Complex Aalysis Sprig 2001 Homework I Solutio 1. Coway, Chapter 1, sectio 3, problem 3. Describe the set of poits satisfyig the equatio z a z + a = 2c, where c > 0 ad a R. To begi, we see from the triagle iequality that 2 a = 2a = (z a) (z + a) z a z + a = 2c ad so the set is empty i case a < c. The case where a = c correspods to equality i the triagle iequality, so z a ad z +a are o-egative multiples of each other. This occurs whe z a or z a. However, oly oe of these rays satisfies the origial equatio, sice the distace from z to a must be greater tha the distace from z to a. From this we see that the ray startig at a directed away from a is the solutio set. We may ow assume a > c. Rewritig the equatio as z a = 2c + z + a, squarig both sides, ad collectig terms gives the equatio Solve for z + a ad square agai to get After gatherig terms, this becomes 4ax = 4c 2 + 4c z + a. (x + a) 2 + y 2 = c 2 + 2ax + a2 c 2 x2. x 2 c2 a 2 c 2 y2 = c 2 which is the equatio of a hyperbola. Oly oe of the two braches satisfies the origial equatio though, sice agai z must be farther from a tha from a. Thus the brach with focus at a is the solutio set. If a is permitted to be complex, the the set of poits is agai a ray or a brach of a hyperbola. This may be see by rotatig a ad z couterclockwise by arg(a). The absolute value is ivariat uder rotatio, so the form of the equatio remais the same for the ew variable. I the ew variable, â lies o the real axis, so the previous aalysis applies. Rotatatio trasforms a ray to a ray, ad the brach of a hyperbola to a brach of a hyperbola. 2. The equatio az + b z + c = 0 ca have solutio set cosistig of a sigle poit, or a lie i the complex plae.
(a) Fid ecessary ad sufficiet coditios o the complex costats a, b, ad c so that the solutio set is a lie. First, if either a = 0 or b = 0, the system has either a uique solutio (whe oe of the two is equal to zero but ot the other), o solutios (whe a = b = 0 ad c 0), or all z C i the case that all coefficiets are zero. Thus we suppose that either a or b is 0. If we separate the real ad imagiary parts of the equatio, writig a = a 1 + ia 2, b + b 1 + ib 2,ad c = c 1 + ic 2 the we obtai two real equatios (a 1 + b 1 )x + (b 2 a 2 )y = c 1 (a 2 + b 2 )x + (a 1 b 1 )y = c 2. This system has a lie of solutios if ad oly if the augmeted matrix of the system [ ] a1 + b 1 b 2 a 2 c 1 a 2 + b 2 a 1 b 1 c 2 has o-zero coefficiet part ad rak equal to 1. The hypothesis that ot both a ad b are 0 esures that the coefficiet part of the augmeted matrix is ot the zero matrix. The matrix will have rak equal to 1 if ad oly if each of the two-by-two miors is equal to zero. Computig each of these determiats gives the coditios a 2 1 b 2 1 + a 2 2 b 2 2 = 0, (a 1 + b 1 )c 2 + c 1 (a 2 + b 2 ) = 0, ad c 1 (a 1 b 1 )c 2 (b 2 a 2 ) = 0. The first says that a = b ad the secod two form the real ad imagiary parts of the complex equatio a c = c b. Thus these give the ecessary ad sufficiet coditios for the solutio to be a lie, subject also to either a or b is 0. We could istead make a similar argumet usig complex liear algebra istead of real liear algebra. If we take the cojugate of the give equatio, we obtai a secod liear equatio i the variables z ad z. Now if the first equatio has ifiitely may solutios, the so does the secod. We form the augmeted matrix of this system. Now the variables z ad z are ot idepedet, ad the full system with idepedet variables z 1 ad z 2 will have more solutios, but there will still be ifiitely may solutios exactly whe the rak of the augmeted matrix is 1. The augmeted matrix is [ a b c b ā c ] ad it will have rak 1 (uder our hypothesis that either a or b is 0) exactly whe each of the two by two determiats is equal to zero. This first gives that aā b b = 0, so a = b is a ecessary coditio. The other codtios are a c c b = 0 ad b c āc = 0, which are cojugates of each other ad so oly oe of them is eeded. Thus ecessary ad sufficiet coditios that the solutio set be a lie are that a = b 0 ad a c bc = 0. (b) Fid the solutio i the case it is a sigle poit.
