Complex Analysis Study Guide

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1 Complex Analysis Study Guide. Equivalences Of Holomorphicity For a domain D C, f(z) = u + iv is holomorphic in D if and only if f z = : (a) If and only if u and v satisfy the Cauchy-Riemann Equations = f z = ( 2 x + i ) (u+iv) = ( u y 2 x v ) + i ( v y 2 x + u ) y (b) If and only if f(z) C (D). First assume f C (D). u x = v y and u y = v x. f(z) f(z ) f(z + h) f(z ) u(x + h, y ) + iv(x + h, y ) u(x, y ) iv(x, y ) lim = lim = lim z z z z h,h R h h h u(x + h, y ) u(x, y ) = lim + i h h On the other hand ( lim h f(z) f(z ) lim = lim z z z z [ ( u(x, y + h) u(x, y ) = lim + i lim h i h h ) v(x + h, y ) v(x, y ) h h,h R Equating the real and imaginary parts for f (z ), we get = u x (x,y ) f(z + ih) f(z ) ih )] = ( i) u y v(x, y + h) v(x, y ) h u x = v y, u y = i v x + i v x (x,y ) (x,y ). + v y evaluated at (x, y ). By part (a), f is holomorphic. Now assume f is holomorphic on D. Fix z D. For z near z we may define γ(t) = ( t)z + tz and γ : [, ] D. f(z) f(z ) = f(γ()) f(γ()) = Dividing both sides by z z yields = Let ɛ >. As f z f z (z )dt + γ f z dz = f z (γ(t))dγ dt dt = f z (γ(t))(z z )dt. f(z) f(z ) f = z z z (γ(t))dt [ ] f f (γ(t)) z z (z ) dt = f z (z ) + is continuous, there is a δ > such that f f (w) z z (z ) < ɛ [ ] f f (γ(t)) z z (z ) dt. whenever w z < δ. Note that γ(t) z = t z z z z for t [, ], therefore f f (γ(t)) z z (z ) < ɛ (x,y ).

2 for z z < δ. Fixing z z < δ, we have [ ] f f (γ(t)) z z (z ) dt f f (γ(t)) z z (z ) dt ɛdt = ɛ. It follows that f(z) f(z ) lim = f z z z z z (z ). 2. Interaction of Holomorphic Functions (a) O(D), the set of all holomorphic functions on D forms an algebra. Let f, g, h O(D) and a, b C. To be an algebra, the following conditions must hold: (f + g) h = f h + g h. f (g + h) = f g + f g. (af) (bg) = (ab)(f g). As the algebra multiplication is standard commutative multiplication, it suffices to show that if f and g are any holomorphic function on D, then fg is as well. So O(D) is in fact an algebra. (fg) z = f z g + f g = g + f =. z (b) Moreover, if f : D E is holomorphic and g : E C is holomorphic, then g f is holomorphic. As f is holomorphic, f(x, y) = u f (x, y) + v f (x, y) and g(x, y) = u g (x, y) + v g (x, y). and As g is holomorphic, and as f is holomorphic So so g f is holomorphic. u(x, y) + iv(x, y) = g f = g(u f, v f ) = u g (u f, v f ) + iv g (u f, v f ). u x = x (u g(u f, v f )) = u g u f u f x + u g v f v f x v y = v g u f u f y + v g v f v f y. u g u f = v g v f, u g v f = v g u f u f x = v f y, u f y = v f x u x = v y, u y = v x, (c) If u is harmonic on a domain E, f : D E is holomorphic, then u f is harmonic in D. Let z E. As E is a domain, there exists a r > such that D(z, r) E. As D(z, r) is simply connected, on this set we can find some holomorphic g such that Re(g) = u. On D(z, r), g f is holomorphic, so its real part is harmonic. Notice Re(g f) = u f. 3. Harmonic Conjugates and Antiderivatives 2

3 (a) If f = u + iv is holomorphic, then u and v are harmonic. v is called the harmonic conjugate of u. By the Cauchy-Riemann Equations, u x = v y 2 u x u y 2 = x u x + y and u y = v x. Thus u y = x As f C (D), order of differentiation can be changed, so 2 u x u y 2 = v x y The computation to show v is harmonic is identical. v y + y v x y =. ( v ). x Lemma. If f, g are C functions on a simply connected set D and if f y = g on D, then there is x a function h C 2 (D) such that h h = f, x y = g on D. For (x, y) D, set h(x, y) = x a f(t, b)dt + y b g(x, s)ds, where (a, b) D. As D is simply connected, these integrals exist. By the Fundamental Theorem of Calculus h (x, y) = g(x, y). y Again by the Fundamental Theorem of Calculus and since g C (D), h (x, y) = f(x, b) + x x y b y g(x, s)ds = f(x, b) + b = f(x, b) + f(x, y) f(x, b) = f(x, y). y g(x, s)ds = f(x, b) + x b f(x, s)ds y (b) Given a harmonic function u on a domain D, if D is simply connected u has a harmonic conjugate v such that F = u + iv is holomorphic in D. Give a counterexample if D is not simply connected. Let f = u y, g = u f. As u is harmonic, we have x y = g x on D. Since f, g C (D), by Lemma, there exists a v C 2 (D) such that v x = f = u y, v y = g = u x. By the Cauchy-Riemann Equations, F = u + iv is holomorphic. Let D be the punctured unit disc. u(x, y) = ln(x 2 + y 2 ) is harmonic on D. Moreover, u along with the function v(x, y) = 2 tan (y/x) satisfies the Cauchy-Riemann equations. However, v is not defined everywhere on D, so u + iv is not holomorphic on D. (c) If f is holomorphic in a domain D, prove that if D is simply connected then there exists a holomorphic function F such that F (z) = f(z) on D. Give a counterexample if D is not simply connected. 3

