f(w) w z =R z a 0 a n a nz n Liouville s theorem, we see that Q is constant, which implies that P is constant, which is a contradiction.

Transcription

1 Theorem [Liouville s Theorem] Every bouded etire fuctio is costat. Proof. Let f be a etire fuctio. Suppose that there is M R such that M for ay z C. The for ay z C ad R > 0 f (z) f(w) 2πi (w z) 2 dw. f(w) w z R For w { w z R}, f(w) M. Thus, (w z) 2 R 2 R 2 f (z) M L({ z w R}) 2π R2 2π M R 2 2πR M R. Sice this holds for ay R > 0, we get f (z) 0 for ay z C. Thus, f is costat. Theorem Suppose that there exists costat C, M > 0 ad k N such that Prove that f is a polyomial of degree k. C z k, if z M. Proof. We left this as a homework problem. Hit: Use the Cauchy itegral formula to estimate f (k+) (z) usig the Jorda curve J { z R} with R M ad R > z. The let R. Theorem [Fudametal Theorem of Algebra] Ay o-costat polyomial has a root i C. Proof. Let P (z) a 0 + a z + a z be a o-costat polyomial with a 0. Suppose P has o zero i C. The Q(z) /P (z) is holomorphic i C. We will show that Q is bouded o C. This requires a careful estimatio. We fid that P (z) a z + a a z + + a 0 z, as z. Thus, there is R > 0 such that P (z) a z /2 for z > R. This the implies that Q(z) 2 a z 2 a R if z > R. Sice Q is cotiuous o the compact set D(0, R), it is bouded o this set. So there is M 0 > 0 such that Q(z) M 0 if z R. Let M 2 max{m 0, a R }. The Q(z) M for ay z C. So Q is a bouded etire fuctio. Applyig Liouville s theorem, we see that Q is costat, which implies that P is costat, which is a cotradictio. Theorem Suppose f is cotiuous o a domai U, ad satisfies that for ay closed path γ i U, γ f 0. The f is aalytic. Proof. From a theorem we studied, f has a primitive F i U. Such F is holomorphic, which is also aalytic. So f F is also aalytic i U. 43

2 4.2 Lauret Series We will cosider the series of the form z, where (z ) Z is a sequece of complex umbers. The series coverges if ad oly if the followig two series both coverge: z, ad the sum z is defied to be z z, z z + The Lauret series (cetered at 0) is of the form a z, where a, Z, are complex umbers. It coverges iff the followig two series both coverges: a z, a z. The first is a power series. The secod ca also be trasferred to a power series: a z z. a (/z). Suppose the radius of a z is R +, ad the radius of a w is R. The a z coverges whe z < R + ad /z < R, i.e., /R < z < R +. Suppose that /R < R +. Let R R +, r /R, ad let A be the aulus {r < z < R}. Let f + (z) a z, g (w) a w, ad f (z) g (/z). The f + is holomorphic i D(0, R + ) D(0, R) ad g is holomorphic i D(0, R ). So f is holomorphic i { z > /R } { z > r}, ad f f + + f is holomorphic i A. Moreover, we have f +(z) a z ad g (w) a w. Usig chai rule, we get f (z) g ( z ) z 2 Thus, the derivative of f f + + f is a z f (z) a z. k ka k z k. 48

