PRELIM PROBLEM SOLUTIONS

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PRELIM PROBLEM SOLUTIONS THE GRAD STUDENTS + KEN Cotets. Complex Aalysis Practice Problems 2. 2. Real Aalysis Practice Problems 2. 4 3. Algebra Practice Problems 2. 8. Complex Aalysis Practice Problems 2. Complex 2. #9.2 Let D be a domai which cotais i its iterior the closed uit disk z. Let f(z be aalytic i D except at a fiite umber of poits z,..., z k o the uit circle z where f(z has first order poles with residues s,..., s k. Let the Taylor series of f(z at the origi be f(z a z. Prove that there exists a positive costat M such that a M. Proof. From the give iformatio, the fuctio g defied by g(z f(z s j z z j is holomorphic o all of D. I particular, its Taylor series b z coverges to g with a radius of covergece r >. The which implies that lim sup b r <, lim sup b <, which i tur implies that there exists some costat B > such that b < B for all. If we solve for f(z i the formula for g(z above, we get f(z g(z + b z + b s j z z j s j z j s j z + j b z + (z j p z p p z a z. s j z j z z j

2 THE GRAD STUDENTS + KEN The a b B + B + s j z + j s j z j + s j. Thus, the coclusio holds with M B + k s j. Complex 2. #2.5 Let A be a simply-coected ope set i C, ad let α be a closed, Jorda, rectifiable curve i A with iterior I (α. Suppose that f is a holomorphic fuctio o A such that the restrictio f α is oe-to-oe. Prove that f has at most oe zero i I (α. Proof. Note that f has o poles. First, suppose that f has o zero o α. The the argumet priciple implies that Z Z P 2πi α f (z dz (f(α,, f(z where Z is the umber of zeroes of f i I α. P is the umber of poles of f i I α. ad (f(α, is the widig umber of f(α aroud zero. Sice the restrictio f α is oe-to-oe, the f(α caot cross itself. Thus the widig umber ca at most be oe ( oe if is iside f(α ad zero if is ot iside f(α. Therefore, Z or Z. Thus f has at most oe zero i I α. Fially, we cosider the case whe f has a zero z o α. (to be cotiued... Complex 2. #3.7 Fid the set of all possible orietatio-preservig coformal maps from A {z C : < I(z < π} to B {z C : I(z > }, ad prove that o other maps are possible. Proof. First, observe that for ay fixed y (, π, the straight horizotal lie {t + iy : t R} gets mapped by f(z e z to {e t e iy : t R}, which is a ope ray from the origi to at agle y. Thus, sice the map is, f is a orietatio-preservig coformal mappig of A to B. Observe that the orietatio-preservig coformal maps from B to itself are of the form h(z az + b cz + d, where a, b, c, d R ad ad bc ca be assumed to be. Thus, ay map of the form h f is a orietatio-preservig coformal map from A to B. Coversely, give ay orietatio-preservig coformal map φ : A B, the φ f : B B is a orietatio-preservig coformal map from B to itself, so that φ f must have the form of such a h above, but the φ h f i the form we obtaied before. I summary, all orietatio-preservig coformal maps from A to B are of the form F (z aez + b ce z + d, with a, b, c, d R ad we may assume ad bc.

PRELIM PROBLEM SOLUTIONS 3 Remark: If the word orietatio-preservig is deleted, we would eed to iclude the possibility of the atiholomorphic maps, which have the form G(z aez + b ce z + d, with a, b, c, d R ad we may assume ad bc. Complex 2. #4. Let f(z e u2 u z du. (a Show that f(z is aalytic i C [, ]. (b Show that f(z may be cotiued aalytically across the ope segmet (,. (c Show that the aalytic cotiuatios of f from above (, ad from below (, are differet. What is their differece o the cut (,? Proof. (a Let z x + iy C \ [, ]. Observe that the itegrad, as a fuctio of z x + iy, is smooth as a fuctio of x ad y. Further, its partial derivatives with respect to x ad y are bouded ad cotious ad thus itegrable whe z C \ [, ]. Therefore, we may differetiate uder the itegral sig to evaluate d dz f(z ( 2 x + i d e u2 f(x + iy du, y dz u z sice u z is holomorphic i z. Therefore, f is holomorphic o C [, ]. (b Observe that if C is the upper half of the uit circle, orieted couterclockwise ad if C deotes the closed orieted curve that is the uio of C ad [, ], the the residue theorem yields { e w2 w z dw 2πie z2 z I(C z outside of C, C where I(C deotes the iterior of C. Thus, e u2 f(z u z du 2πie z2 e w2 C e w2 C w z dw w z dw z I(C z outside of C Note that both the top ad the bottom of the right had side give two differet aalytic cotiuatios of f to the iside of C, ad ote that i both cases the formula is valid at all poits away from C. The upper formula gives a cotiuatio from the upper half plae to poits at or below [, ], ad the lower formula gives a cotiuatio from the lower half plae to poits at or above [, ]. (c From the formula above, the differece betwee the aalytic cotiuatios from below ad above at the poit x [, ] is 2πie x2. Complex 2. #4.3 Let F (z e zx + e x dx. (a Determie the set of z for which the itegral coverges. (b Show that F ca be aalytically cotiued, ad fid the largest possible domai of its aalytic cotiuatio.

