Dupuy Complex Analysis Spring 2016 Homework 02

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1 Dupuy Complex Aalysis Sprig 206 Homework 02. (CUNY, Fall 2005) Let D be the closed uit disc. Let g be a sequece of aalytic fuctios covergig uiformly to f o D. (a) Show that g coverges. Solutio We have g (z) = g (ζ) (ζ z) 2 (ζ z) 2. sice limit of the itegral is the itegral of the limit for uiformly coverget fuctios. (b) Coclude that f is aalytic. Solutio If g f ad g g with f cotiuous, the lim g = f. 2d Solutio By Morera s theorem, it is eough to show that f(z)dz = 0 for all closed γ circles i the domai. But γ f(z)dz = lim γ g (z)dz = Here is a first example of a aalytic cotiuatio from the wild. (a) Show that the Riema Zeta fuctio ζ(z) := coverges for Re z > ad is aalytic o this domai. (You eed to use the aalytic covergece theorem, which states that a uiform limit of aalytic fuctios is aalytic. This is just a slight geeralizatio of the previous problem.) Solutio Let D H = {z C : Re z > } be compact. Sice this is bouded away from the lie Re z =, all z D have Re z > σ >. This meas z = x < σ. The series = σ coverges by the p-test, hece by the M-test = s coverges absolutely ad uiformly o all such regios D. Sice the uiform limit of aalytic fuctios is aalytic we kow that ζ(s) is aalytic. (b) (Whittaker ad Watso, 2.8, problem 0) i. Show that whe Re s >, = s = s + = Solutio It is eough to show that = lim N = ( s z [ s + ( s ( + ) s s ( + ) s s ) = (s ) whe Re s >. We observe that ( ( + ) s ) s = as N, (N + ) s sice Re s >. )]

2 ii. Show that the series o the right coverges whe 0 < Re s <. (This meas the series above gives us access to the iterestig part of the Riema-Zeta fuctio. Hit: + x s dx = (+) s+ s s+ s ) Solutio By the hit = [ s + ( s ( + ) s )] N s = + ( s x s )dx so it is eought to show the series o the right coverges as N whe 0 < Re s <. For x [, + ] we have s x s = s x y s dy s Re s, so the series coverges by the p-test. It, of course, coverges absolutes ad uiformly o compact subsets ad hece is aalytic o this regio. 3. (New Mexico, ot sure which year) Let f be aalytic o C. Assume that max{ f(z) : z = r} Mr for a fixed costat M > 0, ad a sequece of valued r goig to ifiity. Show that f is a polyomial of degree less tha or equal to. Solutio By Cauchy s itegral formula for derivatives we have f () (z) =!. (ζ z) + Let R > 2 z so that γ R f () (z)! M(f, R) 2πR 2π (R z ) + M(f, R)! R 0 as R (R/2) + where the vaishig follows from the hypotheses o M(f, R). Sice f () (z) = 0 for all z, f(z) must be a polyomial of degree. 4. (a) Prove the Riema Extesio Theorem: Let U C be a regio cotaiig a poit z 0. Let f hol(u \ {z 0 }). If f is bouded o U show that there exists a uique f hol(u) such that f U\{z0 } = f hol(u \ {z 0 }). Solutio By aalytic cotiuatio, it is eough to exted the fuctio i a small eighborhood aroud z 0. Let r > 0 be such that the closure of D := D r (z 0 ) is cotaied i U. Cosider f(z) := /(ζ z). D The itegral is defied by boudedess of f, is holomorphic i all D sice we ca switch itegratio ad summatio. For z z 0 we have /(ζ z) = Res( ; ζ = z) + Res( D ζ z ζ z ; ζ = z 0) = f(z) + z 0 z Res(; ζ = z 0) = f(z) + a (f; z 0 ). z 0 z We ca show the residue vaishes: sice f is bouded i a eighborhood of z 0 we have a (f; z 0 ) = sup r 0 as r 0. γ r ζ z 0 =r 2