4 Let f = u + iv. Let g = u, h = v, then by the Cauchy-Riemann equations we have g y = h x. By Lemma there exists a real C 2 function f such that f x = g = u, f y = h = v. ĝ Now let ĝ = v, ĥ = u. Again by the Cauchy-Riemann equations, we get there exists a function f 2 such that f 2 x = ĝ = v, f 2 y = ĥ = u. y = ĥ x so by the Lemma Let F = f + if 2. F is C 2 and by the above equations F satisfies the Cauchy-Riemann equations, so F is holomorphic. Finally, z F = ( 2 x i ) (f +f 2 ) = ( f y 2 x + f ) 2 + i ( f2 y 2 x f ) = y 2 (u+u)+ i (v +v) = f. 2 Let D be the punctured unit disc. Let f(z) = /z. f is holomorphic on D. Assumethere exists a holomorphic function F such that F (z) = f(z). Let γ = D(, ). By calculation f(z) = i. γ However, by the Fundamental Theorem of Calculus, f(z) = F (γ()) F (γ()) =. So no F can exist. 4. Cauchy s Theorem, Morera s Theorem and Cauchy s Integral Formula (a) Cauchy s Theorem D is a bounded domain in C with piecewise C boundary. If f is holomorphic in D and f C(D), then f(z)dz =. By Stokes Theorem, f(z)dz = d(f(z)dz) = D D D D ( ) f f dz + z z dz dz = D γ f dz dz + dz dz =. z (b) Morera s Theorem Let D be connected. If f C(D) and for any simply closed piecewise C curve γ in D we have f(z)dz =, then f is holomorphic in D. γ Fix z D. Define a function F : D C as follows. Given z D choose a piecewise C curve φ : [, ] C such that φ() = z, φ() = z. Set F (z) = dw. φ It is not yet clear that F is well-defined. Let τ be any piecewise C curve such that τ() = z, τ() = z. Let µ be the closed curve φ ( τ). µ is piecewise C. By assumption we have = dw = dw dw. µ φ Let F = U + iv. Fix z = (x, y) D and φ going from z to (x, y). Choose h R small enough so that (x+t, y) D for t h. Let l h (t) be the line segment connecting (x, y) and (x + h, y). Let φ h = φ l h. h F (x + h, y) F (x, y) = dw dw = dw = f(z + s)ds. φ h φ l h 4 τ

5 Taking the real part of this equation yields U(x + h, y) U(x, y) h Letting h on both sides yields that = h h Re f(z + s)ds = h h Ref(z + s)ds. Similar calculations yield that U (z) = Ref(z). x U y = Imf, V x Thus F is holomorphic. Thus F = f is holomorphic. = Imf, V y = Ref. (c) Cauchy s Integral Formula Let D be a bounded open domain in C with piecewise C boundary. Let f(z) be holomorphic in D and f C(D). Then i D dw = f(z), z D. w z i D w z dw = i D ɛ w z dw + i D(z,ɛ) w z dw where D ɛ = D\D(z, ɛ) for some ɛ > such that D(z, ɛ) D. By Cauchy s Integral Theorem we have i D w z dw = + i D(z,ɛ) w z dw = f(z + ɛe iθ ) i ɛe iθ ɛie iθ dθ = f(z + ɛe iθ )dθ ɛ f(z)dθ = f(z). (c ) Generalized Cauchy s Integral Formula Let D be a bounded open domain in C with piecewise C boundary. Let f(z) be holomorphic in D and f C(D). Then k! i D (w z) k+ dw = f (k) (z), z D. For z D, f (k) (z) = k z k i D w z dw = i D (d) Mean Value Property If f is holomorphic in D(z, R), then = k! i D k ( ) z k dw = ( ) k w z i D z k dw w z dw. (w z) k+ for < r < R. f(z ) = f(re z+riθ )dθ 5

6 By Cauchy s Integral Formula f(z ) = f(z) dz = f(z + re iθ ) i z z =r z z i re iθ rie iθ dθ = using the substitution z = z + re iθ. (d ) If u is harmonic in D(z, R), then u(z ) = u(z + re iθ )dθ. f(re z+riθ )dθ, Since u is harmonic and D(, R) is simply connected, there exists a harmonic function v such that f = u + iv is holomorphic. By the Mean Value Property, u(z )+iv(z ) = f(z ) = f(re z+riθ )dθ = u(re z+riθ )dθ+ i v(re z+riθ )dθ. Equating the real parts of both sides yields the desired result as u and v are real-valued. 5. Applications of Cauchy s Integral Formula (a) Liouville s Theorem Any bounded entire function is constant. Let z C, R >. Let f < M. f (z ) = i w z =R (w z ) 2 dw. f (z ) f(z + Re iθ ) Re iθ 2 Rie iθ dθ M R So f (z ) = for all z C. So f is constant. (b) Linear and Sublinear Growth. Let f(z) be entire. f(z) i. If f is sub-linear growth (i.e. lim = ), then f is a constant. z z dθ = M R R. By Cauchy s Integral Formula f (z) = i w =R (w z) 2 dw f (z) w =R (R z ) 2 dw ( f(re iθ ) ) R 2 R (R z ) 2 dθ. ( f(re iθ ) ) As R, by assumption. So f (z) =. So f is constant. R ii. If f(z) is polynomial growth (i.e. there is some positive integer n such that f(z) c n ( + z n )), then f(z) is a polynomial. Let z C and R >> 2 z +. f (n+) (z) = (n + )! i 6 w =R dw. (w z) n+2

7 f (n+) (n + )! (z) So f is a polynomial of degree at most n. f(re iθ ) R (n + )! (R z ) n+ dθ R z (c) If f is holomorphic in a domain D, then f is analytic in D. c n ( + R n ) R R (R z ) n+ dθ. R z Let z D, r > such that D(z, r ) D. By Cauchy s Integral Formula, for < R < r, z D(z, R), f(z) = i w z =R Note that z z w z <. So f(z) = i w z =R w z dw = i w z =R w z ( ) n z z dw = n= w z w z (z z ) dw = i w z =R n= i w z =R since the sum converges uniformly on D(z, R). By Cauchy s Integral Formula f(z) = n= i w z =R (w z ) n+ dw(z z ) n = n= ( ) n z z dw w z w z f (n) (z ) (z z ) n. n! So this sum converges uniformly on D(z, R) and has radius of convergence at least r and the radius of convergence is given by lim sup n a n. n (d) Uniform Limits of Holomorphic Functions Let f j : D C for j =, 2,... be a sequence of holomorphic functions on an open set D in C. Suppose that there is a function f : D C such that, for any compact subset K of D, the sequence f j f uniformly on K. Then f is holomorphic on D. Let z D be arbitrary. Choose r > such that D(z, r) D. Since {f j } converges to f uniformly on D(z, r) and since each f j is continuous, f is also continuous on D(z, r). For any z D(z, r), f(z) = lim j f j(z) = lim j i w z =r f j (z) w z dw = i lim w z j =r f j (z) w z dw = i w z =r The interchange of integral and limit is justified by the fact that, for z fixed, f j (w)/(w z) converges to /(w z) uniformly for w in the compact set {w : w z = r}. (d ) If f j, f, D are as in the Theorem above, then for any integer k, uniformly on compact sets. As we have shown that f j (z) = i w z =r ( ) k ( ) k f j (z) f(z) z z uniformly, we similarly have that f (k) j (z) = k! f j (w) k! dw i (w z) k+ i as f is holomorphic. w z =r f j (w) w z dw dw = f(z) i w z =r w z w z =r (w z) k+ dw = f (k) (z) (w z )( z z w z ) dw. f(z) w z dw. 7