4 THE GRAD STUDENTS + KEN 2. Real Aalysis Practice Problems 2. Real 2. #2.4 Let a be a positive real umber. Defie a sequece (x by x, x + a + x 2,. Fid a ecessary ad sufficiet coditio o a i order that a fiite limit lim x should exist. Proof. We have x, x a, x 2 a + a 2. Clearly each x j is positive. Note that iitially (x j is strictly icreasig. Also, x + x ( a + x 2 ( a + x 2 x 2 x 2 (x x (x + x, which is positive if (x x is positive. By iductio x + > x for all. Next, suppose the limit of the sequece L exists. The lim x + lim ( a + x 2, L a + L 2, so that L 2 L + a, so that L ± 4a 2, so that a ecessary coditio o a is that < a 4. Now, if < a 4, the observe that x, x 2. The for, if x 2, the x + a + x 2 4 + 4 2. By iductio, x 2 for all, ad so (x is a bouded icreasig sequece ad thus coverges. Therefore, < a 4 is a ecessary ad sufficiet coditio. Real 2. #5.3 Show that there exist costats a ad b such that, for all itegers N, N 2 N a < b. N Real 2. #5. Let f ad g be cotiuous fuctios o R such that f (x + f (x ad g (x + g (x for all x R. Prove that Proof. Note that lim L f (x g (x dx f (x g (x dx k ( y f g (y dy k k f (x dx g (x dx. ( y f g (y dy Sice f is cotiuous o [, ], it is uiformly cotiuous, ad thus for every ε >, there exists δ > such that x y < δ ad x, y [, ] implies that f (x f (y < ε. We choose Z > such that < δ. Let k ( k M f g (y dy. k k

The L M k k k ε k PRELIM PROBLEM SOLUTIONS 5 k ( y f g (y dy k k ( k k k k ( ( y f f ( y f g (y dy. f ( k k ( k f g (y dy k k g (y dy g (y dy k k ε g (y dy k This ca be made arbitrarily small by choosig large eough. O the other had, k ( k M f g (y dy But k f k ( k k k g (y dy k f ( k ( k f f (x dx ( k g (y dy. as (a right Riema sum, so that lim M f (x dx g (x dx. The, by the above, lim L also exists ad Real 2. #6.6 Let lim L lim M F (g(t : f (x dx exp( txg(x dx. g (x dx. (a Prove that if g is cotiuous ad bouded o (,, the F (g is cotiuous ad bouded o (,. (b Prove that if g is cotiuous ad bouded o (,, the it is ot ecessarily true that F (g is cotiuous ad bouded o (,. (c Prove that if g is cotiuous, bouded ad improper Riema itegrable o [,, the F (g is cotiuous, bouded ad improper Riema itegrable o [,. (d Prove or disprove the coverses of (a ad (c. (e Show that eve if g is cotiuous, bouded ad improper Riema itegrable o [,, the F (g is ot ecessarily Riema itegrable o [,. Real 2. #. Discuss the umber of solutios i (x, y to for (u, v sufficietly close to (,. u x + y 2 v y + xy

