ECM3703: Complex Analysis

Transcription

1 ECM373: Complex Analysis April 5, 23 Module Convener: Dr. Robin Chapman Transcriber: Oliver Bond Contents Review of Complex Numbers 3. Basics Argand diagram Complex Functions The Exponential Function Complex Logarithms Main Course 2 2. Open and closed sets Domains Convergence of sequences Continuity Holomorphic Functions Theorem (Cauchy-Riemann Equations) Paths and path integrals Theorem (Fundamental Theorem of Calculus) Contours Corollary (ML-Bound) Cauchy s Theorem

2 2.9 Star Domains Cauchy s Integral Formula Corollary (Cauchy s Formula for the nth Derivative) Liouville s Theorem Corollary (Generalised Liouville Theorem) Power Series Theorem (Taylor s Theorem) Zeros of holomorphic functions Corollary (Identity Theorem) Laurent Series Theorem (Laurent s Theorem) Singularities Calculus of Residues Integrals of the form 2π F (cos t, sin t)dt (where F is a nice 2-variable function) The Semicircle Method Slice of pie Method Rectangle Method Rouché s Theorem Maximum Modulus Theorem Winding numbers Introduction to Complex Analysis Complex analysis is the study of functions involving complex numbers. It has real applications, for example, evaluating integrals like dx + x 2 π but contour integration easily gives us cos x + x 2 dx π e. Also we can derive results such as 2

3 n n n n 2 π2 6 n 4 π4 9 n 6 π6 945 etc. Also complex analysis has applications to many other branches of mathematics from number theory to fluid mechanics. For example, the Riemann ζ-function explains the distribution of prime numbers. Review of Complex Numbers. Basics The set of complex numbers, C is defined by C { x + iy x, y R, i 2 }. Addition and subtraction are all easy, for example, if z 3 i and w + 4i then z + w (3 i) + ( + 4i) 4 + 3i z w (3 i) ( + 4i) 2 5i zw 3 i + 2i 4i i Division is a bit trickier, but z w z ( ) w so division boils down to evaluating x+iy. The trick is similar to rationalising the denominator : x + iy x iy (x + iy)(x iy) x iy x 2 + y 2 x x 2 + y 2 y x 2 + y 2 i. Consider x 2 + y 2 for x, y R. Since x 2 and y 2, x 2 + y 2 also, and x 2 + y 2 > unless x 2 and, which can only happen if x and y. So we can always divide by a complex number as long as it s not zero. 3

4 Example z 3 i, w + 4i. 3 i (3 i) ( 4i) + 4i ( + 4i) ( 4i) 3 i 2i 4i i i. Jargon For z x + iy with x, y R, x R(z) - the real part of z y I(z) - the imaginary part of z x iy z z - the complex conjugate of z. Also, and (x + iy) + (x iy) x 2 R(z) z + z 2 and I(z) z z 2 zz (x + iy)(x iy) x 2 + y 2 which is a non-negative real number. Let z x 2 + y 2 zz be the absolute value (or modulus) of z. Note that x 2 x 2 + y 2 x 2 x 2 + y 2 x x 2 + y 2 i.e. R(z) z and I(z) z. 4

5 .2 Argand diagram Regard x + iy as the point (x, y) in the Cartesian plane. I 3 + 2i + i + i R Addition and subtraction in C correspond to addition and subtraction of position vectors. By Pythagoras theorem, z x 2 + y 2 is the distance from to z in the Argand diagram! Lemma If z, w C then (z ± w) z ± w (zw) z w (z/w) z /w if w. Idea of Proof Write z x + iy and w u + iv. Note that ( i) 2 i 2 in a formula is valid as long as all variables are real. Corollary If z, w C then zw z w. Proof zw 2 (zw)(zw) zw z w zz ww z 2 w 2 zw z w (taking square roots) 5

6 Corollary (Triangle Inequality) If z, w C then z + w z + w Proof (Estermann) If z + w then this is definitely true! Otherwise assume z + w. Then z z + w + w z + w ( ) ( ) z w R + R z + w z + w Also note that ( ) z R z + w ( ) z w z + w and R w z + w z + w so z z + w + w z + w z + w z + w (multiplying both sides by z + w ). The triangle inequality says that the length of a side of a triangle is at most the sum of the lengths of the other two sides. Remember it, as it will be used a lot in this module! z + w z + w I w z z R Also, another very useful inequality is z + w z w 6