3 Theorem Let r < R [0, ]. Suppose that f is holomorphic o A {z : r < z < R}. The f has a Lauret expasio: where a z, z A, a dz, Z, r < t < R. (4.) 2πi z t z+ Proof. First, from Cauchy s theorem, the value of each a does ot deped o t. Let z 0 A. Pick s < S (r, R) such that s < z 0 < S. Let ε mi{ z 0 s, S z 0 }/2 > 0. Let J { z S}, J 2 { z s}, ad J 3 { ε}. The J 2 ad J 3 lie iside J. The fuctio z z 0 is holomorphic o J, J 2, J 3, ad the domai bouded by these circles. From Cauchy s Theorem ad Cauchy s itegral formula, Now we expad z z 0 J dz usig J 2 dz J 3 dz 2πif(z 0 ). /z z 0 /z z0 z +, z J. /z 0 z/z 0 J k0 z k+ 0 z k k z 0 z +, z J 2. The first holds because z 0 /z < for z J. The secod holds because z/z 0 < for z J 2. Thus, 2πif(z 0 ) dz dz J J 2 z 0 z0 dz + z+ z + dz. J 2 If the ifiite sums exchage with the lie itegrals, we have 2πif(z 0 ) ( J ) z + dz z0 + ( J 2 ) z + dz z0 2πia z 0. (4.2) It remais to show that the ifiite sums exchage with the lie itegrals. Note that, for z J, z 0 z 0 z + f J R +, 49

4 ad from z 0 /R <, we fid that For z J 2, ad from z 0 /r >, we fid that z 0 f J R + f J R z 0 z 0 z + f J2 r +, z 0 f J2 r + f J 2 r k From compariso priciple, we justify the step i (4.2). ( z0 ) <. R ( r ) k <. z 0 Similarly, if f is holomorphic i A {r < < R}, the f has a Lauret series expasio i A: a ( ), where a dz, Z, r < t < R. 2πi z z 0 t ( ) + Usually, we do ot use (4.) to compute the coefficiets. O the cotrary, we use other methods to fid the coefficiets, ad use (4.) to compute the itegral. For example, from the power series of e z, we fid that e /z z /! 0 z ( )!, z 0. the coefficiets for e /z are a 0 whe > 0, ad a /( )! whe 0. The series coverges i the aulus {0 < z < }. The we ca compute z r e /z z + dz 2πia 2πi( )!, 0. Other examples appear i the above proof. We used two ways to expad z z 0 : /z z 0 /z z0 ; (4.3) z+ /z 0 z/z 0 z k k0 z k+ 0 k z0. (4.4) z+ 50

5 The first coverges o {z : z > z 0 }. The secod coverges o {z : z < z 0 }. Suppose a z z + b z z 2, ad z < z 2. The f is holomorphic o C \ {z, z 2 }. To fid the Lauret series of f i { z > z 2 }, we expad both /(z z ) ad /(z z 2 ) usig (4.3). To fid the Lauret series of f i { z < z }, we expad both /(z z ) ad /(z z 2 ) usig (4.4). To fid the Lauret series of f i { z < z < z 2 }, we expad /(z z ) usig (4.3), ad expad /(z z 2 ) usig (4.4). Example. Let Fid the Lauret series for f: (a) i { z < 2}; (b) i (z 2)(z 5). {2 < z < 5}; (c) i { z > 5}. First, we may write ad b /3. We have Thus, i { z < 2}, i {2 < z < 5}, a z 2 + /3 z 2 /6 z/2 6 /3 /3 /z z 2 2/z b z 5. Solvig a liear equatio, we fid that a /3 ( z 2) z, z < 2; 6 2 3z /3 z 5 /5 z/5 5 /3 z 5 /(3z) 5/z 3z ( 2 ) 2, z > 2; z 3z+ ( z 5) z 5, z < 5; 5 ( 5 ) 5, z > 5. z 3z+ ( )z ; 2 3z + + z 5 5 ; ad i { z > 5}, 2 /3 + 5 /3 z +. If we kow the Lauret series for f: a z, the the Lauret series for z m is a z +m, the Lauret series for f(/z) is a z, ad the Lauret series for f(z m ) is a z m. 5

