Theorem [Mean Value Theorem for Harmonic Functions] Let u be harmonic on D(z 0, R). Then for any r (0, R), u(z 0 ) = 1 z z 0 r

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1 2. A harmonic conjugate always exists locally: if u is a harmonic function in an open set U, then for any disk D(z 0, r) U, there is f, which is analytic in D(z 0, r) and satisfies that Re f u. Since such f is infinitely many times complex differentiable, we see that u is infinitely many times real differentiable in D(z 0, r). Since D(z 0, r) U can be chosen arbitrarily, we see that every harmonic function is infinitely many times real differentiable. 3. If U is simply connected, we may use the following method to find a harmonic conjugate of u. Here is an example. Let u(x, y) x 2 + 2xy y 2. Then u xx + u yy So u is harmonic in R 2. We now find a harmonic conjugate of u. If v is a harmonic conjugate, then v y u x 2x+2y. Thus, v 2xy +y 2 +h(x), where h(x) is a differentiable function in x. From u y v x, we get 2y 2x 2y + h (x). So we may choose h(x) x 2. So one harmonic conjugate of u is 2xy + y 2 x 2. Theorem [Mean Value Theorem for Harmonic Functions] Let u be harmonic on D(z 0, R). Then for any r (0, R), u(z 0 ) 2π u(z 0 + re iθ )dθ; 2π 0 u(z 0 ) πr 2 u(z)dxdy. z z 0 r Proof. Since D(z 0, R) is simply connected, there is f holomorphic on D(z 0, R) such that u Re f. From the Mean Value Theorem for holomorphic functions, the two formulas hold with f in place of u. Then we can obtain the formulas for u by taking the real parts. Corollary With the above setup, if u attains its maximum at z 0, then u is constant in D(z 0, R). Proof. We have seen a similar proposition, which says that if f is holomorphic in D(z 0, R), and f attains its maximum at z 0, then f is constant in D(z 0, R). A similar proof can be used here. Here is another proof. Let f be analytic such that u Re f. Then e f is also analytic, and e f e u. Since u attains its maximum at z 0, e f also attains its maximum at z 0. An earlier proposition shows that e f is constant, which implies that u log e f is constant. Theorem [Maximum Principle for Harmonic Functions] Let u be harmonic in a domain U. (i) Suppose that u has a local maximum at z 0 U. Then u is constant. (ii) If U is bounded, and u is continuous on U, then there is z 0 U such that u(z 0 ) max{u(z) : z U}. (iii) The above statements also hold if maximum is replaced by minimum. 52

2 Proof. (i) From the above corollary, there is r 0 > 0 such that u is constant in D(z 0, r 0 ). We can not apply the uniqueness theorem to u because u is not harmonic. However, f : u x iu y is holomorphid on U and vanishes on D(z 0, r 0 ). Applying the uniqueness theorem to f and using that U is connected, we conclude that f is constant 0 on U. Thus, u x u y 0 on U. Using the connectedness of U, we can then conclude that u is constant on U. (ii) Since U is bounded, U is compact. Since u is continuous on U, it attains its maximum at some w 0 U. If w 0 U, we may let z 0 w 0. If w 0 U, then (i) implies that u is constant in U. The continuity then implies that u is constant in U. We may take z 0 to by any point on U. (iii) Note that u is also harmonic, and when u attains its maximum, u attains its minimum. Corollary Suppose u and v are both harmonic in a bounded domain U and continuous on U. Suppose that u v on U. Then u v on U. Proof. Let h u v. Then h is harmonic in U, continuous on U, and h 0 on U. From the above theorem, h attains its maximum and minimum at U. So h has to be 0 everywhere, i.e., u v in U. The above corollary says that, if u is harmonic in a bounded domain U and continuous on U, then the values of u on U are determined by the values of u on U. Remark. The smoothness, mean value theorem and the maximum principle also hold for harmonic functions in R n for n 3. But the technique of complex analysis can not be used. Homework. Chapter VIII, : 7 (a,b,c,e).. Find all real-valued C 2 differentiable functions h defined on (0, ) such that u(x, y) h(x 2 + y 2 ) is harmonic on C \ {0}. 53