7 whose proof is a question on the problem sheet. Let z and z C. Consider w z/ z. then w. We claim there is some θ R such that w cos θ + i sin θ. If w is in the first quadrant: I w θ cos θ sin θ R In general to get to the other 3 quadrants, we need θ outside the interval [, π 2 ]. For example, if θ 3π 4 then cos θ + i sin θ 2 + i 2, and if θ π 4 say, then cos θ + i sin θ 2 i 2. If θ >, cos θ + i sin θ is obtained by starting at the point and rotating an angle θ anticlockwise. If θ < we get cos θ + i sin θ by starting at and rotating clockwise through an angle θ. A corollary to this is that each nonzero complex number has the form z re iθ where r z and e iθ is defined to be cos θ + i sin θ. This is called the polar form of z. Lemma For θ, φ R, e i(θ+φ) e iθ e iφ. Proof The LHS is cos(θ + φ) + i sin(θ + φ). The RHS is (cos θ + i sin θ) (cos φ + i sin φ) cos θ cos φ sin θ sin φ + i (cos θ sin φ + sin θ cos φ). These are the same (they are addition formulae for trigonometric functions!) 7

8 Corollary ( re iθ) ( se iφ) rse i(θ+φ). We call θ in z re iθ an argument of z. Note that arguments are not unique, as e 2πi cos 2π + i sin 2π + i e i. Similarly, e i(θ+2π) e iθ e 2πi e iθ. This can also be extended to e i(θ+2nπ) e iθ if n Z. In general, e iθ e iφ θ φ 2nπ with n Z. Arguments are only defined modulo 2π. Define the principal argument of z to be the unique θ satisfying π < θ π with z z e iθ. We write the principal argument as Arg(z). For example, we have Arg() Arg(i) π 2 ( Arg( ) ) π Arg 2 + i 2 3π ( ) 4 Arg 2 i 2 π 4. Also we have Arg(z + w) Arg(z) + Arg(w) + 2mπ where m Z, and m depends on z and w. In the product ( re iθ) ( se iφ) rse i(θ+φ), although θ and φ may be the principal arguments of z and w, θ + φ may not be the principal argument of zw. But θ+φ differs from Arg(zw) by a multiple of 2π, e.g. ( )( ) and Arg( )+Arg( ) 2π Arg(). z I w θ + φ θ φ R z + w 8

9 .3 Complex Functions Complex analysis deals with functions of complex numbers. We consider functions of the form f : D C where D C. Example f(z) z 2 +. Note that f(±i) since f can take the value over C. Polynomials: f(z) a + a z + a 2 z a n z n are all functions from C to C. Rational functions: f(z) z or f(z) z 2 + etc. In general, f(z) g(z) is a rational function when g and h are polynomials. h(z) f(z) z is a function from C\{} to C. f(z) z z 2 is defined on C\{i, i} and is a function from C\{i, i} to C. + Theorem If g(z) a n z n + a n z n a z + a with a n, and a, a 2,..., a n C, then the equation g(z) has at most n complex roots. Proof We need to use the following result. Lemma (Remainder Theorem) If g(z) is a polynomial over C and b C, then g(z) (z b)h(z) + g(b) where h(z) is a polynomial. Proof By long division of polynomials, g(z) (z b)h(z) + c where c C. Substitute z b to get g(b) h(b) + c, i.e. c g(b). If g(z) had no roots then we win. Suppose z is a root. Then g(z ) and so by the remainder theorem, g(z) (z z )h(z) where h is a polynomial. As g(z) a n z n +... then h(z) a n z n If h(z 2 ) then we can remove another factor (z z 2 ) and so on. Eventually g(z) (z z ) (z z 2 )... (z z n ) p(z) where p(z) is a polynomial and p(z) has no solutions. p(z) a n z n k +... (with k n). 9

10 g(z ) z z or z z 2. or z z k or p(z) z {z,..., z k }. Therefore g(z) has k n distinct solutions. The only polynomials p(z) over C with no solution to p(z) are constant polynomials (Fundamental Theorem of Algebra). In general, every polynomial over C does factorise as a (z z ) (z z 2 )... (z z n ). Given a rational function f(z), f(z) is defined for all but finitely many complex values. If f(z) g(z) h(z) then h(z) except for finitely many values of z..3. The Exponential Function Let z x + iy, with x, y R. Define exp(z) e z e x+iy e x (cos y + i sin y). The formula e x n xn n! should generalise to e z n zn n!. For this to be the case, e yi (iy) n n n! + iy y2 2! iy3 + y4 3! 4! +... ) ) ( y2 2! + y4 4! (iy iy3 3! +... cos y + i sin y.