8 (e) Fundamental Theorem of Algebra Let p n (z) = a n z n + + a, a n be a polynomial of degree n. Then p n (z) must have a zero in C. Assume toward contradiction that p n (z) on C. Then is entire. There exists R such p n (z) that for z R, 2 a n z n p n (z) 2 a n z n. So p n (z) 2 a n z 2 n a n. As is continuous it has a maximum on the compact set D(, R). Let M be this maximum. p m (z) Therefore p n (z) M + 2 a n for z C. Therefore p n (z) is constant, implying p n(z) is constant, contradicting that n. Therefore p n (z) has a zero in C. (f) Uniqueness Theorem If f(z) is holomorphic in an open connected domain D and if f(z) = on an open subset of D, then f(z). Let z be an accumulation point of Z(f). n f Claim: z n (z ) = for all n Z +. Assume toward the contrary that this is not the case. Then f there is some n such that n z (z ). Then, on some disc D(z, r) D, we have n Hence the function g defined by f(z) = g(z) = ( ) j (z z j f(z z ) j ). j! j=n j=n ( ) j f(z ) (z z ) j n z j! is holomorphic on D(z, r). Notice that g(z ) by our choice of n. As z is an accumulation point, there exists a sequence {z k } Z(f) such that z k z. Note that g(z k ) = for all z k. By the continuity of g this implies g(z ) =. So f (n) (z ) = for all n. As f(z) = a n (z z ) n for z D(z, r ), where r = dist(z, D). By definition of a n, we n= have that f(z) on D(z, r ). If D(z, r ) D, then there exists some z D(z, r ), z z such that there is a r > such that D(z, r ) D and D(z, r ) (D D(z, r )) since D is open. So we may continue inductively as D is connected to find {z i } such that D(z j, r j ) = D j= and f(z) on each D(z j, r j ). So f(z) on D. Note that if f(z), then Z(f) must be discrete. 8

9 (f ) Factorization of a Holomorphic Function. Let f(z) be holomorphic in D, f on D. If z D such that f(z ) =, then there exists some k Z + and a holomorphic function g where g(z ) such that f(z) = (z z ) k g(z). Since f(z) on D, there is some k such that k is the first positive integer such that f (k) (z ). So f (j) (z ) f(z) = (z z ) j, z D(z, r ). j! Let g(z) = j=k f(z) = (z z ) k j= j= g(z) is holomorphic in D(z, r) and, when z z, g C(D). So g is holomorphic in D and f (j+k) (z ) (z z ) j (j + k)! f (j+k) (z ) (z z ) j if z D(z, r) (j + k)! f(z) (z z ) k if z D\D(z, r) f(z) (z z ) k is holomorphic on D\{z }. So and g(z ) = f (k) (z ) k! f(z) = (z z ) k g(z), z D. 6. Isolated Singularities and Laurent Series (a) A Laurent Series for a meromorphic function f(z) about an isolated singularity z is given by f(z) = n= a n (z z ) n, where a n = f(z) dz, where γ is a positively oriented simple closed curve enclosing i γ (z z ) n+ z in the annulus of convergence of f(z). The annulus of convergence of a Laurent series is given by A(z, r, R) = D(z, R)\D(z, r), where r = lim sup n n a n and R = lim sup n n a n. Let r < s < z z < s 2 < r. Now w z =s 2 w z f(z) = i w z =s 2 j= w z =s 2 (z z ) j (w z ) k dw = w z dw i w z =s w z dw = w z =s 2 w z =s 2 w z z z w z j= w z dw. w z dz = (z z ) j (w z ) k dw = w z =s 2 j= (z z ) j dw (w z ) j+ 9

10 where the geometric series converges since z z /s 2 <. As this is independent of w, we may switch order of integration and summation to obtain ( w z dw = ) dw (z z (w z ) j+ ) j. w z =s 2 w z =s j= w z =s 2 For s < z z, a similar argument justifies that ( w z dw = j= w z =s ) dw (z z (w z ) j+ ) j. Thus f(z) = j= ( i w z =r ) dw (z z (w z ) j+ ) j. To find the annulus of convergence, converges on D(z, R) where R is given by n= R = converges when z z R = lim sup n n a n a n (z z ) n n= lim sup n n a n. a n (z z ) n = n= So our sum converges on r z z R = A(z, r, R). (b) Isolated Singularities. Let f be a meromorphic function in a domain D. a n (z z ) n z z n = r r = lim sup a n. R n i. Removable: z D is a removable singularity of f if and only if lim f(z) C if and only if z z (Riemann Lemma) lim (z z )f(z) = if and only if f(z) = a n (z z ) n. z z (Riemann Lemma for Removable Singularities) Let { (z z ) g(z) = 2 f(z) if z z if z = z As f is meromorphic, there exists some δ > such that g is holomorphic in D(z, δ)\{z }. n= g(z) g(z ) lim = lim (z z )f(z) = z z z z z z by assumption. So g (z ) =. Therefore g is holomorphic in D(z, δ) and g(z ) = g (z ) =. g(z) is therefore analytic in D(z, δ) and for z D(z, δ), g(z) = n= g (n) (z ) (z z ) n = n! n=2 g (n) (z ) (z z ) n = n! a n (z z ) n = (z z ) 2 n=2 n= a n+2 (z z ) n.

11 Therefore, when z z, (z z ) 2 f(z) = (z z ) 2 a n+2 (z z ) n = f(z) = a n+2 (z z ) n. n= Thus we have that lim f(z) = a 2. So z is a removable singularity of f. z z Assume z is a removable singularity. Then lim f(z) = c so lim (z z )f(z) = c =. z z z z If f(z) = n= a n (z z ) n, then lim z z (z z )f(z) = lim removable. ii. Pole: z D is a pole of f if and only if lim that lim z z (z z ) k+ f(z) = if and only if f(z) = z z n= n= a n (z z ) n+ =. So z is f(z) = if and only if there is some k such z z a n (z z ) n. n= k Since lim f(z) =, there exists some δ > such that f(z) is holomorphic in D(z, δ)\{z } z z and f(z) on D(z, δ)\{z }. Let g(z) = f(z). Then g(z) is holomorphic in D(z, δ)\{z } and g(z) on D(z, δ)\{z }. By the Squeeze Theorem, lim (z z )g(z) =. By z z Riemann s Lemma, z is a removable singularity for g. So g is holomorphic in D(z, δ) and lim g(z) = lim z z z z f(z) =. g(z) = a n (z z ) n for z D(z, δ) and let k be the number n= such that a k is the first coefficient not equal to. Note that k. We have that g(z) = (z z ) k n= a n+k (z z ) n = (z z ) k h (z), where h is holomorphic in D(z, δ) and h(z ) = a k. Therefore f(z) = g(z) = (z z ) k h (z) = h(z) (z z ) k. As g(z) on z D(z, δ)\{z }, h (z) in D(z, δ). So h is holomorphic and h(z). So for z D(z, δ)\{z } f(z) = h(z) (z z ) k, so lim (z z ) k+ f(z) = lim (z z )h(z) =. z z z z iii. Essential Singularity: z D is an essential singularity of f if and only if lim f(z) does z z not exist if and only if (Casorati-Weierstrass) f(d(z, r)\{z }) is dense in C for any r > if and only if f(z) = a n (z z ) n where there are infinitely many n < such that a n. n= Suppose there is some r > such that f(d(z, r)\{z }) is not dense in C. So there is some w C and ɛ > such that f(z) w ɛ for all z D(z, r)\{z }. Let g(z) =. f(z) w g is holomorphic in D(z, r)\{z }. So z is an isolated singularity of g. Moreover, g(z) = f(z) w ɛ = lim (z z )g(z) = z z