6 THE GRAD STUDENTS + KEN Proof. Let f (x, y ( x + y 2, y + xy. Suppose that f (x, y is very close to (,, so there exists a very small ε >, ε <. such that x + y 2 < ε ad y + xy < ε. I particular, this meas that (x, y is betwee the parabolas x y 2 ε ad x y 2 + ε ad is withi the hyperbolic boudaries of y + x ε. I particular, this meas that (x, y is withi a 2ε-ball of oe of the three itersectio poits of the parabola x y 2 ad the double lie x (y +, i.e. the poits i f {(, } {(,, (,, (, }. (To see the computatio: if f (x, y (,, the x+y 2, y +xy. The y or x. I the first case, x. I the secod case, y ±. We will choose ε to be very small, very soo. What we kow ow is that if f (x, y (u, v is withi ε of (,, the (x, y is withi 2ε of oe of the three poits. Observe that f is C. We compute f (x, y ( 2y y x +, ad f (, ( (, f 2 (, (, f 2 (,, all of which have ozero determiats ad so are ivertible. By the iverse fuctio theorem, there exists a eighborhood U of (, ad a eighborhood V of (, such that f : U V is a homeomorphism ad that (f U : V U is differetiable ad cotiuous. Similarly, there exists a eighborhood U 2 of (, ad a eighborhood V 2 of (, such that f : U 2 V 2 is a homeomorphism ad that (f U : V 2 U 2 is differetiable ad cotiuous, ad there exists a eighborhood U 3 of (, ad a eighborhood V 3 of (, such that f : U 3 V 3 is a homeomorphism ad that (f U : V 3 U 3 is differetiable ad cotiuous. Next, choose ε > such that the 2ε-balls aroud f {(, } are cotaied iside f (V V 2 V 3. The we have show that if (u, v is i the ball of radius ε of (,, there exist exactly 3 poits i f {(u, v}. Real 2. #.4 Let A R 3 be the set defied by x 3 y + y 3 z 2 2xz 4 2. Prove or disprove that there exists δ > ad a curve α : ( δ, + δ A such that α(t (t, g(t, h(t with g ad h differetiable. Real 2. #3.3 The fuctio h is periodic of period 4 ad satisfies h(t {, t <, t < 4 (a Fid the correspodig Fourier series for h. (b Prove or disprove that the Fourier series coverges uiformly to h o [, 2]. (c Prove or disprove that the Fourier series coverges uiformly to h o [2, 3]. Proof. (a The Fourier series of h is H(t a + ( π a cos 2 t + ( π b si 2 t,

ad the coefficiets satisfy a 4 a 2 2 b 2 4 4 4 (b Observe that if we let PRELIM PROBLEM SOLUTIONS 7 h(t dt 4 ; ( π h(t cos 2 t ( π cos 2 t dt dt { ( k ( π h(t si 2 t dt S N (t a + N 2k + for k Z π ; 2l for l Z π 2k + for k Z 2 π 2(2l + for l Z. 2(2m for m Z ( π a cos 2 t + N ( π b si 2 t, The this fuctio evaluated at x is S N ( 2 (the average of the oe sided limits. If S N did coverge uiformly to h, the the limit fuctio would be cotiuous, ad the S N ( would have to coverge to, a cotradictio. Thus S N does ot coverge to h uiformly o [, 2]. (c Because h is cotiuously differetiable o the iterval [2, 3], the covergece is uiform there. Additioal Problem Let f : R R satisfy ( f is cotiuous o [,. (2 f (x exists for all x. (3 f (. (4 f is icreasig. For x >, defie g (x f(x x. Prove that g is icreasig. Proof. Observe that g exists o (,, ad g (x f (x x f (x x 2. So it suffices to show that f (x x f (x. Note that by the FTC f (x x f (t dt xf (x sice f is icreasig. Thus, f (x x f (x f (x x xf (x. Alterately, we could use the mea value theorem: For ay x >, there exists c such that < c < x ad f (x f f (x f ( (c f (x x x. The sice x >, we have xf (x f (x, so f (x x f (x. Additioal Problem Prove that 2 si (tx lim t x 2 dx. x

