Ω be a bounded domain in C, and let {f j }, j N be a sequence of analytic functions on Ω such that f j (z) da(z) Prove that {f j } is a normal family in Ω Ω

13 3 Proof: By definition a normal family implies that for every compact subset K of Ω, there exists a subsequence {f jk } that converges uniformly some f in K. Fix a compact set K Ω. Let r >, now we have K D(x, r) as an open cover for K. Since K is compact there exists finite set {x i }, such that n K D(x i, r) x K i= Now for each x j, we have the by the mean value theorem for holomorphic functions f j (x i ) πr f j (x, y) da D(x i,r) by Hölders inequality πr µ(d(x i, r)) f j (x, y) Since µ(d(x i, r)) = πr and f j (x, y) by the hypothesis. Now this implies that f j (x i ) is bounded for all j and x i, hence it is uniformly bounded. Therefore by Montel s theorem there exists a subsequence {f jk } such that f jk converges uniformly on K. Thus {f j } is a normal family in Ω Problem: (a) State the Montel Theorem for normal family. Montel s Theorem: Let F be a family of holomorphic functions on an open set U C. Suppose that for each compact set K U, there is M = M(K) such that f(z) M for all z K and all f F. Then for every {f α } F, there is a subsequence {f αk } that converges uniformly on compact subsets of U to a holomorphic limit, in otherwords, F is a normal family. (b) Let F be a set of holomorphic functions on the unit disk D(, ) so that Show that F is a normal family. π sup <r< f(re iθ ) dθ <. Solution: Let K U be compact. Since K is compact and contained in D(, ), there is < R < such that K D(, R) D(, ). Define ɛ > as follows: If z K and f F, then f(z) = ɛ = dist ( K, U). = πi πi w z =ɛ w =R+ɛ f(w) w z dw f(w) w z dw, where the second line is by Cauchy s integral theorem. Hence we have f(z) f(w) π w z dw. w =R+ɛ Now if w = R + ɛ and z K, we have w z ɛ since z R. Hence w z ɛ. Thus, f(z) f(w)dw πɛ = = πɛ w =R+ɛ π (R + ɛ) πɛ f((r + ɛ)e iθ ) (R + ɛ)dθ π f((r + ɛ)e iθ ) dθ (R + ɛ). πɛ

14 4 Therefore f(z) is uniformly bounded on K. The same bound holds for all f F. Since this is for any K U, by Montel s theorem F is a normal family. Harmonic Functions Problem: Let U be a bounded, connected, open subset of C, and let f be a nonconstant continuous function on U which is holomorphic on U. Assume that f(z) = for z on the boundary of U. (a) Show that is in the range of f. Solution: By the max mod principle, the maximum of the function takes its max on the boundary, hence we know that f(u) D. If / f(u) then let g(z) = f(z), then we have g(z) = f(z) = which implies that f(z) = for all z U, which is a contradiction to the open mapping theorem. Hence is in the range of f. (b) Show that f maps U onto the unit disk. Solution: Let α D, set B α = z α az. Consider h(z) = B α f(z), then h(z) = on U which implies that h(u), by part (a), which implies that a f(u). Hence the image of f is the unit disk. Problem: Let U : C R be harmonic. Prove or disprove each of the following. (a) If u for all z C, then u is constant on C. Proof: Suppose u in C, then u on D(, R) for any R >, so by Harnacks inequality we have R z R + z u() u(z) u(), z D(, R) R + z R z taking R, we have u() u(z) u(). Therefore u(z) = u() hence u is constant. (b) If u = for all z =, then u(z) = for all z C. Proof: Suppose u = for all z D(, ), now consider D(, ), then by the maximum/minimum modulus principle we have Now we have max u(z) = max u(z) = min u(z) = u z D(, ) z D(,) z D(,) z D(,) = u(z) = π π u(p + re ıθ ) dθ This implies that u = on D(z, r), for all z D(, ) and for all r >. u(z) = on C (c) If u = for all z R, then u(z) = for all z C. Solution: The statement is not true, consider u(z) = u(x + ıy) = y, then u and u(z) = for all z R but u Problem: Let u be a harmonic function on R that does not take zero value (i.e. u(x), x R ). Show that u is constant. Proof: u(x, y) implies that u(x, y) is either strictly positive or strictly negative. It suffices to consider u(x, y) as strictly positive, (otherwise consider u(x, y)). Then there exist f(z) holomorphic on C, such that f(z) = f(x, y) = u(x, y) + ıv(x, y), where u(x, y) is the given harmonic function and