The solutio set is a sigle poit if ad oly if the determiat of the coefficiet part of the matrix above is o-zero. That is, aā b b [ 0. ] I this case, the c solutio is obtaied by multiplyig the the colum vector by the iverse of [ ] c 1 ā b the coefficiet matrix,. The solutio is z = aā b b b a b c cā. aā b b 3. The ext two questios pertai to the correspodece betwee the exteded complex plae ad the uit sphere i R 3 give i Chapter 1, sectio 6. (a) Fid the mappig of the sphere to itself iduced by the followig two mappigs of the (exteded) complex plae to itself: iversio, i which z 1 for z 0, z 0 ad 0 ; ad iversio i the uit circle, where z for z 0, with 1 z 0 ad 0. The correspodece assigs to (x 1, x 2, x 3 ) S 2 the poit z = x 1+ix 2 1 x 3 for 1 x 3, 1 ad seds (0, 0, 1) to the poit. Followig this by iversio, (1 x 3 )(x 1 ix 2 ) x 2 1 +x2 2 = (1 x 3)(x 1 ix 2 ) 1 x 2 3 = 1 x 3 z x 1 +ix 2 = = x 1 ix 2 which correspods to (x 1, x 2, x 3 ), i the case x 3 ±1. For the special cases of the orth ad south poles, it is evidet that S is set 0 which maps to uder iversio, which maps back to N, while N is set to S. Iversio i the uit circle differs from the precedig by followig iversio with complex cojugatio. Thus (x 1, x 2, x 3 ) is mapped to x 1+ix 2 which correspods to (x 1, x 2, x 3 ) i the uit sphere. (b) Show that o-zero complex umbers z ad z correspod to diametrically opposed poits o the sphere if ad oly if z z = 1. Suppose that o-zero complex umbers z ad z satisfy z z = 1. The 1 = z z ad by the precedig part if z correspods to (x 1, x 2, x 3 ) ad z correspods to (ˆx 1, ˆx 2, ˆx 3 ) the 1 correspods to (x z 1, x 2, x 3 ) while z correspods to ( ˆx 1, ˆx 2, ˆx 3 ), sice cojugatio chages the sig of the secod compoet while multiplyig by 1 chages the sig of the first two compoets. The equatio 1 = z z the implies that (x 1, x 2, x 3 ) = ( ˆx 1, ˆx 2, ˆx 3 ) which says that x j = ˆx j for all j, ad so the poits o the Riema sphere correspodig to z ad z are diametrically opposite. Now suppose that the poits correspodig to z ad z are diametrically opposite (ad that z ad z are o-zero). If (x 1, x 2, x 3 ) correspods to z, the z = x 1+ix 2 1 x 3 while z = x 1 ix 2. Takig the product, z z = x 1 + ix 2 1 x 3 x 1 + ix 2 1 + x 3 = x2 1 x 2 2 1 x 2 3 = x2 1 x 2 2 x 2 1 + x 2 2 = 1. 4. Coway, chapter 3, sectio 1, problem 7.
Show that the radius of covergece of the series covergece at z = 1, 1, ad i. ( 1) z(+1) is 1, ad discuss We may evaluate the radius of covergece by usig the formula 1 = lim sup a R k 1 k or by fidig directly a R so that the series coverges whe z < R ad diverges whe z > R. The latter is slightly easier here it seems, but either is hard. First if z > 1 the there is a α > 0 so that z 1 + α ad hece z (+1) 1 + ( + 1)α ad so z (+1) ( + 1)α ad so the terms do ot ted to zero ad the series diverges. O the other had, the series coverges abosolutely for z < 1 by compariso with the geometric series. Thus R = 1. To test covergece at the specific values of z, we recall Dirichlet s test for covergece, which is a geeralizatio of the alteratig series test. It states that if {a k } ad {b k } are real sequeces such that the partial sums of the series a k are bouded ad the sequece {b k } is positive, mootoe descreasig with limit equal to zero, the the series ak b k coverges. (The alteratig series case arises whe a k alterates betwee ±1.) Now sice ( + 1) is always eve, ( 1) (+1) = 1 ad so the series takes the same value at z = ±1. This is =1 ( 1) =1 which is a coverget alteratig series. At z = i, ( 1) +(+1)/2. usig that i 2 = 1 ad that ( + 1) is eve, the series becomes =1 Now if = 2j, + ( + 1)/2 = 2j)j + 1) + j so ( 1) +(+1)/2 = ( 1) j = ( 1) /2, whereas if = 2j + 1 the + ( + 1)/2 = (2j + 1)(1 + j + 1) = 2(j + 1) 2 + j so ( 1) +(+1)/2 = ( 1) j = ( 1) ( 1)/2. Thus for eve or odd, the power of 1 reduces to ( 1) 2 where x is the greatest iteger fuctio. The sigs the chage at the eve itegers alteratig betwee plus ad mius. The partial sums of the series ( 1) +(+1)/2 are bouded, ad 1 is mootoe decreasig with limit zero, ad so the origial series coverges by Dirichlet s test. There were some ice solutios submitted which avoided this asty aalysis of the powers of i, but oly looked at the subsequece of eve (or odd) idexed partial sums. To complete the problem that way, it would be ecessary to show that the subsequece of partial sums of the other parity also coverges, ad to the same limit. Sice S +1 S = a +1 ad the terms ted to zero, this is clear. 5. Coway, chapter 3, sectio 2, problem 9. Suppose that z, z G = C {z : z 0}, z = r e iθ, z = re iθ with π < θ, θ < π. Prove that if z z the θ θ ad r r. Sice z z z z, if z z the z z which says that r r. Now z must satisfy oe of the coditios, I(z) > 0 or R(z) > 0 or I(z) < 0, ad so lies i the upper half plae, the right half plae, or the lower half plae. (Perhaps two of these may be satisfied.) Each of these is a ope set, ad so if z z the for sufficietly large each z also lies i the same half space, ad so does cos θ +i si θ = e iθ = z r which coverges to cos θ + i si θ = e iθ = z. Now i each half space, θ ad θ r may be uiquely determied by formulas ivolvig the arctaget fuctio, which is
cotiuous. For example, i the half space R(z) > 0, θ = ta 1 y if z = x reiθ = x + iy. Sice the arctaget is cotiuous, ad R(z ) > 0 for large if R(z) > 0, we obtai 1 y ta x ta 1 y provig that θ x θ. Similar computatios work i I(z) > 0 ad I(z) < 0. 6. Coway, chapter 3, sectio 2, problem 14. Suppose that G is a coected ope set i C ad f : G C is aalytic. Show that if f is real valued, the f is costat. Let f = u + iv. Sice f is real valued, v = 0 o G. The v v ad are also idetically x y zero o G, ad the by the Cauchy-Riema equatios so are u x u ad y The f (z) = u x + iv x = 0 i G, so by Coway 3.2.10, f is costat o G.