6 4.3 Isolated Sigularities Suppose f is holomorphic i U, z 0 U, but there is r > 0 such that D(z 0, r)\{z 0 } U. The we say that z 0 is a isolated sigularity of f. The f has a Lauret expasio i {0 < z z 0 < r}: where a ( ), (4.5) a dz, Z, t (0, r). (4.6) 2πi z z 0 t ( ) + Case : a 0 for all N. The (4.5) becomes the usual power series a (z z 0 ), which coverges to a holomorphic fuctio i { < r}. Thus, if we defie f(z 0 ) a 0, the f is holomorphic i U {z 0 }. I this case, we call z 0 a removable sigularity. Case 2: Not all a, N, equal to 0, ad there are oly fiitely may ozero a. We may fid m N such that a m 0 ad a 0 for > m. I this case, we call z 0 a pole of f of order m. We fid that z 0 is a removable sigularity of g(z) : ( ) m, ad g(z 0 ) a m 0. A pole of order is called a simple pole. Case 3: There are ifiitely may ozero a, N. I this case, we call z 0 a essetial sigularity of f. For ay m N, z 0 is still a (essetial) sigularity of ( ) m. Defiitio The series a ( ) is called the pricipal part at z 0 of f. Suppose there is m Z such that a m 0 ad for all < m, a 0. This meas that z 0 is either a removable sigularity or a pole, ad f is ot costat 0 ear z 0. I this case, we say that the order of f at z 0 is m, ad write ord z0 f m. We see that ord z0 f m if ad oly if there is a holomorphic fuctio g i D(z 0, r) with g(z 0 ) 0 such that ( ) m g(z). If m 0, z 0 is removable. If m, z 0 is a zero of f after removig the sigularity, ad we say that z 0 is a zero of f of order m. A zero of order is called a simple zero. Sice a f () (z 0 )!, z 0 is a zero of order m iff f (k) (z 0 ) 0 for 0 k m ad f (m) (z 0 ) 0. If m < 0, z 0 is a pole of f of order m. Note that if f ad g are holomorphic at z 0, ad if f(z 0 ), g(z 0 ) 0, the h fg ad h 2 f/g are both holomorphic at z 0, ad h (z 0 ), h 2 (z 0 ) 0. So we have ord z0 (f f 2 ) ord z0 f + ord z0 f 2, ord z0 (f /f 2 ) ord z0 f ord z0 f 2. Examples. Sice ord 0 z ord 0 si z, we have ord 0 /z ord 0 / si z, which implies that 0 is a simple pole of /z ad / si z. From ord 0 si z/z ord 0 si z ord 0 z 0, we see that 0 is a removable sigularity of si z/z. After removig the sigularity 0, we exted si z/z to a etire fuctio. 52

7 Defiitio Let U be a ope set. Suppose that S U has o accumulatio poit i U. If f is holomorphic i U \ S, ad each z 0 S is a pole of f, the we say that f is meromorphic o U. A meromorphic fuctio may be costructed by the quotiet of two holomorphic fuctios. Suppose f ad g are holomorphic i a domai U such that g is ot costat 0. Let Z deote the set of zeros of g. The Z has o accumulatio poit i U. Let h f/g. The h is holomorphic i U \ Z. Every z Z is either a removable sigularity or a pole of h depedig o whether ord z f ord z g. By extedig h to the removable sigularities, we obtai a meromorphic fuctio o U. Examples. ta z si z cos z cos z ad cot z si z are meromorphic i C. For ta z, sice the zeroes of cos z are kπ + π/2, k Z, which are simple because cos z si z 0 at kπ + π/2, ad sice si(kπ +π/2) 0, we fid that every kπ +π/2 is a simple pole of ta z. Similarly, cot z is also a meromorphic fuctio i C, whose poles are kπ, k Z, ad every pole is simple. The quotiet of two polyomials is also a meromorphic fuctio i C, which is called a ratioal fuctio. Now we describe the behavior of f ear a isolated sigularity of each kid. Theorem If f is bouded i D(z 0, r) \ {z 0 }, the z 0 is a removable sigularity of f. Proof. Suppose M o D(z 0, r) \ {z 0 }. From (4.6), we see that, for ay t (0, r), a 2π f z z 0 tt L({ t}) Mt, Z. If, the t 0 as t 0, which implies that a 0 for. 53