15 5 v(x, y) is the harmonic conjugate of u(x, y). Now consider the following: e f(z) e f(z) e u(x,y) < So e f(z) is a bounded entire function. Hence by Liouville s theorem e f(z) must be constant, which implies that f(z) is constant and thus that R(f(z)) = u(x, y) is constant Problem: Let u be a positive harmonic function on the right half plane {R(z) > }, and lim u(r) =. r + Prove that then lim u(re ıθ ) = for all θ ( π r +, ) π. Solution: Since u is harmonic on R = {z : R(z) > }, there exists a function v which is conjugate to u, i.e, the function f(z) = u + ıv is holomophic on R. Now f(z) is continuous on this set R and lim f(r) =. Now let θ ( π r +, ) π and let ρ(z) = e rıφ. Now consider g(z) = f ρ ρ, then we have lim r + g(z) = lim r + (f ρ)z = lim r + ρ(z) = Hence we have lim u ρ(z) = lim u(reıθ ) = r + r + Problem: (a) Suppose a continuous function u : C R has the following property: u(x + ıy) = (u(x + a + ıy) + u(x a + ıy) + u(x + ı(y + a)) + u(x + ı(y a))) 4 for all a R. Does it imply that u is harmonic? Solution: Yes, u(z) is harmonic and u(z) is in the following the set A = {u : C R : u is continuous}, check conditions. A is the space of the following polynomials a(xy 3 yx 3 ) + b(x 3 3xy ) + c(y 3 3x y) + d(x y ) + ex + fy + g, where the letters are complex numbers. (b) Suppose a continuous function u : C R has the following property: u(x + ıy) = (u(x + a + ıy) + u(x a + ıy) + u(x + ı(y + a)) + u(x + ı(y a))) 4 for all a C. Does it imply that u is harmonic? Solution: write z = x + ıy and write a in it s polar form, then for any a C we have the following: u(z) = 4 (u(z + aeiθ ) + u(z + ae i(θ+π) + u(z + re i(θ+π/) ) + u(z + re i(θ+3π/) ) by integrating both sides with respect to θ from toπ we have πu(z) = π u ( z + ae ıθ) dθ so u(z) has the mean value property, hence u(z) is harmonic Problem: Let u and v be real-valued harmonic functions on the whole complex plane such that Find the relation between u and v. u(z) v(z), z C Solution: Let h(z) = v(z) u(z), then h(z) = v(z) u(z) on C. Let R >. Then by the Harnack inequality, if z < R, as h(z) is real-valued harmonic on D(, R) C, h(z) on C, h() R z R + z h(z) h() R + z R z. Now fix z C. Then for all R > z, the above holds, and so R z R + z lim h() h(z) lim h() R R + z R R z.

16 6 Hence we have h() h(z) h(), which implies that h(z) = h(), z C. Thus v(z) u(z) = v() u() v(z) = u(z) + v() u() v(z) = u(z) + α for some α C Remark: Harnack s inequality: D(, R). Then, for any z D(, R) Let u be a nonnegative, harmonic function on a neighborhood of u() R z R + z u(z) u() R + z R z. Problem: Prove or disprove each of the statements: (a) If f is a function on the unit disk D such that f (z) is analytic on D, then f itself is analytic. Solution: This statement is false. Let f(z) = z, then f (z) = z which is holomorphic on D(, ), but f(z) is not holomorphic at z =. (b) If f(z) is a continuously differentiable function on D, and if f (z) is analytic on D, then f(z) itself is analytic. Proof: Let f(z) = f(x, y) = u(x, y) + ıv(x, y) where u, v are harmonic. Then f (z) = u (z) v (z) + ıu(z)v(z). Define g(z) and h(z) as follows: g(z) := R(f (z)) = u (z) v (z) h(z) := I(f (z)) = u(z)v(z) Now since f (z) is holomorphic we know that f (z) satisfies the Cauchy-Riemann equations, i.e., Computing the above we have this implies that () u Computing the other equality we have which implies that () u g x = h y g y = h x g x = u u v v x x = h y = u v u + v y y ( u x v ) ( u = v y y + v ) x g y = u u v v y y = h v u = u v x x x ( u y + v ) ( v = v x y u ) x Solving the above system of equations in () and () implies that for all z D(, ) either, u + v = or u x = v y, u y = v x The former case implies that u = v =, which implies f(z) = and thus f(z) is analytic in D(, ). The latter case implies f(z) satisfies the Cauchy-Riemann equations, and thus f(z) is analytic in D(, ). Either way f(z) is analytic in D(, ). Suppose a function f : D(, ) C is continuous and holomorphic in D. Suppose also that for any z D we have Rf(z) = (If(z)). Prove that f(z) is constant.