12 By Riemann s Lemma z is a removable singularity of g. Therefore lim g(z) = c. If c =, z z then lim f(z) = lim w + z z z z g(z) =. So z is a pole of f. If c, then lim f(z) = lim w + z z z z g(z) = w + c. So z is a removable singularity of f. Both of these are a contradiction. Therefore w must be a limit point of f(d(z, r)\{z }). If f(d(z, r)\{z }) is dense in C for any r >, then there exists a sequence {z j } D(z, r)\{z } converging to z such that f(z j ) and a sequence {w j } D(z, r)\{z } converging to z such that f(w j ). So lim f(z) does not exist. z z (c) Residue Theorem Suppose D is an open simply connected set in C and that z,..., z n are distinct points of D. Suppose that f : D\{z,..., z n } C is a holomorphic function and γ is a closed, piecewise C curve in D\{z,..., z n }. Then i γ f(z)dz = j= ( ) Res(f(z); z = z j ) dz. i γ z z j Let S j (z) be the singular part of f at z = z j, i.e. if f(z) = n= So = a n (z z j ) n. Then f(z) j= k= Note: 2 i f(z) = i γ a (j) k i γ γ = j= i j= ( d(z z j ) n+ n + n= a n (z z j ) n, then S j (z) = S j (z) is holomorphic in D. By Cauchy s Theorem j= γ f(z) S j (z)dz =. j= S j (z)dz = i γ 2 γ k= +a (j) z z j dz is an integer. a (j) k i γ j= i γ (z z j) k dz + a (j) z z j dz = j= k= γ a (j) k (z z j) k dz ) dz z z j ( ) Res(f(z); z = z j ) dz. i γ z z j (c ) Computing Residues Let f be a function with a pole of order k at z. Then ( ) k ( Res(f(z); z = z ) = (z z ) k f(z) ) (k )! z. z=z By the Residue Theorem, for some r >, Res(f(z); z = z ) = i z z =r 2 f(z)dz = f(z)(z z ) k i z z =r (z z ) k.

13 By the proof above, f(z)(z z ) k has a removable singularity at z, so it may be extended to a holomorphic function on D(, r). By Cauchy s Integral Formula 7. Evaluating Improper Integrals (a) (f(z)(z z ) k ( ) k ) ( i z z =r (z z ) k dz = f(z)(z z ) k (k )! z. z=z R(cos θ, sin θ)dθ, R is a rational function. We use the parameterization, e iθ = z. As z =, z =, so we have z cos z = ( z + ), sin z = ( z ). 2 z 2i z As z = e iθ, dz = ie iθ dθ, so dz = dθ. Our integral can then be written iz ( ( R z + ), ( z )) 2 z 2i z iz dz. z = (b) P (x) dx, Q(x) on R, deg(q) deg(p ) + 2. Q(x) If z,..., z n R 2 + are the roots of Q(z), then consider the curve where R is large enough so that {z j } D R. P (x) Q(x) dx = P (z) D R Q(z) dz = i n k= ( ) P (z) Res Q(z) ; z = z k. (c) P (x) dx, Q(x) on (, ), deg(q) deg(p ) + 2. Q(x) Let Q(x) = + a x n + + a k x n k, a j. Let n = gcd(n,..., n k ). Consider the curve where R is large enough so that all roots of Q(x) with argument between and /n are in D R. i k= ( ) P (z) R Res Q(z) ; z = z P (x) k = = D R Q(x) dx + P (z) P (re i/n C R Q(z) dz + ) Q(re i/n ) ei/n dr 3

14 R ( e /n) P (x) Q(x) dx + where z,..., z k are the zeros of Q(z) in D R for R large enough. ( ) (d) R(x) cos(x)dx = Re R(x)e ix dx, where R(x) is a rational function. Fact: f(x)e ix dx = i Res ( f(z)e iz ) ; z = z k, for zk R 2 + so long as f(re iθ ) as k= r uniformly for θ (, π). (e) R(x)x α dx, α ((, ) (, )) Q, R a rational function. For degr 2, consider the curve i k= ( ) P (z) Res Q(z) ; z = z k = R r R(x)x α dx+ R(z)z α dz+ C R r R R(xe i )x α e iα dx R(x)x α dx. C r (f) r +,R + R(x) ln xdx. Consider the curve R(x)x α dx e iiα R(x)x α dx = ( e iα) R(x)x α dx. D r,r R(z) ln zdz = 8. Argument Principle R r R(x) ln xdx+ r R R(x)(ln x +iπ)dx R(z) ln zdz+ C r R(z) ln zdz. C R (a) Argument Principle Let f be a meromorphic function on a domain D, f continuous on D, D has piecewise C boundary. If F has neither poles nor zeros on D, then i D f (z) f(z) dz 4

15 (b) Rouché s Theorem Let D be a bounded domain with piecewise C boundary. Let f, g be holomorphic in D and continuous on D. If f(z) + g(z) f(z) + g(z), z D, then #Z D (f) = #Z D (g). (c) Hurwitz s Theorem Let {f n }, f be holomorphic functions in a domain D C. f n (z) is nonzero for z D. If f n f uniformly on any compact subset of D, then either f(z) or f(z) for z D. Corollary If f n f uniformly on D and f(z) on D, then there exists some N such that f n (z) for n N, z D. Let z D. As D is open, there exists r > such that D(z, r) D. Let δ = min{ f(z) : z D(z, r)} >. By uniform continuity, for z D(z, r), there exists some N such that for any n N, We also have Therefore f(z) f n (z) < δ 2. δ 2 < f(z) f(z) + f n(z). f(z) + ( f n (z)) < f(z) + f n (z). So #Z D(z,r)(f) = #Z D(z,r)(f n ). So, in particular, as f(z ), f n (z ). As z was arbitrary, f n (z) for z D. (d) Gauss-Lucas Theorem Let p(z) be a polynomial. Then all the zeros of p (z) lie in the convex hull of the zero set of p(z). n Let p(z) = a (z a j ). j= If z is a zero of p and p(z), then p (z) p(z) = n j= z a j. j= z a j =. Multiplying top and bottom of each term by z a j respectively, we get Rewriting this, we have j= j= z a j z a j 2 =. z a j 2 z = j= z a j 2 a j. Taking the conjugate of both sides, we have z is a weighted sum with positive coefficients that sum to one. So z is in the convex hull of the roots of p. Note that if p(z) = also, then as z is a root of p it is in the convex hull of the zero set of p(z) already. 5