3 Hece, f(z) = /(ζ z) = f(z) for z z 0 D ad f(z) is defied ad aalytic at z = z 0, so we are doe. (b) Recall that a morphism of topological spaces f : X Y is proper if ad oly if the iverse image of every compact set is compact. Show that a aalytic map f : C C is proper if ad oly if for all z j we have f(z j ). Solutio We will show the cotrapositive: There exists some sequece z j with f(z j ) ot goig to ifiity, if ad oly if there exists a compact subset K C such that f (K) is ot compact. Suppose that we have a z j with f(z j ) bouded. There exists a coverget subsequece f(z jk ) w 0. For K a compact subset aroud w 0 we kow that f (K) is ot compact as z jk f (K) ad z jk. Coversely, suppose there exists some compact subset K such that f (K) is ot compact. Sice f (K) is ot compact it is ot bouded (the iverse image of a closed set is always closed). Let z j be a ubouded sequece of poits i f (K). We ca arrage so that z j by goig to a subsequece if ecessary. We kow that f(z j ) is bouded which provides the example we wated. (c) Show that the oly proper maps f : C C are polyomials. (see page 27 of McMulle, you eed to cosider the fuctio g(z) = /f(/z) ad show that g(z) = z g 0 (z) where g 0 (z) is aalytic ad o-zero. This will allows you to coclude g(z) > c z for some which will allows you to coclude behavious about the growth of f(z) as z. ) Solutio Let f : C C be propert. Cosider the fuctio g(z) = /f(/z). Note that g hol(c \ {0}). By properess, if z j 0, g(z j ) 0. This meas g(z) is bouded i a eighborhood of zero so by the Riema extesio theorem g(z) is actually etire with a fiite order zero at z = 0. By the local structure of aalytic fuctios we ca write g(z) = z g 0 (z) where g 0 (0) 0. We coclude g(z) c z ear z = 0 where c = c r = if z <r g 0 (z) for some r sufficietly small. Now we have So for z > /r we have f(z) = /g(/z) f(z) < / g(/z) < z /c, hece, by the modified versios of Liouville s Theorem for polyomial growth (Problem 3), we are doe. 5. (New Mexico, ot sure which year) Let f ad g be etire fuctios satisfyig f(z) g(z) for z 00. Assume that g is ot idetically zero. Show that f/g is ratioal. Solutio Sice g(z) is aalytic, its zeros are isolated. Let {z,..., z } = {z C : z < 00 ad g(z) = 0}. Let m j = ord zj (g). Let p(z) := (z z j ) m. j= By costructio p(z)/g(z) has o poles whe z < 00. Similarly, sice f(z)/g(z) has ot poles where z > sice f(z) g(z). This meas that p(z)f(z)/g(z) is etire ad p(z)f(z)/g(z) p(z) for z > 00. By the modified versio of Liouville s theorem for polyomial growth (Problems 3) the fuctio p(z)f(z)/g(z) = q(z) C[z]. This implies f(z)/g(z) = q(z)/p(z) ad we are doe. 3

4 6. Prove Goursat s theorem. Let γ be a simple cotour. If f : γ + C is holomorphic (but whose derivative is ot ecessarily cotiuous) the = 0. Solutio See the appedix of Greee ad Kratz. γ 7. Suppose that f(z) = a 0 + a (z z 0 ) + a 2 (z z 0 ) 2 + has a fiite radius of covergece. Let g(z) = a + a + (z z 0 ) + a +2 (z z 0 ) 2 +. Show that g(z) has the same radius of covergece as f(z) at z 0. (Hit: do t thik about this too much) Solutio Note that a 0 +a (z z 0 )+a 2 (z z 0 ) 2 + = (a 0 + +a (z z 0 ) )+(z z 0 ) (a +a + (z z 0 )+a +2 (z z 0 ) 2 + ). This shows that coverges if ad oly if coverges. a 0 + a (z z 0 ) + a 2 (z z 0 ) 2 + a + a + (z z 0 ) + a +2 (z z 0 ) Let f(z) = =0 a z ad let R be the radius of covergece (which is possibly ifiite). Let S N (f)(z) = N =0 a z. Show that for all r < R ad all z C with z < r we have where M(f, r) = max z =r f(z). f(z) S N (f)(z) M(f, r) z N+ r z r N Solutio By expadig Cauchy s formula i a geometric series ad trucatig we get ( N ( ) ) z S N (z) = ζ ζ =0 ( ( ) ) N+ z = ζ z ζ = ζ z + ( ) N+ z ζ z ζ = f(z) + E N (z) where E N (z) := ζ z ( ) N+ z ζ is the error term i the trucated power series approximatio. Estimatig E N (z) usig a circle of radius r gives which proves the result. E N (z) M(f, r) 2π r z ( ) N+ z 2πr = r M(f, r) z N+ r z r N 9. (UIC, Sprig 206) Describe all etire fuctios such that f(/) = f( /) = / 2 for all Z. Solutio The fuctio must be z 2. Note that f(z) z 2 has a accumulatio poit of zeros hece must be idetically zero. 4

5 0. Let U C be a coected ope set. Cosider U C with the subspace topology (ope subset of U are the itersectio of ope subsets of C with U ad closed subset are closed subset of C itersected with U). Show that the oly subset of U which are ope, closed ad oempty is U itself. Solutio Suppose that A U is ope ad closed. The A c is both ope ad closed ad A A c =. We also have U = A A c which is the uio of two proper ope sets. By defiitio of coected, we have U = A. 5