8 THE GRAD STUDENTS + KEN Proof. The itegral is Additioal Problem Let f : [, ] R. Suppose that ( f C [, ] (2 f ( f ( 2 si (tx x 2 x dx Prove that either there exists x (, such that f (x ad f (x have the same sig, or f is costat. Proof. Observe that the give f (x satisfies g (x f (x 2 C [, ], g ( g (. We see by the mea value theorem that there exists c (, such that g (c g ( g (. If g ( > g (, we are doe, because the g (c 2f (c f (c >, ad so f ad f have the same sig at c. If g ( g (, the g (c. If there is o poit x (, at which g (x >, the g (x for all x (,. If there exists x (, such that g (x <, the i fact this is true for a iterval aroud x, so that g ( g ( g (x <, a cotradictio, so i this case we must have g (x for all x. Therefore there exists c (, such that g (c f (c f (c >, or else g (x is costat (ad thus f (x is costat. Additioal Problem Does there exist a cotiuous fuctio f such that ( f : (, [, ] is oto? (2 f : [, ] (, is oto? (3 f : (, [, ] is - ad oto? (4 f : [, ] (, is - ad oto? Proof. ( yes. eg. let f (x 2 + 2 si (2πx. (2 o. Sice the cotiuous image of a compact set is compact, f ([, ] must be compact, but (, is ot compact. (3 o. Suppose you have such ad f. The there exist a, b (, such that f (a, f (b. Without loss of geerality, suppose a < b. By the itermediate value theorem, for all x (,, there exists y (a, b such that f (y x. Thus, f : (a, b (, is oto. Thus, f caot be -, sice for ay z (, a, f (z f (x for some x [a, b]. (4 o. Same reaso as (2. 3. Algebra Practice Problems 2. Algebra 2. #2.2 Prove that ay group G of order 6 is isomorphic to Z 6 or S 3. Proof. (Case If G is abelia, the by the FTFGAG, G Z 6 (sice 6 2 3 ad gcd (2, 3, so Z 6 Z2 Z 3. (Case If G has a elemet of order 6, the it is cyclic ad is therefore isomorphic to Z 6. (Case 2 Otherwise, G must have elemets of order (the idetity, so every other elemet has order 2 or 3, by Lagrage s theorem. By Cauchy s Theorem, there exist elemets of order two ad three. Let H be the subgroup of geerated by a elemet g of order three. This subgroup is ormal sice [G : H] G H 2. Let a G such that a2 e, so a / H sice all elemets of H have order

PRELIM PROBLEM SOLUTIONS 9 or 3. The ah H. Sice G H ah, the elemets of G are { e, g, g 2, a, ag, ag 2}. The oly multiplicatios that are ot yet determied are aga, agag, agag 2, ag 2 a, ag 2 ag, ag 2 ag 2, etc. aga ca t be e because ga ca t be a a Cases: (A If aga g, the ga ag, ad the group is abelia. (cotradictio to Case 2 assumptio. (B so we must have aga g 2. The agag e, agag 2 g, ag 2 a agaaga g 2 g 2 g, ag 2 ag g 2, ag 2 ag 2 g 2 g e. Also ga aaga ag 2, gag a, gag 2 ag, g 2 a g ( ag 2 ag, g 2 ag ag 2, g 2 ag 2 a. To prove G S 3, we determie a map φ : G S 3 by φ (g (, 2, 3, ad φ (a (, 2. We check that φ (aga φ (a φ (g φ (a (, 2 (, 2, 3 (, 2 (, 3, 2 (, 2, 3 2 φ (g 2 φ ( g 2. Therefore, φ is a homomorphism. Sice S 3 is geerated by (, 2 ad (, 2, 3, φ is oto, ad sice G S 3, φ is bijective ad thus ad isomorphism. Algebra 2. #6.2 Let R be a rig with idetity, ad let u be a elemet of R with a right iverse. Prove that the followig coditios o u are equivalet: ( u has more tha oe right iverse; (2 u is a left zero divisor; (3 u is ot a uit. Proof. ( (2. If x ad y are right iverses of u, ad if is the multiplicative idetity. If ux uy the ux uy, so u (x y. So if x y the u is a left zero divisor. (2 (. Let ux, ad suppose that uz for some z. The ux + uz u (x + z, so u has more tha oe right iverse. (2 (3. Let u be a left zero divisor, so there exists z such that uz. So if there exists v such that uv vu, the v (uz (vu z z z but v (uz v (, a cotradictio. So u is ot a uit. (3 (2. Suppose that u is ot a uit but has a right iverse x. The ux, so uxu u, or uxu u. The u (xu. But xu is ozero because otherwise x would be the iverse of u. Thus, u is a left zero divisor.

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