Solutions to Homework 1

Transcription

1 Solutios to Homework MATH 36. Describe geometrically the sets of poits z i the complex plae defied by the followig relatios /z = z () Re(az + b) >, where a, b (2) Im(z) = c, with c (3) () = = z z = z 2. This is the equatio for the uit circle cetered at the origi. For (2), writig z = x + i y we have < Re(az + b) = Re(az) + Re(b) = Re(a)x Im(a)y + Re(b) which is the set of poits below the lie y = Re(a) Im(a) x Re(b) Im(a) whe a is ot real or purely imagiary. If a is real, the this is just the set of poits o the plae to the right of the vertical lie x = Re(b) Re(a) O the other had if a is purely imagiary, the this is just the set of poits below the horizotal lie y = Re(b) Im(a) For (3), we eed the imagiary part of z to equal c. Writig z = x + i y, we see that this is clearly the horizotal lie passig through c, that is, y = c. With ω = se iϕ, where s ad ϕ, solve the equatio z = ω i where. How may solutios are there? Writig z i polar form z = z e iθ, we have z e iθ = se iϕ

2 we obtai two equatios, oe for moduli ad the other for argumets. z = s (4) θ = ϕ + 2kπ, k (5) Whece, the set of solutios is give by z = s / e i(ϕ/+2kπ/), k There are oly distict solutios sice e iθ has a period of 2π. Ideed, e iϕ+2mπ = e iϕ, m Whe z is uit modulus ( z = ), these are called the th roots of uity. Let z, w : zw. Prove that w z wz (6) for z ad w, with equality if z = or w =. We wat to show that w z wz w z 2 < wz 2 w 2 z w w z + z 2 ( wz)( w z) ( z 2 )( w 2 ) which is true ad holds with equality if either z = or w =. Prove that for a fixed w i the uit disc, the mappig F : z w z wz (7) satisfies. F maps the uit disc to itself ad is holomorphic 2. F iterchages ad w 3. F(z) = if z = 4. F : is bijective. 2

3 We have already prove above that F maps the uit disc to itself. Sice it is a ratio of polyomials it is clear that it is holomorphic. That F iterchages ad w is clear by explicit computatio. That F maps the uit circle to itself is also clear from the previous proof, where we showed that z = = F(z) =. We have already show that F is oto. To show that it is also ijective we show that F F = id w w z wz w w z wz = w( wz) w + z wz ww + wz = z Show that i polar coordiates, the Cauchy-Riema equatios take the form u r u r θ = v r θ = v r Use these equatios to show that the logarithm fuctio defied by (8) (9) log z := log r + iθ, where z = re iθ, π < θ < π () is holomorphic i the regio r > ad π < θ < π. The Cauchy-Riema equatios for u(x, y), v(x, y) are u x = v y, u y = v x The trasitio matrix for polar coordiates ca be obtaied by applyig chai rule to x = r cos θ, y = r si θ We obtai cos θ si θ si θ cos θ ur r u θ = ux u y Similarly, cos θ si θ si θ cos θ vr r v θ = vx v y So that the Cauchy-Riema equatios ow read cos θ si θ vy si θ cos θ = = si θ cos θ v x cos θ si θ ur r u θ vr r v θ Whece, ur r θ = vr r v θ 3

4 That is, u r = r v θ, r u θ = v r as desired. To check log z is holomorphic, we write it i polar coordiates log z = log r + iθ so that u = Re(z) = log r ad v = Im(z) = θ. The, u r = r = r v θ, r u θ = = v r 2. Suppose f is a holomorphic fuctio o a ope set Ω. Prove that if ay of the followig hold, f is a costat.. Re(f ) is costat 2. Im(f ) is costat Either of the coditios alog with the Cauchy-Riema equatios imply that f (z) = (u x + iv x ) + (v y iu y ) = Determie the radius of covergece of the power series with coefficiets:. a := (log ) 2 2. a :=! 3. a := Usig the ratio test, we have for () R = lim log log( + ) = lim + = For (2), For (3), R = lim! +! = lim + = R = lim 2 /(4 + 3) ( + ) 2 /( ) = 4 3. Let f be a power series cetered at the origi. Prove that f has a power series expasio aroud ay poit i its disc of covergece. 4

5 Suppose f (z) := a z ad let z < R. Set z = z + (z z ). The biomial expasio of z gives whece, f ca be writte as f (z) = z = k= = k = k >k z k (z z ) k z k (z z ) k Prove that. z does ot coverge o the uit circle 2. z / 2 coverges o the uit circle Sice z =, the first sequece z diverges, so the series caot coverge. The secod coverges absolutely z 2 = 2 < = = 4. Cosider the fuctio f defied o by if x, f (x) = e /x 2 if x >. Prove that f is ifiitely differetiable o, ad that f () () =,. Coclude that f does ot have a covergig power series expasio ear the origi. We prove usig mathematical iductio that f () (x) = r(x)e /x 2, where r(x) is some ratioal fuctio. This is clear for =, ad assumig true for it is clearly true for +. Now the expoetial term guratees that all derivatives vaish at the origi f () () =,. Whece, we caot write f as a power series ear the origi. 5

6 Let be a circle cetered at the origi with positive orietatio. Evaluate z dz for. Now, do this for a circle ot cotaiig the origi. Parametrize as z(t) = re iθ, θ 2π The, dz = ire iθ ad suppose z dz = 2π ir + e i(+)θ dθ () 2π = ir + e i(+)θ dθ (2) = r + e 2πi(+) (3) + = (4) Whe =, we have 2π dz z = ire iθ dθ = 2πi reiθ Now suppose does ot cotai the origi. The we may parametrize it by For, we agai obtai For =, we have 2π dz z = z(t) = z + re iθ, θ 2π z dz = ire iθ z + re dθ = l(z iθ + re iθ ) 2π = 6