17 7 Proof: Let f(z) = f(x, y) = u(x, y) + ıv(x, y), where u and v are harmonic functions. From the hypothesis we know that u = v. Computing the partials we have adding and factoring we have u xx = vv xx + v x u yy = vv yy + v y = = u xx + u yy = v(v xx + v yy ) + (v xv y) = (v xv y) this implies that v x = v y = for all z D, hence v and u are constant on D. Then by uniquness of the taylor expansion for f(z) = u(x, y) + ıv(x, y), f(z) must be constant as well. Conformal Mappings Important Conformal maps Aut(C) = {f(z) : f(z) = az + b, a } ıθ z a Aut(D(, ) = {f(z) : f(z) = e az, a D(, ), θ [, π]} Aut(D(, ) {}) = {f(z) : f(z) = ze ıθ, θ [, π]} Aut(C { }) = {f(z) : f(z) = az+b cz+d, ad bc } Biholo({z : I(z) > }, D(, )) = z ı z+ı (Cayley transform) Biholo(D(, ), {z : I(z) > }) = ı +z z (inverse Cayley transform) Biholo({z : I(z) > }, {z : I(z) >, R(z) > }) = z (applies for the half disk as well) Two annli {z : r < z < r }, {z : s < z < s } are conformally equivalent iff r /r = s /s Biholo({z : < I(z) < ı, D(, )) = e z Biholo({z : π/4 < R(z) < π/4, D(, )) = tan(z) Biholo({z : / < R(z), D(, )) = z Biholo({z : < I(z), { < I(z), < R(z) < }) = sin (z) Problem: Find explicitly a conformal mapping φ which maps the strip {z C : 3 < R(z) < } to the unit disk. Solution: Let Ω = { z C : 3 < R(z) < }, define φ on Ω by φ : z 3 ( z ) 3 Then φ (Ω) = Ω := {z C : < R(z) < }. Now define φ on Ω by φ : z ıπz Then φ (Ω ) = Ω := {z C : < I(z) < π}. Now define φ 3 on Ω by φ 3 : z e z Then φ 3 (Ω ) = Ω 3 := {z C : < I(z)}. Finally define φ 4 on Ω 3 by φ 4 : z z ı z + ı Then φ 4 (Ω 3 ) = D(, ). Hence the composition maps Ω conformally onto D(, ) φ := φ 4 φ 3 φ φ = eıπ( 3 z ) ı e ıπ( 3 z ) + ı Problem: Find explicitly a conformal mapping of the domain to the unit disk. U = {z C : z <, R(z) >, I(z) > }

18 8 Solution: First consider the map φ (z) = z this takes the quarter disk into the upper half disk. Then the map φ (z) = ı + z z takes the half disk into the upper half plane and finally the cayley transform φ 3 (z) = z ı z + ı which takes the upper half plane into the unit disk. Therefore the map (φ 3 φ φ )z = maps the quarter disk U conformally to the unit disk ( + z) ı( z) ( + z) + ı( z) Problem: Find explicitly a conformal mapping of G onto the unit disk, where { G = z = x + ıy; z < and y > / } Solution: Consider the map φ (z) defined as φ (z) = z + i + z + i. This sends ( i) to, i to, and ( i) to. Then let G := φ (G), so we have G = {z = re ıθ : < r <, π/4 < θ < π} Now φ maps G into G conformally. Next consider the map φ (z) defined by Then let G := φ (G ), so we have φ (z) = e π 4 i z G = {z = re ıθ : < r <, θ3π/4} Then φ maps G into G conformally. Now define φ 3 (z) by Then let G 3 := φ 3 (G ), so we have φ 3 (z) = z 4 3 G 3 = {z = x + ıy : y > } Then φ 3 maps G into G 3 conformally. Finally define φ 4 as the Cayley transform, which maps the upper half plane into the unit disk. So define φ(z) as ( ( z+i+ )) 4 e π 4 i 3 z+i i φ(z) = (φ 4 φ 3 φ φ )(z) φ(z) = ( ( z+i+ )) 4. e π 4 i 3 z+i + i Then φ(z) maps the set G into the unit disk conformally. Schwarz Reflection Priciple Problem: Let L C be the line L = {x + ıy : x = y}. Assume that f is an entire function, such that for any z L, f(z) L. Assume that f() =. Prove that f(ı) =. Proof: Because and ı have symmetry about L, we want to consider the Schwarz reflection principle. First consider the following change of coordinates: ( φ(z) = ze ıπ/4 p := φ() =, ) ( ), p := φ(ı) =,.