16 (e) Open Mapping Theorem If f is holomorphic and nonconstant in a domain D, then f : D C is an open mapping. Let O be open in D. Let w f(o). Then there exists a z O such that f(z ) = w, i.e. f(z ) w =. By the Uniqueness Theorem, there exists a δ > such that D(z, δ) O and f(z) w on D(z, δ)\{z }. Let ɛ = min{ f(z) w : z z = δ} >, which exists by compactness and choice of δ. It suffices to show D(w, ɛ) f(o), that is, for any w D(w, ɛ), f(z) w = has a solution in D(z, δ), then D(w, ɛ) f(d(z, δ)) f(o). It suffices to show, for w D(w, ɛ), Set #Z D(z,δ)(f(z) w) = i D(z,δ) (f(z) w) f(z) w dz = i D(z,δ) g(w) = f (z) i D(z,δ) f(z) w dz. f (z) dz >. f(z) w By the Argument Principle, g(w) is integer-valued on D(w, ɛ). From the definition, g(w) is continuous on D(w, ɛ) as ɛ = min{ f(z) w : z D(z, δ)}, so for z such that z z = δ, f(z) w >. Moreover, f(z ) = w. So g(w ). By continuity, g(w) g(w ). So D(w, ɛ) f(o). So f is an open mapping. (f) Maximum Modulus Theorem Let f be holomorphic in a domain D. If there is some z D such that f(z) f(z ) for all z D, then f is constant. Let z be as above and assume toward contradiction f be nonconstant. Then f is an open mapping, so there exists some δ > such that D(f(z ), δ) f(d). However, there exists a point on D(f(z ), δ), for instance, with modulus greater than f(z ). This contradicts our choice of z. So f must be constant. Corollary If f is holomorphic in a bounded domain D and f is continuous on D, then the maximum of f occurs on D. Minimum Modulus Theorem If f is holomorphic in a domain and f on D, then if there is some z D such that f(z ) f(z) for z D, then f(z) is constant. Define g(z) = z and apply the Maximum Modulus Theorem to g(z). 9. Conformal Maps (a) Schwarz Lemma Let f : D(, ) D(, ) be holomorphic and f() =. Then (i) f(z) z for z D(, ) (ii) If f () = or f(z ) = z for z D(, )\{}, then f(z) = e iθ z, θ [, ). (i) Let g(z) = f(z). By Riemann s Lemma, g is holomorphic in D(, ), g(z) = z z f(z) z for z D(, ). For r <, g(z) r for z D(, r) by the Maximum Modulus Principle. Letting r, we have g(z), i.e. f(z) z. Note also that this implies that g : D(, ) D(, ). (ii) If z and f(z) = z, then g(z ) =, which by the Maximum Modulus Theorem implies g(z) =, i.e. g(z) = e iθ, θ [, ). So f(z) = e iθ z. If f () =, then f () = lim z f(z) f() z = lim z f(z) z =, which implies g() =. Again by the Maximum Modulus Theorem, g(z) = e iθ, so f(z) = e iθ z. 6

17 (b) Schwarz-Pick Lemma Let f : D(, ) D(, ) f(z) f() (i) f()f(z) z f(z) f(a) (ii) Let a D(, ). Then f(a)f(z) z a az for z D. f() f(z) (i) Let g(z) =. Then g : D(, ) D(, ). Note g() =, so by Schwarz Lemma, f()f(z) g(z) z, i.e. f(z) f() z for z D(, ). f()f(z) (ii) Let a D(, ) and define φ a (z) = a z az. Then φ a() = a. Note that f φ a : D(, ) D(, ). By the above argument, f φ a (z) f φ a () z for z D, f φ a ()f φ a (z) or after simplification f φ a (z) f(a) z for z D. f(a)f φ a (z) As this holds for all z D(, ), we may plug in φ a (z) = φ a (z) into this function. Therefore f φ a φ a (z) f(a) f(a)f φ a φ a (z) = f(z) f(a) f(a)f(z) φ a(z) = a z az for z D. Corollary If f : D(, ) D(, ), then f (z) f(z) 2 z 2 for z D(, ) By Schwarz-Pick, after rearrangement f(z) f(a) z a f(a)f(z) az. Letting a z, we have as f(z), z <. f (z) f(z) 2 z 2 = f(z) 2 z 2 Corollary If there is a z D(, ) such that f (z ) = f(z ) 2 z 2, then f(z) f(z ) f(z )f(z) = z z z z. Proof follows from the Maximum Modulus Theorem. (c) Image of Boundary under Biholomorphic Maps Let D, D 2 be domains in C. Let f : D D 2 be biholomorphic. Then f( D ) D 2. 7

18 (d) Characterization of Aut(D(, )) Let φ Aut(D(, )), a D(, ) such that φ(a) =. Let g(z) = φ(z) az. Then g : D(, ) D(, ) is holomorphic, g(z) on D and g(z) a z as z. By the Maximum and Minimum Modulus Theorem, min{ g(z) : z = r} g(z) max{ g(z) L z = r} for z D(, r). Letting r, we have g(z) for z D(, ). So g(z) = e iθ for θ [, ). So φ(z) = e iθ φ a (z) = e iθ a z az. (e) Möbius Transfoms A Möbius Transfom is a rational function of the form f(z) = az + b, where ad bc. For any cz + d two circles Γ, Γ 2 C, there exists a Möbius Transfom T that maps Γ Γ 2 where z, z 2, z 3 Γ, z i z j, w, w 2, w 3 Γ 2, T (z j ) = w j. We use the real line as an intermediate step. Consider first S(z) = z z ( ) z2 z / z z 3 z 2 z 3 Then So S : Γ R. Consider now Then S(z ) = S(z 2 ) = S(z 3 ) = S(w) = w w ( ) w2 w / w w 3 w 2 w 3 S(w ) = S(w 2 ) = S(z 3 ) = Define T (z) = S S. Then T (z ) = S S(z ) = S () = w T (z 2 ) = S S(z 2 ) = S () = w 2 T (z 3 ) = S S(z 3 ) = S ( ) = w 3 (f) Conformal Maps I. R 2 + D(, ) 8