3 Chapter 4 Calculus of Residues 4. Laurent Series The Laurent series (centered at 0) is of the form a n z n, where a n, n Z, are complex numbers. It converges iff the following two series both converges: a n z n, a n z n. The first is a power series. The second can also be transformed into a power series: a n z n a n (/z) n. n Suppose the radius of a n z n is R +, and the radius of a n w n is R. Then a nz n converges when z < R + and /z < R, i.e., /R < z < R +. Suppose that /R < R +. Let R R +, r /R, and let A be the annulus {r < z < R}. Let f + (z) a nz n, g (w) n a nw n, and f (z) g (/z). Then f + is holomorphic in D(0, R + ) D(0, R) and g is holomorphic in D(0, R ). So f is holomorphic in { z > /R } { z > r}, and f f + + f is holomorphic in A. Moreover, we have f +(z) na nz n and g (w) n na nw n. Using chain rule, we get f (z) g ( z ) z 2 na n z n n k ka k z k. 54

4 Thus, the derivative of f f + + f is f (z) na n z n. Theorem 4... Let r < R [0, ]. Suppose that f is holomorphic on A {z : r < z < R}. Then f has a Laurent expansion: a n z n, z A, where a n dz, n Z, r < t < R. (4.) 2πi z t zn+ In the proof we will use the Laurent series expansion of a particular function z z 0, where z 0 C \ {0} is fixed. Let r z 0. Note that f is holomorphic in the disc { z < r} and the annulus {r < z < }. In the disc, we have z/z 0 <, so z ( 0 z ) n z n z 0. z/z 0 z 0 This means that a n 0 if n < 0; a n /z0 n+ if n 0. In the annulus, we have z 0 /z >, so /z z 0 /z z ( z0 ) n z z n 0 z n+ z n+ 0 m z m z0 m+ That is, a n 0 if n 0, a n /z0 n+ if n < 0. The above method can be used to derive the Laurent series of more complicated functions. Here is an example. Let (z 2)(z 4). Then f is holomorphic on three annuli: { z < 2}, {2 < z < 4}, and {4 < z }. We will find the Laurent series expansion of f in each annulus. For this purpose, we first express f as a z 2 + b z 4, where a, b C are to be determined. After a reduction of fractions, we find that a /2 and b /2. Now we have z 2 z n z, z < 2; (4.2) 2 2n z 2 /z 2 z 2 n z n+ 55 z n, z > 2; (4.3) 2n+.

5 z 4 z 4 /z 4 z z 4 4 n z n+ z n, z < 4; (4.4) 4n z n, z > 2. (4.5) 4n+ Then the Laurent series expansion of f in { z < 2} equals to ( /2) (4.2) plus /2 (4.4); that in {2 < z < 4} equals to ( /2) (4.3) plus /2 (4.4); and that in {4 < z } equals to ( /2) (4.3) plus /2 (4.5). Proof of the above theorem. First, from Cauchy s theorem, the value of each a n does not depend on t. Let z 0 A. Pick s < S (r, R) such that s < z 0 < S. Let ε min{ z 0 s, S z 0 }/2 > 0. Let J { z S}, J 2 { z s}, and J 3 { ε}. Then J 2 and J 3 lie inside J. The function z z 0 is holomorphic on J, J 2, J 3, and the domain bounded by these circles. From Cauchy s Theorem and Cauchy s formula, J Now we expand z z 0 dz using J 2 dz J 3 dz 2πif(z 0 ). /z z 0 /z z0 n z n+, z J. /z 0 z/z 0 J k0 z k+ 0 z k k z n 0 z n+, z J 2. The first holds because z 0 /z < for z J. The second holds because z/z 0 < for z J 2. Thus, 2πif(z 0 ) dz dz J J 2 z n 0 z0 n dz + zn+ z n+ dz. J 2 If the infinite sums exchange with the integrals, we have 2πif(z 0 ) ( J ) z n+ dz z0 n + ( J 2 ) z n+ dz z0 n 2πia n z n 0. (4.6) It remains to show that the two series inside the integrals converge uniformly on the curves. Note that, for z J, zn 0 z 0 n z n+ f J R n+, 56