11 Properties e z+w e z e w follows from e x+y e x e y x, y R. (This is a homomorphism from (C, +) to (C\{}, ).) e z+2πi e z (as e z+2πi e z e 2πi e z e z ), so exp is a periodic function on C with period 2πi. e x+iy e x e iy ex cos y + i sin y e x cos 2 y + sin 2 y e x e x. So e z e R(z). For y R, e iy cos y + i sin y and e iy cos( y) + i sin( y) cos y i sin y. So cos y eiy + e iy 2 and sin y eiy e iy 2i Now define, for z C, Then cos z eiz + e iz 2 and sin z eiz e iz. 2i cos ix e x + e x 2 cosh x sin ix e x e x ( 2i e x + e x ) i 2 i sinh x..3.2 Complex Logarithms Given w C, what are the solutions of exp z w? We need w as exp(z) (because exp(z) exp( z) ). For any nonzero w, exp(z) w is soluble but with infinitely many solutions! Write w re iθ in polar form for y R. Set z x + iy, with x, y R. Then exp(z) e x e iy should equal re iθ. e x e x e iy re iθ r >. So x has to equal log r. Then we get e iy e iθ, i.e. y θ + 2nπ where n Z. So e z re iθ has solutions z log r + i (θ + 2nπ) where n Z. Any of these solutions is a logarithm of re iθ w, that is, if w C\{} then a logarithm of w is a solution to exp z w. One logarithm of w is log w + iarg(w). We call this the principal logarithm and write it as Log(w). As before we can deduce that

12 log(zw) Log(z) + Log(w) + 2πni (n N). Example Log( 3 4i) 4 φ 5 3 Here, Log( 3 4i) log 3 4i θi ( π log 5 i 2 + arctan 3 ). 4 2 Main Course We want to be able to study differentiable functions on subsets of C. 2. Open and closed sets An open disc is a set of the form where a C and r R +. D(a, r) {z C z a < r} A closed disc is one of the form and a punctured open disc is one of the form D(a, r) {z C z a r} D (a, r) {z C < z a < r}. 2

13 I I I r a D(a, r) R r a D(a, r) R a D (a, r) R Let A C. Each point a C is either an interior point of A or an exterior point of A, or a boundary point of A. a is: an interior point of A is there is no r > with D(a, r) A (for example, a ) an exterior point of A if there is r > with D(a, r) A (for example a 2 ) a boundary point of A if a is neither an interior nor exterior point of A (for example, a 3 ) a 2 a a 3 A subset A C is open if all a A are interior points of A, i.e. no element of A is a boundary point of A. A subset A C is closed if every a C which is a boundary point of A is an element of A, i.e. A contains all its boundary points. Claim. D(a, r) is open 2. D(a, r) is closed. 3

14 Proof. Let b D(a, r). We need to prove b is an interior point of D(a, r), i.e. we need to find s > with D(b, s) D(a, r). b a Try s r b a. Let z D(b, s). Then z b < s. We need to prove that z D(a, r), i.e. that z a < r. then z a (z b) + (b a) z b + b a (by the triangle inequality) < s + b a r So z D(a, r) and D(b, s) D(a, r). We should have proved that s > but as b D(a, r) then b a < r and so s r b a >. 2. To show D(a, r) is closed, we need to show every boundary point of D(a, r) is an element of D(a, r). It suffices to prove that if b / D(a, r) then b is not a boundary point of D(a, r) (which is the contrapositive), so that b a r. So we shall show that each b / D(a, r) is an exterior point. a b D(a, r) We claim there is s > with D(b, s) D(a, r). Take s b a r. Note that as b / D(a, r) then b a > r. Skipping the details, we again use the triangle inequality. 4