19 II. {z C : Re(z), Im(z) > } D(, ) III. {z C : θ < arg(z) < θ + α} D(, ) IV. D(, ) R 2 + D(, ) 9

20 V. Given two circles C, C 2 in C that intersect at P, Q, where P C. Let α denote the interior angle between C and C 2 at P. Let S be the area bounded by C and C 2. S D(, ) VI. {z C : a < Im(z) < b} D(, ) 2

21 VII. D(, ) D( /2, /2) D(, ) (g) Normal Families (i) Let D be a domain in C and let {f n } be a sequence of continuous functions on D and f is continuous on D. f n is said to converge to f normally in D, if, for any compact K D, f n f uniformly on K. (ii) Let D be a domain and F be a family of continuous functions on D. F(D) is a normal family if, for any sequence {f n } F, there is a subsequence {f nk } and f C(D) such that f nk f normally on D. (iii) Montel s Theorem Let D be a domain in C, F(D) is a family of holomorphic functions on D. F is a normal family if and only if F is uniformly bounded on compact subsets of D. (h) Riemann Mapping Theorem Let D be a simply connected domain in C, D C. Then, for any given point p D, there is a unique biholomorphic map φ on D such that φ : D D(, ), φ(p ) =, φ (p ) >. Uniqueness of Riemann Mapping Theorem Let φ, φ 2 : D D(, ), φ j (p ) =, φ j (p ) > for j =, 2. It suffices to show f = φ 2 φ : D(, ) D(, ) is the identity. Note first that f() = φ 2 φ () = φ 2(p ) =. Second note that f () = φ 2(φ ()) ( φ ) () = φ 2 (p ) φ (φ ()) = φ 2(p ) φ (p ) >. So f(z) = e iθ z by Schwarz s Lemma. As f () >, θ =, so f is the identity, as desired. Lemma Let f be a one-to-one holomorphic function f : D D(, ), f(p ) =, f (p ) >, f not onto, then there exists a holomorphic g : D D(, ), g one-to-one, g(p ) =, g (p ) > f (p ). As f is not onto, there is a w / f(d). The function f(z) w is nonzero on D. As D is wf(z) simply connected, we can define h(z) 2 = f(z) w, where h(z) is a holomorphic function. wf(z) Let g(z) = h (z ) h(z) h(z ) h (z ) h(z )h(z). 2

22 Notice that g(z ) =, g : D D(, ) and g is one-to-one. Also, g (z ) = h (z ) h(z ) 2 = + w 2 w f (p ) > f (p ).. Reflection Principle (a) Let f(z) be holomorphic on D, D R 2 +, (a, b) D R. If f(z) is continuous on D (a, b) and f(z) is real-valued for z (a, b), then f(z) can be extended to a holomorphic function on D e = D (a, b) D where D = {z : z D}. Corollary Let D be a domain contained in D(z, r) so that Γ = {z + re iθ : θ < θ < θ 2 }, Γ D D(z, r). Assume f is holomorphic on D, continuous on D Γ and f is real valued on. Infinite Products (a) (b) Γ. Then f extends to a holomorphic function on D Γ D where D = {z + r2 z z : z D}. ( + z j ) converges to z if and only if j= ( + z j ) converges absolutely if j= and only if ln( + z j ) converges. j= z n converges, for z n. j= ln( + z j ) converges in C. j= ( + z j ) converges absolutely if (c) Weierstrass Factorization Theorem Let D be a simply connected domain in C, let f(z) be meromorphic on D. Then f(z) = e g(z) h (z) h 2 (z) where g, h, h 2 are holomorphic on D, where Z D (h ) = Z D (f) counting multiplicity and Z D (h 2 ) = P (f) counting order. Example Prove that if D is a simply connected domain and f(z) is holomorphic and nonzero on D, then f(z) = e g(z), where g(z) is holomorphic on D. Solution: As f(z) is nonzero on D, the function f (z) f(z) j= is holomorphic on D. Let F (z) = f (z) f(z). Consider the function e F (z) f(z). We claim this is a constant function. As it is holomorphic, it suffices to check its derivative is zero. ( e F (z) f(z)) = e F (z) F (z)f(z) + e F (z) f (z) = e F (z) ( F (z)f(z) + f (z)) ( = e F (z) f ) (z) f(z) f(z) + f (z) = e F (z) () =. Therefore e F (z) f(z) = e iα, or after rearrangement f(z) = e F (z)+iα. As F (z) is holomorphic on D, setting g(z) = F (z) + iα yields the desired result. (d) Mittag-Leffler Theorem Let D be a domain in C, {z n } a sequence in D without a limit point in D. Let S n = k= N n a kn (z z n ) k for some N n. Then there is a meromorphic function f on D such that f S n is holomorphic at a neighborhood of z n. 22

23 2. Harmonic Functions (a) Harmonic Function u : D R is harmonic if u = in D. (b) Dirichlet Boundary Value Problem For a continuous φ on D(, ), consider the conditions { u(z) = if z D(, ) u = φ on D(, ) This has the unique solution u(z) = z 2 ze iθ 2 φ(eiθ )dθ. We first show that u(z) is harmonic. We begin by writing Therefore, for z D(, ), u(z) = z 2 ze iθ 2 = eiθ e iθ z + e iθ e iθ z. e iθ φ(e iθ ) e iθ z dθ + e iθ φ(e iθ ) e iθ z dθ φ(e iθ )dθ. The first integral is a holomorphic function (so it is harmonic) on D(, ) since e iθ /(e iθ z) is holomorphic in z on D(, ). The second integral is harmonic, since its derivative with respect to z is. The final integral is a constant, so it is also harmonic on D(, ). So u is harmonic. It remains to show that u is continuous at the boundary, i.e. for z D(, ). Notice first that lim u(z) = φ(z ) z z z 2 ze iθ dθ =, 2 therefore φ(z ) = z 2 ze iθ 2 φ(z )dθ where z D(, ). Consider now the function E(z) = z 2 ze iθ 2 ( φ(e iθ ) φ(z ) ) dθ. As φ is continuous, for any ɛ >, there exists some δ > such that for θ θ < δ we have φ(e iθ ) φ(e iθ ) < ɛ. Therefore, θ θ δ E(z) z 2 ze iθ 2 2Mdθ + z 2 ze iθ 2 φ(e iθ ) φ(z ) dθ θ θ <δ where M = max z D(,) {φ(z)}. Notice ze iθ δ.33 e iθ z e iθ. So as desired. E(z) z 2 ze iθ 2 ɛdθ M π 2(δ/6) 2 ɛ δ/6 + ɛ, θ θ δ z 2 ze iθ dθ + ɛ, 2 23