6 and from z 0 /R <, we find that z 0 n f J R n+ f J R From comparison principle, we see that z J 2, and from z 0 /r >, we find that zn 0 z 0 n z n+ f J2 r n+, z 0 n f J2 r n+ f J 2 r ( z0 ) n <. R z n 0 z n+ converges uniformly over z J. For k ( r ) k <. z 0 From comparison principle, we see that proof is now finished. z n 0 z n+ converges uniformly over z J 2. The Theorem The Laurent series expansion of the above f is unique. Proof. We leave this as a homework problem. Remark. We will not use (4.) to calculate the coefficients a n. Instead, we will find the a n using other methods, and then use (4.) together with the uniqueness of the Laurent series expansion to calculate the value of integrals z t dz for n Z. z n+ For example, the Laurent series expansion of e /z is (/z) n n! 0 z n ( n)!. So a n 0 if n > 0 and a n /( n)! if n 0. A similar example is e /z2 ( /z2 ) n n!. Then we have e /z dz 2πia 2πi. z Similarly, if f is holomorphic in A {r < < R}, then f has a unique Laurent series expansion in A: a n ( ) n, where Homework. Chapter V 2: 4, 8 Additional: a n dz, n Z, r < t < R. 2πi z z 0 t ( ) n+ 57

7 . Suppose that f is holomorphic in A {r < z < R}, where 0 r < R. Suppose that there are two series of complex numbers (a n ) n Z and (b n ) n Z such that a n z n b n z n, z A. Show that a n b n for all n Z. This means that the Laurent series expansion is unique. Hint: It suffices to show that if f 0, then a n 0 for all n. Use a nz n a nz n to construct a bounded entire function. 2. Suppose f is holomorphic in {r < z < R}. Prove that for any s (r, R), f (z) z n dz n dz, n Z. zn+ z s 4.2 Isolated Singularities Suppose f is holomorphic in U, z 0 U, but there is r > 0 such that D(z 0, r)\{z 0 } U. Then we say that z 0 is an isolated singularity of f. Then f has a Laurent expansion in {0 < z z 0 < r}: where z s a n ( ) n, (4.7) a n dz, n Z, t (0, r). (4.8) 2πi z z 0 t ( ) n+ Case : a n 0 for all n N. Then (4.7) becomes the usual power series a n(z z 0 ) n, which converges to a holomorphic function in { < r}. Thus, if we define f(z 0 ) a 0, then f is holomorphic in U {z 0 }. In this case, we call z 0 a removable singularity. Case 2: Not all a n, n N, equal to 0, and there are only finitely many nonzero a n. We may find m N such that a m 0 and a n 0 for n > m. In this case, we call z 0 a pole of f of order m. We find that z 0 is a removable singularity of g(z) : ( ) m, and g(z 0 ) a m 0. A pole of order is called a simple pole. Case 3: There are infinitely many nonzero a n, n N. In this case, we call z 0 an essential singularity of f. For any m N, z 0 is still a (essential) singularity of ( ) m. Suppose there is m Z such that a m 0 and for all n < m, a n 0. This means that z 0 is either a removable singularity or a pole, and f is not constant 0 near z 0. In this case, we say that the order of f at z 0 is m, and write ord z0 f m. We see that ord z0 f m if and only if there is a holomorphic function g in 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 58

8 of f after removing the singularity, and we say that z 0 is a zero of f of order m. A zero of order is called a simple zero. Since a n f (n) (z 0 ) n!, z 0 is a zero of order m iff f (k) (z 0 ) 0 for 0 k m and f (m) (z 0 ) 0. If m < 0, z 0 is a pole of f of order m. Note that if f and g are holomorphic at z 0, and if f(z 0 ), g(z 0 ) 0, then h fg and h 2 f/g are both holomorphic at z 0, and h (z 0 ), h 2 (z 0 ) 0. This means that ord z0 f ord z0 g 0 implies that ord z0 (fg) ord z0 (f/g) 0. Now if ord z0 f m and ord z0 g n, then there are F and G, which are holomorphic at z 0 with F (z 0 ), G(z 0 ) 0, such that ( ) m F (z) and g(z) ( ) n G(z). Then we get g(z) ( ) m+n F (z)g(z), /g(z) ( ) m n F (z)/g(z). Thus, we have ord z0 (f g) ord z0 f + ord z0 g, ord z0 (f/g) ord z0 f ord z0 g. Examples.. We have ord z0 0 for any z 0 C, ord 0 z ord 0 sin z (because the derivative of z and sin z does not vanish at 0). Thus, ord 0 /z ord 0 / sin z, which implies that 0 is a simple pole of /z and / sin z. From ord 0 sin z/z ord 0 sin z ord 0 z 0, we see that 0 is a removable singularity of sin z/z. After removing the singularity 0, we extend sin z/z to an entire function. 2. Since the Laurent series expansion of e /z at 0 is 0 zn ( n)!, there are infinitely many n < 0 such that a n 0. So 0 is an essential singularity of e /z. Definition Let U be an open set. Suppose that S U has no accumulation point in U. If f is holomorphic on U \ S, and each z 0 S is a pole of f, then we say that f is meromorphic on U. A meromorphic function may be constructed by the quotient of two holomorphic functions. Suppose f and g are holomorphic in a domain U such that g is not constant 0. Let Z denote the set of zeros of g. Then Z has no accumulation point in U. Let h f/g. Then h is holomorphic in U \ Z. Every z Z is either a removable singularity or a pole of h depending on whether ord z f ord z g. By extending h to be analytic on the removable singularities, we obtain a meromorphic function on U. Examples. The quotient of two polynomials is called a rational function, which a meromorphic on C. The functions tan z sin z cos z cos z and cot z sin z are meromorphic in C. For tan z, since the zeroes of cos z are kπ + π/2, k Z, which are simple because cos z sin z 0 at kπ + π/2, and since sin(kπ + π/2) 0, we find that every kπ + π/2 is a simple pole of tan z. Similarly, cot z is also a meromorphic function in C, whose poles are kπ, k Z, and every pole is simple. Homework. For each of the following functions, find all of its singularities, and determine the type of each singularity. If a singularity is a pole, also find the order of this pole. (a) cos(/z); (b) sin z cos z ; (c) z sin z. 59