15 Examples C is both open and closed. B {z C z and I(z) > } is neither open nor closed. For example, and i are boundary points of B. B so B isn t closed, but i B so B isn t open. i B 2.2 Domains A domain is a connected open set. A subset A C is bounded if A D(a, r) for some a C, r >. C is unbounded, as we can t have C D(a, r) as a + 2r / D(a, r). Definition An open set U C is connected if for any 2 points a and b in U, I can join a to b in a finite sequence of straight line segments contained within U. (This is only appropriate for open sets.) a b Examples D(a, r) is connected. Let b, c D(a, r). One can join b and c directly by a straight line segment which itself lies within D(a, r). U {z C R(z) } is not connected. U consists of points off the imaginary axis. Any sort of path made up of line segments from to must cross the y-axis so can t lie entirely inside U. 5

16 Consider A(a, r, s) {z C r < z a < s, r, s R, < r < s}, the annulus with centre a and the region between the two circles z a r and z a s. This is connected. Draw a polygon inside A(a, r, s) encircling the centre. For z, w A(a, r, s) we can join z and w to the polygon by straight line segments, so to get z to w, we walk from z to the polygon, walk around the polygon, then walk to w. So the annulus is connected. z w 2.3 Convergence of sequences Let (z n ) be a sequence of complex numbers. We say (z n ) converges to z or z n z or lim n z n z if either R (z n ) R (z) and I (z) I (z) z n z. (These are equivalent.) n z n converges to z if n z n converges to z as N (same as in real analysis). 6

17 We claim that if z < then n zn converges to z. N n z n zn+ z so N z z n z N+ z n as z <. As with real series, n z n is divergent when z n. If z > then z n. The calculus of limits works over C in exactly the same way as over R. Lemma (Absolute Convergence) If n z n converges then so does n z n. Outline of Proof Apply the real version to R (z n ) and I (z n ), etc. We say n z n is absolutely convergent if n z n converges. Example For all z C, n zn n! is absolutely convergent since z n n! n n z n n! is a convergent series over R. Proof that n zn n! e z Note that (z + w) n n! n n n! n k k n w k k! n k ( ) n z n k w k (by the binomial theorem) k z n k w k (n k)!k! ( ) m z m m! (letting m n k). 7

18 At ( ), we interchanged the order of summation. This is legitimate as the double series is absolutely convergent. In particular, if x, y R, then (x + iy) n n n! n x n n! (iy) n n n! e x (cos y + i sin y) e x+iy. 2.4 Continuity Definition Let f : A C where A C. We say that f is continuous at a A if all sequences (a n ) A with a n a satisfy f(a n ) f(a). We say f is continuous on A if f is continuous at all a A. Examples Obviously constant functions are continuous and f(z) z is continuous. Theorem Continuity holds under addition, subtraction, multiplication, division and composition, subject to the usual caveats (dividing by zero/well-defined composition, etc.) Proof Same as for real analysis. Remark Polynomials are continuous on all of C, rational functions are continuous where defined. Important Example f(z) exp z is continuous. Proof Suppose a n a. We will show that e an e a. Note that e an e a e an a e a e bn where b n a n a. It suffices to show that e bn. e bn k k b k n k! b k n k! 8

19 As b n, b n eventually. That is, there exists N N with b n n N. Hence e bn b n k! k b n k! k b n k! k b k n b k n b n (e ) as b n for n N. As b n, e bn so that e bn as required. Hence exp(z) is continuous on all of C I R Here is a complex plot, generated with MATLAB, to illustrate the continuity of exp z. The brightness illustrates the absolute value (so z is black, and z leads to white), and the arguments are represented by hue (on a colour scale) where an argument of is red, and an argument of π is cyan. Notice that there are no discontinuities (e.g. changing colours/shading along lines or points) so this is visual confirmation that exp z is continuous, and you can see that it is actually periodic (with period 2πi - the colour of a point will the exactly the same as that 2πi above it). 9

20 Important Example g(z) Log(z) is not continuous. Let (a n ) be a sequence of points convergine to a from below the real axis (e.g. a n i n ). I a a n a 3 a 2 R a Now g(a) Log( ) iπ but the principal arguments of the a n are negative and tend to π as n so that lim Log(a n) iπ Log(a). n As in real analysis we can deal with limits of functions, i.e. lim z a f(z) b means that f(a n ) b for all sequences (a n ) with a n a such that a n a. 2