19 9 Now consider the function h(z) = (φ f φ )z then h(z) maps the real line to the real line. since (f φ )(R) = f(l) L. So for z = x R we have h(z) = h(z). and since h(z) = h(z) on the real line we have Which implies that φ () = f(ı) = h(p) = (φ f φ )p = (φ f) = φ() = = h(p) = h(z) = (φ f φ )p = (φ f)ı Problem: Let f(z) be holomorphic in the upper half plane U = {z = x + iy : y > } and continuous on U. Assume f(x) = ix 3 for all x (, ). Find all such f(z). Solution: Let g(z) = f(z) iz 3, then g(z) is holomorphic on U. Define U as follows: U = {z = x + iy : < x <, y > } Then g(z) is holomorphic on U and continuous on U (, ), furthermore lim z (,) I(g(z)) =, z U Hence by the Schwarz reflection principle for holomorphic functions, we have g(z) z U ĝ(z) = g(x) = f(x) ix 3 z = x (, ) g (z) z U is holomorphic on U = {z = x + iy : < x <, y R}. Now {z U : ĝ(z) = } (, ), which has an accumulation point in V. Thus by uniqueness, ĝ on U. This implies g on U, and hence U {z U : g(z) = }, which has an accumulation point on U. Which again implies, by the uniqueness theorem, g on U. Therefore f(z) = iz 3 on U Remark: Schwarz reflection principle for holomorphic functions: Let V be a connected open set in C such that U R = V (real axis) = {x R : a < x < b} for some a, b R. Set U = {z V : I(z) > }. Suppose that F : U C is holomorphic and that lim I(F (z)) =, z z x U for each x U R. Define Û = {z C : z U}. Then there is a holomorphic function G on U Û V R such that G U = F. In particular, F (z) z U G(z) = lim z x R(F (z)) z U, x U R F (z) z Û Problem: Let f(z) be analytic and satisfy f(z) z in strip α R(z) α. Prove the function is a constant function of x [α, α ]. h(x) = f(x + ıy)dy Proof: Let x, x [α, α ] such that x < x, R > and consider the following contour Γ. γ := x + ıt t [ R, R] γ := t t [x + ır, x + ır] Γ = γ 3 := x ıt t [ R, R] γ 4 := t t [x ır, x ır]

20 Now since f(x) is analytic we know that γ f(z) dz = ı R R Γ f(z) dz =. Now computing γ we have f(x + ıt) dt ı Also for γ 3 and with the change of variable y = t we have R f(z) dz = ı f(x ıt) dt ı γ 3 R Now for γ and z lying on the line γ we have f(z) dz γ x+ır x +ır x+ır x +ır f(t) dt z f(x + ıt) dt as R = z (x x ) as R f(x + ıy) dy as R For γ 4 and z lying on the line γ 4 we obtain the same result as in γ. This implies that = ı f(x + ıt) dt ı f(x + ıt) dt f(x + ıt) dt = Now this is for all x, x [α, α ]. Therefore h(x) must be constant. Problem: Prove that the product ( z n ( z )) + exp n k= converges uniformly on compact sets to an entire function. f(x + ıt) dt Solution: Denote the above product p(z). Fix r > and consider D(, r). By the triangle inequality we have ( zn z ) + exp n e z/n + z n now the last series converges everywhere on C to e z. So all the needs to be shown is the first series e z/n is finite on compact disks. Now there exists ρ > such that for all z D(, ρ) we have e z c ρ z. This implies Hence we have z s.t. z < ɛ e z/ z n n c ρ n e z/n < cρ Therefore the product p(z) converges Problem: Let f : C C be an entire function such that Prove that f must be constant. z, z D(, ρ ɛ) n f(z + ) = f(z), f(z) e z, z C Solution: Consider A = {z C : R(z) < }, and g(z) = f(z) f(/) cos(πz) for z A