9 Now we describe the behavior of f near an isolated singularity of each kind. We will always assume that z 0 is a singularity of f, and f is holomorphic on D(z 0, r) \ {z 0 }. Theorem z 0 is a removable singularity of f f is bounded in D(z 0, r ) \ {z 0 } for some r (0, r). Proof. Since the extended f is continuous at z 0, there is r (0, r) such that f(z 0 ) < for z D(z 0, r), which implies that f(z 0 ) + in D(z 0, r ) \ {z 0 }. Suppose M on D(z 0, r ) \ {z 0 }. From (4.8), we see that, for any t (0, r ), a n 2π Mt n L({ t}) Mt n, n Z. If n, then t n 0 as t 0, which implies that a n 0 for n. Theorem z 0 is a pole of f lim z z0. Proof. z 0 is a pole of f z 0 is a zero of /f lim z z0 / 0 lim z z0. Here that lim z z0 / 0 implies z 0 is a zero of /f follows from the above theorem: we first conclude that z 0 is a removable singularity of /f using the boundedness of /f near z 0, and then use the limit to see that the extended value of /f at z 0 is 0. Recall that S C is dense in C if S C, which is equivalent to the following: for any w 0 C and r > 0, D(w 0, r) S. Theorem z 0 is an essential singularity of f for any t (0, r), f(d(z 0, t) \ {z 0 }) is dense in C. Proof. We first prove the part. Assume that f(d(z 0, t)\{z 0 }) is dense in C for any t (0, r). If z 0 is a removable singularity, then lim z z0 exists. So there is t (0, r) such that f(d(z 0, t) \ {z 0 }) is contained in a disc, so it can not be dense in C. If z 0 is a pole, then lim z z0. Then there is t (0, r) such that f(d(z 0, t) \ {z 0 }) { z > }, which also can not be dense in C. So z 0 must be an essential singularity. Then we prove the part. Assume that z 0 is an essential singularity, but f(d(z 0, t) \ {z 0 }) is not dense in C for some t (0, r). Then there exist w 0 C and ε > 0 such that w 0 r for every z D(z 0, t) \ {z 0 }. Let g(z) w 0. Then g is holomorphic and bounded in U. So z 0 is a removable singularity of g. Since w 0 + g(z) for z U, we see that z 0 is either a removable singularity (if g(z 0 ) 0) or a pole (if g(z 0 ) 0) of f, which is a contradiction. Remark. Actually, it is known that the f(d(z 0, t) \ {z 0 }) in the above theorem is either the whole C or C without a single point. Using a homework problem, one can show that, if e /z, then for any r > 0, f(d(0, r) \ {0}) C \ {0}. 60

. Then g is holomorphic and bounded in U. So z 0 is a removable singularity of g. Since f(z) = w 0 + 1

. Then g is holomorphic and bounded in U. So z 0 is a removable singularity of g. Since f(z) = w 0 + 1 Now we describe the behavior of f near an isolated singularity of each kind. We will always assume that z 0 is a singularity of f, and f is holomorphic on D(z 0, r) \ {z 0 }. Theorem 4.2.. z 0 is a removable

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