21 I R There is a slit on the negative real axis; the colours are not the same on either side of the line, so the arguments on either side do not reach the same limit. 2.5 Holomorphic Functions Definition Let f : A C (with A C). We say that f is differentiable at a A if exists. This limit is called f (a) provided it exists. f(z) f(a) lim z a z a Definition A function on an open set U which is differentiable at every point of U is called a holomorphic function on U. Theorem As in real analysis, if f and g are differentiable then so are f ± g, fg, f/g, f g, g f subject to the usual caveats, and their derivatives are given by the standard formulae from calculus (product/quotient/chain rules). The proof is the same as in real analysis. Examples ( ) z 2 + has derivative d dz 2z z 2 + ( e z2 has derivative d dz e z2) 2ze z2 etc. (z 2 +) 2. 2

22 Theorem exp(z) is differentiable and it is its own derivative. Proof so e z e a z a ea ( e z a ) z a e z e a lim z a z a e z a ez lim z a z a e w lim (setting w z a) w w But e w w w n w n n2 n! w n 2 n! and by an argument similar to that which was used for the continuity of exp(z) we get that ( e w ) lim w w whence d dz (ez ) e z for all z C. Corollary d (cos z) sin z, etc. dz Proof d dz (cos z) ieiz ie iz 2 eiz + e iz 2i sin z, etc. 22

23 Example Let g(z) z. Let a C. Then g(z) g(a) z a z a z a (z a) (z a) so that provided the limit exists. But this limit does not exist! Let a n n g (z a) (z) lim z a (z a) z lim w z as n. Now (a n ) a n /n /n. But if b n i n as n we have (b n ) b n i/n i/n. So g(z) is not differentiable anywhere! 2.5. Theorem (Cauchy-Riemann Equations) Let where x, y, u(x, y), v(x, y) R. If f is holomorphic on an open set U then f(x + iy) u(x, y) + iv(x, y) u x v y and v x u y. So the real and imaginary parts of a holomorphic function f almost detemine each other; R(f) u and I(f) v are deeply entwined with each other. Proof Let z x + iy U. Then f f(z) f(z ) (z ) lim z z z z f(z + h) f(z ) lim h h 23

24 . Specialise h R. Then f [u(x + h, y ) + iv(x + h, y ) u(x, y ) iv(x, y )] (z ) lim h h u(x + h, y ) + u(x, y ) v(x + h, y ) v(x, y ) lim + i lim h h h h u + i v x x (x,y ) 2. Specialise h ik with k R. Then (x,y ) f [u(x, y + k) + iv(x, y + k) u(x, y ) iv(x, y )] (z ) lim k ik v(x, y + k) v(x, y ) u(x, y + k) u(x, y ) lim i lim k k k k v i u. y y (x,y ) (x,y ) Compare both results (equating real and imaginary parts) to give u x (x,y ) v y (x,y ) and v x (x,y ) u y (x,y ). A converse If f(x + iy) u(x, y) + iv(x, y) is a function on an open set U, then if f satisfies the Cauchy- Riemann equations and u v x and y are continuous on U then f is holomorphic. The proof is omitted here, but see Chen s notes - to which there is a link on the website. Example f(z) e z In this case, f(x + iy) e x (cos y + i sin y) with u e x cos y and v e x sin y. Then confirming f to be holomorphic. u x ex cos y v y ex cos y v x ex sin y u y ex sin y Example f(z) z In this case, f(x + iy) x iy with u x and v y. But u v x and y u x so f isn t holomorphic. 24

25 Example f(z) z In this case, f(x + iy) x 2 + y 2 with u x 2 + y 2 and v. Therefore u x x u and x 2 + y2 y y x 2 + y 2 but also v x v y. The Cauchy-Riemann equations can t hold unless x y, but then u v x and y Therefore f isn t differentiable anywhere! would not exist. Application A holomorphic real-valued function has to be constant. Since v, then by the Cauchy- Riemann equations, u x v y and u y v x so u is constant. Claim Log(z) is holomorphic on the set C\ {x x R, x }. Proof For convenience, we ll prove this for {z x + iy x > }. For x >, Log(x + iy) log x + iy + iarg(x + iy) log x 2 + y 2 + i arctan y x. In this case, u 2 log(x2 + y 2 ) and v arctan y x. 25