21 Now at g(z) is bounded on A, since z = / is a removable singularity. Hence by Louville s theorem g(z) = c, this implies that f(z) = f(/) + c cos(πz) f(z) = f(/) + c cos(πz) e z c = Which implies that f(z) = f(/), hence f(z) is constant Problem: Let h : C R be harmonic and non-constant. Show that h(c) = R Solution: Suppose that h(c) [k, ) then, h(z) k [, ). Now consider and entire function f(z) O(C) st R(f(z)) = h(z) then e f(z) OC and e f = e h. so e h is constant, which is a contradiction, therefore h(c) = R. Problem: Let f be holomorphic in D(, ) and continuous in D(, ). Suppose that f(z) 6 for z D(, ) and f() =. Prove that f has at most 4 zeros in D(, ). Solution: Consider the following: Now if < we have Hence we have ln(f()) + n ln z = π k= ln() Thus f(z) has at most 4 zeros inside D(, ) π ln ln n ln 4 ln() k= ln f(e ıθ ) dθ ln(6) Problem: Denote A = {r < z < R}, where < r < R <. TRUE OR FALSE: For every ɛ > there exists a polynomial p(z) such that { sup p(z) }, z A < ɛ Solution: The statement is true. Let ρ = R + r and consider the following: z p(z)z < ɛ ɛ > Fix ɛ > now this implies that sup z =ρ z sup p(z)z < ɛρ = δ z =ρ Now let f(z) = z p(z), then f() =, since f(z) is a polynomial f(z) = u(z) + ıv(z), for some harmonic functions u(z) and v(z). Now f(z) = u(z) + v(z), so if f(z) δ, then u(z) < δ; however on D(, ρ) we have u() =, which is a contradiction to the maximum principle, since u() = and u(z) < δ Infinite productions and their applications Convergence of infinite products < k= Remark: Let E = z, E p = E exp ( + ) < k= ( p k= ) z k k! ( + ) < k=

22 Lemma: Let { } such that be a sequence with no accumulation point. If p j are positive integers such that ( ) pj r < r > Then the product k= ( ) z E pk converges uniformly on compact subsets of C to an entire function. Proof: Now r < hence pk k= function. k= E z p p k z r k pk+ E z p k <, thus ( ) z E pk converges normally to an entire k= Weierstrass Factorization: Let f(z) be an entire function, and suppose f(z) vanishes at of order m. Let { } be the non-zero zeros of f(z). Then there exists an entire function g(z) such that ( ) z f(z) = z m e g(z) E k k= k= Mittag-Leffler Theorem: Let U C open, and let {a j } be a set of distinct elements with no accumulation point in U. Suppose for all j, V j is a neighborhood of a j such that a j / V k for k j and suppose that m j is meromorphic on V j with only pole a j. Then there exists a meromorphic function m(z) on U such that m m j is holomorphic on V j for all j which has no poles other then those at a j Second version: Let U C open, and let {a j } be a set of distinct elements with no accumulation point in U. Let s j be a sequence of Laurent polynomials, i.e. s j = n= p(j) a j n(z a j ) n Then there exists a meromorphic function on U whose principle part at each a j is s j and has no other poles. Jensen s formula: Let f(z) be holomorphic in a neighborhood of D(, r). Suppose that f(), let { } be the zeros according to their multiplicities, then n(r) ln(f()) + ln r = π ln ( f re ıθ ) dθ π Where n(r) is the number of zeros of f(z) counting multiplicities. j= a j Problem: Show that if f(z) is a nonconstant holomorphic function on D(, ) with f() and { } as the roots then ( ) < k= Solution: Let M = sup{ f(z) : z D(, r)}, where < r <, by Jensen s formula we have n(r) ln f() + ln = π ln f(re ıθ ) dθ ln(m) π n=k