24 (c) Harnack s Inequality Let u be a nonnegative harmonic function on D(, R). Then for z D(, R), R z R + z u() u(z) R + z R z u(). Without loss of generality, we may assume u is continuous on the boundary. Recall Now u(z) = From these two above equations, we have Similarly, note that u(re iθ ) R2 z 2 Re iθ z 2 dθ. R 2 z 2 Re iθ z 2 R2 z 2 (R z ) 2 = R + z R z. u(z) R + z R z which yields the other inequality. (d) Corollary u(re iθ )dθ = R + z R z u(). R 2 z 2 Re iθ z 2 R2 z 2 (R + z ) 2 = R z R + z, Let u be a positive harmonic function in a domain D. For any compact K D there exists a constant C K dependent only on K and D such that max{u(z) : z K} C K min{u(z) : z K}. Let > δ = dist( D, K). As K is compact, there exist finitely many points z,..., z n K such that D(z j, δ) covers K. For any z j and z D(z j, δ) K, we have Therefore u(z) + δ δ u(z j) = ( ) 2 + δ Let C K =. Then, we have δ (e) Harnack s Principle u(z) + z z u(z j) + δ δ u(z j), u(z) z + z u(z j) δ + δ u(z j). ( ) 2 + δ δ δ + δ u(z j) max{u(z) : z K} C K min{u(z) : z K}. ( ) 2 + δ u(z). δ Let {u j } be a sequence of harmonic functions on D such that for z D, u j (z) u j+ (z). Then either u j uniformly on compact sets, or there is a harmonic function u on U such that u j u uniformly on compact sets. Without loss of generality u n (z), as otherwise we can replace this sequence with v n (z) = u n (z) u (z). 24

25 Let z D. We have two cases. Case : lim u n(z ) =. n Then, for any K D, by Harnack s Inequality, there is some C K > such that u n (z) C K u n (z ), z K, So lim u n(z) = uniformly on K. n Case 2: dis lim n u n (z ) = c <. In this case, for any K D there exists some C K such that u n (z) C Ku n (z ) C Kc <. Then lim n u n(z) =: u(z) is finite. Let K D. Then there exists a function u on D such that lim n u n(z) = u(z), for z D. Notice if m > n, u m (z) u n (z), so u m (z) u n (z) C K (u m (z) u n (z)) as m, n, m > n. By Cauchy s Test we have uniform convergence. It remains to show u is harmonic in D. Let z D, r < dist(z, D). Consider D(z, r). u n (z + rz) = z 2 ze iθ 2 u n(z + re iθ )dθ, z D(, ). Letting n, we have by uniform convergence, u(z + rz) = z 2 ze iθ 2 u(z + reiθ)dθ is harmonic at z. Therefore u is harmonic on D. (f) Example Show that if f : D(, ) D(, )\{} is a holomorphic function, then max{ f(z) 2 : z /5} min{ f(z) : z /7} and furthermore find a f(z) to show that this inequality is sharp. Solution: As f is holomorphic and nonzero, ln f(z) is harmonic. Consider the harmonic function u(z) = ln f(z) is harmonic on D(, ). We apply Harnack s Inequality to u to obtain Exponentiating this yields z + z ( ln f(z) ) ln f(z) ( ln f() ). + z z e z + z f() f(z) e + z z f(). Notice then we have so for z /7, as a minimal value f(z) e + z z f(), min{ f(z) } e +/7 /7 f() = e 4/3 f(). Similarly, we have z + z ( 2 ln f(z) ) ln f(z)2 + z ( 2 ln f() ), z 25

26 or which on z /5, reduces to f(z) 2 z 2 e + z f(), max{ f(z) 2 /5 2 } e +/5 f() = e 4/3 f(), which gives us the desired result. To show that the bound is sharp, consider the function g(z) = e z +z. Then min{g(z) : z /7} = e +/7 /7 = e 4/3. Similarly, max{g(z) 2 /5 2 : z /5} = e +/5 = e 4/3. (g) Maximum Principle for Harmonic Functions If u is harmonic in D and there is some z D such that u(z ) u(z) for z D then u(z) = u(z ). By the Mean-Value Property so u(z ) = = u(z + re iθ )dθ, r < dist( D, z ), ( u(z + re iθ ) u(z ) ) dθ. As u(z ) u(z + re iθ ) for all r, θ, by continuity, we have u(z ) = u(z + re iθ ) for all r, θ. Therefore, for any z D, as D is open and connected, there exists a path γ z : [, ] D such that γ z () = z, γ z () = z and overlapping discs on points z,..., z k of γ z with radii r i respectively. As we get equality in each disc, by transitivity, we get u(z ) = u(z). (h) Automorphisms of Annuli Define A(;, R) = {z C : < z < R}. A(;, R ) and A(;, R 2 ) are biholomorphically equivalent if and only if R = R 2, where R, R 2 >. If we assume R = R 2 the result holds as the identity map suffices. Conversely, assume there exists a biholomorphic map φ : A(;, R ) A(;, R 2 ). φ is continuous, so φ is proper, therefore φ : A(;, R ) A(;, R 2 ). We claim φ( D(, )) = D(, ) and φ( D(, R )) = D(, R 2 ) or φ( D(, )) = D(, R 2 ) and φ( D(, R )) = D(, ). This holds as the continuous image of a connected set is connected since if otherwise, by connectedness, without loss of generality (the other case is identical) φ( D(, )) D(, ). Then if D(, ) φ( D(, R )), we have φ( D(, R )) is not connected as D(, R 2 ) φ( D(, R )). Assume first the case where φ(z) = for z = and φ(z) = R 2 for z = R. Consider u(z) = log φ(z) log R 2 log R log z. Then u(z) = when z = or z = R. By the Maximum and Minimum Principles, u(z) on A(;, R ). Therefore φ(z) log = = φ(z) =, z log R 2 log R z log R 2 log R or after simplification, φ(z) = e iθ z log R 2 log R. As φ is holomorphic, log R 2 /logr is an integer. As φ is injective, log R 2 / log R =, i.e. R = R 2 as desired. So biholomorphic maps in this case are of the form φ(z) = e iθ z. Assume now that φ(z) = R 2 for z = and φ(z) = for z = R. Consider ψ(z) = φ(r /z). Then ψ(z) = for z = and ψ(z) = R 2 for z = R. As we have seen above, ψ(z) = e iθ z and R = R 2. Therefore φ(z) = e iθ R /z. 26