26 u x 2x 2 x 2 + y 2 x x 2 + y 2 u y y x 2 + y 2 v x + ( y ) 2 ( y ) x 2 x x2 ( x 2 + y 2 y ) x 2 y x 2 + y 2 v y + ( y x ) 2 x2 x 2 + y 2 x x x 2 + y 2 ( ) x The derivatives all agree, so Log(z) is holomorphic for R(z) x >. Then for when x < and y >, u is the same as before and v i [ π arctan y x]. And for when when x < and y <, u is the same and v changes sign from the previous case. 2.6 Paths and path integrals These are analogous to line integrals in the plane. A path is a continuous map : [a, b] C. It is smooth if is differentiable and is continuous. We can write (t) x(t) + iy(t), with x, y : [a, b] R. is continuous (and differentiable) if both x and y are. 26

27 I (b) (a) R Examples. Let z, z 2 C. The line segment from z to z 2 is the path : [, ] C defined by (t) z + t(z 2 z ). Sometimes we denote this as [z, z 2 ]. z I R z 2 2. Let z C, r >, α, β R and α < β. 27

28 Define : [α, β] C by (t) z + re it. is an arc of the circle centre z and radius r. I z + re iβ z R z + re iα If α, β 2π, one traverses the whole circle. If α and β 4π, then the circle is traversed twice, and so on. Both of these paths are smooth, as (t) z 2 z and (t) ire it respectively (which are continuous). Path jargon For the path : [a, b] C: (a) is the start point (b) is the end point If (a) (b) then the path is closed (e.g. a full circle) If (t) (s) t s then the path is simple (i.e. injective). It doesn t intersect itself except it might close up at the end. ( t) is a map : [ b, a] C is the reversal of. 28

29 Examples Simple closed path Simple but not closed Closed but not simple Neither closed nor simple Let f be continuous on an open set U. Let : [a, b] C be a smooth path contained in U. Define f(z)dz b a f((t)) (t)dt (as a mnemonic, z (t) so dz (t)dt), In the last integral, we integrate a complex function over a real variable. Naturally b a (u(t) + iv(t)) dt b a u(t)dt + i b a v(t)dt. Example Integrating f(z) z over, the line segment from to 2 + 2i. So then : [, ] C t + t( + 2i) so (t) + 2i. f(z)dz ( + ( + 2i)t) ( + 2i)dt (( + 2i) + ( 3 + 4i)t) dt [ ( + 2i)t + 2 ( 3 + 4i) t2 ] + 2i + ( 3 + 4i) i. 29

30 Remark There is a connection with line integrals: f(x + iy) u(x, y) + iv(x, y) f(z)dz (u(x, y) + iv(x, y)) (dx + idy) Some properties (f(z) + g(z)) dz f(z)dz + g(z)dz αf(z)dz α f(z)dz α C If is the reversal of, then f(z)dz f(z)dz. f(z)dz f(z)dz + 2 f(z)dz where : [a, b] C, : [a, c] C and 2 : [c, b] C, with a < c < b. Also (t) (t) and 2 (t) (t). 2 Notation: w 2 w f(z)dz means f(z)dz where is the line segment from w to w 2 - this only makes sense if f is defined in a domain containing this segment Theorem (Fundamental Theorem of Calculus) Assume f : U C is holomorphic and U is open. Also assume that f is continuous. Then f (z)dz f ((b)) f ((a)) If is a closed path, then f (z)dz. 3

31 Proof f (z)dz b a b a f ((t)) (t)dt d f ((t)) dt (applying the chain rule in reverse) dt [f ((t))] tb ta (by the real version of the Fundamental Theorem of Calculus) f ((b)) f ((a)) Example 2+2i zdz using the Fundamental Theorem of Calculus. 2+2i zdz 2+2i d dz (2 + 2i)2 2 ( z i i. ) dz Example On U C\(, ], Log(z) has derivative z. So if is a path from w to w 2, which does not touch the negative real axis or, then dz z Log(w 2) Log(w ). 3

32 I w 2 w R However, if is the unit circle: : [, 2π] C t e it then dz 2π z i (t) (t) dt 2π ie it dt eit 2π tdt 2πi. Corollary is not the derivative of any holomorphic function defined on a region containing the unit circle. z 2.7 Contours A contour of (,..., n ) is a sequence of smooth paths arranged end-to-end, i.e. the end-point of j is the start point of j+. It is closed if the end-point of n is the start point of (The last two contours shown are closed.) 32