23 3 where n(r) is the number of roots inside D(, ), which implies that n(r) ln r ln(m) ln(f()) Letting r we have and n=k n(r) ln ln(m) ln(f()) < n=k n(r) ( ) ln < k= Problem: For which real values of ρ and µ does the following product converges normally on C ( n µ ) z Solution: Denote a n = n µ ρ z n, now we have the following ρ a n < ( a n ) < Fix r > then for all z in D(, r) we have n=k n µ z n ρ µ = ρ, µρ > Now let µ = ρ (, ], then we have ( z ) n ρ z = n ρ Therefore if µ = ρ and µ, ρ >, a n < Problem: Let f(z) be an entire functions such that f(z) = for each z =. Find all such entire functions. Solution: Consider D(, ) and first notice that the set of roots of f(z) must be finite. Now let { } be the set of roots of f(z) inside D(, ) and consider the following function. n z φ(z) = z k= Now since φ(z) shares the same roots as f(z) inside D(, ) we have f(z) is holomorphic inside D(, ). φ(z) Furthermore we have f(z) f(z) φ(z) = if z =, this implies that = for all z D(, ). φ(z) Hence f(z) = ωφ(z) for some ω D(, ). Now f(z) is entire which implies that ā k = for all k. Therefore f(z) = ωz n, for any n N are all such entire functions n ρ Problem: Set U = {z = x + ıy : x >, y < x}. Suppose that given a sequence of holomorphic functions f n : D U, where D is the unit disk. Prove that if f() converges then the series f(z) converges normally on D.

24 4 Solution: Consider the translation g n (z) = e ıπ/4 f n (z). Now g n (z) = u n (z) + ıv n (z) where u n (z) and v n (z) are harmonic functions. Also g n () < f() < Now Harnacks principle says that u n (z), v n (z), either converge normally to infinity for all z D(, ), or u n (z), v n (z) are finite for all z D(, ). Now g n () = u n () + ı v n () < by assupmtion. Therefore on D(, ). g n (z) is finite for all z D(, ), and hence f n (z) converges normally Problem: Suppose that f(z) is holomorphic in the unit disk D, continuous on D, and has the following properties: a) f() = a > b) f(z) > a 3 for every z D c) f(z) does not have zeroes in D(, /a) Prove that f(z) must have at least three zeros in D. Solution: Let > r > a and consider D(, r). Denote n(r) as the number of roots inside D(, r). Now n(r) ln(f()) + ln r = π ln ( f re ıθ ) dθ π where { } are the roots of f(z) inside D(, r). So we have n(r) ln(a) + ln r > π dθ = ln(a 3 ) π which implies that k= k= n(r) ln > ln(a) as r k= Hence n(r) >, therefore f(z) must have at least 3 zeros inside D(, ). Problem: Suppose that both series f(z) = a n z n and g(z) = b n z n converge in some neighborhoos of the origin. Assume that for each n N either b n = a n or b n =. In other words, the series for g(z) is obtained by removing some terms from the series for f(z). a) Is it possible that the radius of convergence for g(z) is strictly larger than the radius of convergence for f(z)? Strictly smaller? Solution: The raduis of convergence can be larger. Suppose there exists N N such that b n = for all n > N, then the radius of convergence for g(z) is infinite since g(z) is a polynomial. Otherwise if there does not exist such an N. Consider the seqence b nj = b n if b n = a n then we have lim n sup a n /n = lim n sup b n j /n = lim n sup b n /n

25 5 Hence the radii of convergence are the same. b) Is it possilbe that the domain of holomorphy for g(z) (the largest open connect set where the function g(z) can be extended) is strictly larger that the domain of holomorphy for f(z)? Strictly smaller? Solution: The domain of holomorphy can be larger or smaller. For both cases consider the following function f(z) f(z) = z n = z The domain of holomorphy for f(z) is C {}. Now fix N and let b n = for all n > N, then g(z) is a polynomial of degree N, and hence is entire. For a smaller domain consider the following g(z) g(z) = where b n = for the non n terms in f(z). Then the domain of holomorphy for g(z) is exactly D(, ), which is much smaller then C {} Problem: Let f be analytic in the unit disk D(, ) and continuous on D(, ). Assume that Find all such f. z n f(z) = e z z D(, ) Solution: Let α i be the zeros of f(z) inside D(, ). Then there exists only a finite number of α i D(, ), Otherwise {α i } would have an accumulation point, hence f(z). Consider the Blaschke factors, and the Blaschke product defined by: φ j (z) = z α j ᾱ j z φ(z) = n φ j (z) where n is the number of roots includeing multiplicites. Note that φ(z) = when z =, now consider the following: g(z) = f(z) φ(z)e z g(z) = for z D(, ) Now all sigularities of g(z) are removable, so by Riemanns removeable singularity theorem, there is a holomorphic function ĝ(z) with the same properties as g(z). Now by the maximum modulus principle we have g(z), for all z D(, ). Also, by the minimum modulus principle we have g(z) for all z D(, ). therefore we have g(z) = ω, where ω D(, ). f(z) = ωφ(z)e z j= z D(, )