27 3. Sub-harmonic Functions (a) Definition u C 2 (D) is sub-harmonic if u. (b) Definition 2 u is sub-harmonic if i. u is Upper Semi-Continuous, ii. u(z) u(z + re iθ )dθ for appropriate r. (c) Sub-harmonic Functions For u C 2 (D), these two definitions are equivalent. Assume first that u. By Green s Theorem, for n normal to D at z, u n vdσ v n udσ = uv vuda. So D u(z + re iθ )dθ = = r D(z,r) We have log z z = δ z, so r log z z r r D(z,r) D D n log z z r r dθ log z z r u(z) u log z z da. r u(z) u log z z da = u(z )+ r D(z,r) 2 n u(z + reiθ ) log z z r r dθ Assume now that u(z ) πu(z + re iθ )dθ for all r < dist(z, D). established in what we have shown above, u(z + re iθ )dθ = u(z ) + r u log z z da. Therefore we know that D(z,r ) D(z,r) r u log D(z,r) z z da. u log z z DA u(z ). r By the identity We claim u(z ) for all z D. If not, then fix z such that u(z ) <. Then there exists an r < r such that u(z) < ɛ for z D(z, r ). Then we have r u log z z da ɛ r log da <, z z a contradiction. This proves our claim. (d) Jensen s Inequality If u is subharmonic and φ is convex on the range of u then ( ) φ u(z + re iθ )dθ D(z,r ) φ u(z + re iθ )dθ. We have ( φ ) ( n ) u(z + re iθ )dθ = φ u(z + re iθ k θ k ) k= ( n ) φ u(z + re iθ k ) θ k = k= k= φ u(z+re iθ )dθ. 27

28 (e) Examples of Subharmonic functions log f is subharmonic (f) Hadamard Three-Line Theorem Let f be holomorphic on S(a, b) = {x+iy : a < x < b} and define M(x, f) = max{ f(z) : Re(z) = x}. Assume M(a, f), M(b, f) <. Then if f(z) Be A z on S(a, b), for real numbers A, B, then M(x, f) is finite and log(m(x, f)) is convex. The proof proceeds by cases. Case : Assume f(z) is bounded on S(a, b). Then there is some M < such that f(z) < M. Therefore M(x, f) < M for x [a, b]. It remains to show that log M(x, f) is convex. log M(x, f) is convex if, since we may write x = b x b a a + x a b a b, Notice that this is equivalent to log M(x, f) b x x a log M(a, f) + log M(b, f). b a b a f(x + iy) M(x, f) M(a, f) b x b a M(b, f) x a b a, for y R, x [a, b]. This, in turn, is equivalent to f(z)m(a, f) b z z a b a M(b, f) b a since a z = a x. Define g(z) = f(z)m(a, f) b z b a M(b, f) z a b a. Then g(a + iy) = f(a + iy) M(a, f), g(a + iy) = f(b + iy) M(b, f). Define now g ɛ (z) = g(z) + ɛz for some ɛ >. Then g ɛ(a + iy), g ɛ (b + iy). For x [a, b], y >>. Then g ɛ (x + iy) MM(a, f) b x b a M(b, f) a x b a M min{m(a, f), M(b, f)} ɛ y ɛ y, for y ɛ M min{m(a, f), M(b, f)}. Then, by the Maximum Modulus Principle, g ɛ (z) on S(a, b). Letting ɛ +, we have g(z), which proves our claim. Case 2: f(z) Be A z. Consider the function g ɛ (z) = f(z)m(a, f) b z b a M(b, f) a z b a e ɛz 2. Then we have and similarly For x [a, b], y >> we have g ɛ (a + iy) e ɛ(a+iy)2 = e ɛ(a2 y 2) e ɛa2, g ɛ (b + iy) e ɛb2. g ɛ (x + iy) = f(z) M(a, f) b x b a M(b, f) x a b a e ɛ(x 2 y 2) Be A z min{m(a, f), M(b, f)} e ɛb2 ɛy 2 = Be A(b+ y )+ɛb2 ɛy 2 min{m(a, f), M(b, f)} for y appropriately large. Therefore, g ɛ (z) e ɛb2 for z S(a, b). Letting ɛ + yields f(z) M(a, f) b z b a M(b, f) a z b a for z S(a, b) as desired. 28

29 (g) Hadamard Three-Circle Theorem Let f(z) be holomorphic in A(; r, R) and define M(r, f) = max{ f(re iθ ) : θ [, )},. M(r, f), M(R, f) <, then log M(e t, f) is convex on (log r, log R). Let S = {x + iy : y R, log r < x < log R} and define F to be If F (z) = f(e z ), so F is holomorphic on S and F (z) = f(e z ). Then we have F (log r +iy) = f(re iy ) M(r, f) and similarly F (log R + iy) M(R, f). Then, by the previous theorem, we have 4. Jensen s Formula (a) Jensen s Formula f(r e it ) = F (log r + it) M(a, f) log R log r log r log r log R log r M(b, f) log R log r Let f be holomorphic in D(, R), f(z) for z D(, R) and f(). Then, if a,..., a n are the zeros of f in D(, R) counting multiplicity, then log f() + k= log R a k = log f(re iθ ) dθ. Let f(z) g(z) = ( ). n R(z aj) j= R 2 a jz Then g(z) for z D(, R) and By the Mean-Value Property, Further completing the proof. log g() = g(re iθ ) = log g() = log (b) Application of Jensen s Formula f(re iθ ) n j= R(z aj) R 2 a jz log g(re iθ ) dθ = f() n j= a j R = f(re iθ ). n = log f() + log f(re iθ ) dθ. j= log R a j, Let f : D(, ) D(, ) be holomorphic and f() = 2. What is the best upper bound for the number of zeros of f in D(, /2)? Solution: Let /2 < r < and a,..., a n be all the zeros of f in D(, r) such that f(re iθ ). By Jensen s Formula log f() + log r a j = log f(re iθ ) dθ <, so j= j= log r < log f() = log 2. a j 29

30 Without loss of generality, the zeros of f in D(, /2) are a,..., a k, k n. Then For a j D(, /2), log or after simplification, we have k r n a j log r log 2. a j j= r a j log r /2, so Letting r, we have k. 5. Special Functions (a) The Gamma Function log 2 j= k log r a j k log r /2, j= k log2, for all /2 < r <. log 2r Γ(z) = t z e t dt, which is holomorphic for Re(z) >. Note by integrating by parts, we get and continuing in this method, we get Γ(z) = Γ(z + ), z Γ(z) Γ(z + n + ), z(z + ) (z + n) which defines a meromorphic function on Re(z) > (n+) with simple poles at z =,, 2,..., n for any positive integer n. Note also, that Γ(n) = (n )! for positive integer values n, so Res(Γ(z); z = k) = lim z k (b) The Riemann Zeta Function z k z (z + k) Γ(z+k+) = ( k)( k + ) ( k + k ) ζ(z) = n= n z, ( )k Γ() =. k! for Re(z) >. ζ(z) can be extended to a meromorphic function with a simple pole only at z = and with residue there. ζ(z) also satisfies the following functional equation: ζ(z) = 2 z π z sin ( πz 2 ) Γ( z)ζ( z). 3