33 For a contour (,..., n ) define f(z)dz f(z)dz +... f(z)dz. n The Fundamental Theorem of Calculus also works for contours. In particular, f (z)dz if is a closed contour. Estimations of contour integrals If is a smooth path : [a, b] C, the length of is and the length of a contour (,..., n ) is l() b a (t) dt n l() l ( j ). j Example If is a line segment from w to w 2, that is, : [, ] C t w + (w 2 w )t and (t) w 2 w. Then trivially, l() w 2 w dt w 2 w. Example with w C and r >. : [α, β] C t w + re it Then (t) rie it and (t) r. So l() β α rdt r(β α). 33

34 A whole circle centre, radius r has length 2πr. : [, 4π] C, (t) e it has length l() 4π etc. Lemma If f is complex-valued then b a b f(t)dt f(t) dt. a Proof Let I C b a f(t)dt We claim that there is α e ia with a R, such that αi I. When I, any α works, otherwise I I has absolute value so it equals eia with a R. Then α. I αi β a β a αf(t)dt β u(t)dt + i v(t)dt a where u(t) R [αf(t)], v(t) I [αf(t)]. Since I is real, I b a b a b a b a u(t)dt R [αf(t)] dt αf(t) dt f(t) dt as α. 34

35 2.7. Corollary (ML-Bound) Consider a contour and a continuous function f defined on. Suppose f(z) M, for all z on. Then f(z)dz ML where L l(). Proof Assume is a smooth path : [a, b] C. Then b f(z)dz f ((t)) (t)dt a b M f ((t)) (t) dt (by the Lemma) a b a ML. (t) dt (as f((t)) M) Example Showing that e z z dz eπ where is the arc of the unit circle to the right of the imaginary axis. I i R i 35

36 On, e z e R(z) e z e e and also z z 2 (by the alternate triangle inequality) 2 z e z z e on. So by the ML-bound, e z z dz eπ. 2.8 Cauchy s Theorem Let f be holomorphic on a domain U and let be a closed contour inside U. U By the Fundamental Theorem of Calculus, if f has an antiderivative g, i.e. f g, then f(z)dz g (z)dz But dz z 2π where is the unit circle. The function f(z) is holomorphic everywhere except at z, and this lies inside the contour. Vague Cauchy s Theorem If f is holomorphic on and at all points inside a closed contour, then f(z)dz. 36

37 Cauchy s Theorem for triangles WARNING: The proof of this theorem is the hardest in the course! A triangle T consists of 3 points its vertices, the line segments joining these 3 points, the edges and the interior points. Points lying inside intervals with endpoints on two different edges of the triangle. w 3 T T w 2 w Use T to denote a typical triangle (set of vertices, edge points and interior points) and T denotes the boundary of T (vertices and edge points but not the interior points). We also use T to denote the contour formed by the edges. For definition s sake, assume that T goes round anticlockwise. Theorem (Cauchy s Theorem for Triangles) If f is holomorphic on a domain U and T U is a triangle, then T f(z)dz. Remarks. T f(z)dz w 2 w f(z)dz + w 3 w 2 f(z)dz + w w 3 f(z)dz (integrals along straight line segments). 2. If we knew f was continuous on U, then Cauchy s theorem follows quickly from Green s Theorem. In fact f will be continuous, but proving that relies on this theorem! (To see the Green s theorem proof see William Chem s notes). Lemma Let T be a triangle with vertices w, w 2, w 3. 37

38 w z 3 T z 2 T T 4 2 T 3 w 2 z w 3 Then T f(z)dz 4 j T j f(z)dz. Here we take the boundaries in the anticlockwise direction. T f(z)dz T 4 f(z)dz is the sum of 2 integrals over the sides of the small triangles, 6 of which cancel, e.g. z2 f(z)dz occurs in f(z)dz z 3 T z3 and f(z)dz occurs in f(z)dz z 2 T 2 and these cancel. z3 Also f(z)dz occurs in f(z)dz w T w2 and f(z)dz occurs in f(z)dz z 3 T 2 w2 and these add to give f(z)dzwhich occurs in w T f(z)dz. The remaining 6 integrals combine in this way to give T f(z)dz. Proof of Theorem Write 38

39 T f(z)dz I(T ) 4 I(T j ). j One of these T j satisfies I(T j ) 4 I(T j). Set T () T and T () T. Note that T j has side lengths half of those of T. We get a sequence ( T (m)) of triangles by repeating the argument, with T (m+) T (m), T () T.... T T () T () T (2) The side lengths of T (m) are 2 m those of T and We claim there is a point z with z T (m) for all m. ( I T (m)) I (T ). 4m (The website has a proof that a decreasing sequence of closed bounded non-empty subsets of C has a point in common) Take ε >. The function f is differentiable at z, i.e. Therefore we can pick a radius such that lim z z f(z) f(z ) f (z ) z z. f is holomorphic on D(a, r) f(z) f(z ) f (z ) z z < ε for z D(z, r). On D(z, r), f(z) f(z ) f (z )(z z ) ε z z. Let g(z) f(z) f(z ) f (z )(z z ). For large enough m, T (m) D(z, r). 39

40 T m z z D(z, r) So g(z)dz ( T ε length (m)) max z z T (m) ε2 m P d m (where d m greatest side length of T (m) P greatest side length of T 2m 2m ) ε4 m P 2 but g(z) f(z) + a + bz for some constants a and b. g(z)dz f(z)dz + (a + bz) dz T (m) T (m) T ( (m) I T (m)) d + (az + 2 ) T (m) dz bz2 dz I }{{} (T (m)) so 4 m I ( T (m)) ε4 m P 2 I(T ) εp 2 I(T ). 2.9 Star Domains A star domain is a domain in C with a star centre. A point z is a star centre of U if for all z U the line segment from z to z lines inside U. Examples. Discs are convex; every line segment with endpoints in the disc are in the disc. 4

41 z z 2. An annulus is not a star domain. If z lies in the annulus, let z be a reflection of z about the centre, then the line segment from z to z is not contained in the annulus. z z 3. The slit plane U C\ {x R x } is a star domain. is a star centre. Theorem If f is holomorphic on a star domain U then f g for some g holomorphic on U. Proof Let z be a star centre of U. z U z Define g(z) z z f(w)dw. We need to prove that ( ) g(z + h) g(z) lim f(z). h h As U is a domain, U is open so there is r > with D(z, r) U. In considering this limit I can assume h < r. So z + h D(a, r). 4

42 z U z + h z D(z, r) Let T be the triangle with vertices z, z, z + h. I claim that T U. The line segment from z to z + h lies in D(z, r) and so also in U. For any point z on this line segment, the line segment from z to z lies inside U as U is a star domain, and z is a star centre. Any point inside T lies on a line segment from z to a point on the line segment between z and z + h, so T U. We can apply Cauchy s theorem to T, so that T f(w)dw, i.e. z z+h f(w)dw f(w)dw + z z z+h z f(w)dw g(z + h) g(z) g(z + h) g(z) h g(z + h) g(z) h z+h z z+h f(z) h z f(z) h f(w)dw z+h z (f(w) f(z)) dw (f(w) f(z)) dw Pick ε >. For some s > (s < r if we like) D(z, s) has the property that w D(z, s), then f(w) f(z) < ε. As f is continuous, lim w z (f(w) f(z)). If h < s, then f(w) f(z) < ε for all w on the line segment from z to z + h. So I can take ε for M in the ML-bound, with the length of the segment being h. So z+h (f(w) f(z)) dw ε h z g(z + h) g(z) f(z) h ε holds if h < s. So ε > s > such that the above line holds, and therefore from the definition of the limit, Thus g is differentiable at z and g (z) f(z). ( ) g(z + h) g(z) lim f(z). h h 42

43 Corollary (Cauchy s theorem for star domains) closed contour on U, then f(z)dz. If U is a star domain with f holomorphic on U, and is a Proof There exists g with f g such that f(z)dz g (z)dz by the Fundamental Theorem of Calculus. Example e z3 +z cos(z 2 ) (2 + z 2 ) 2 dz where is the unit circle. D(, 2) The integrand (however horrible looking it is) is holomorphic on D(, 2), which contains the unit circle, and is a star domain. So the integral is zero. Example U C\ {x R x } (the "slit plane") is a star domain with star centre. Now z is holomorphic on U, so z L(z) z dw w. We claim that on U, L(z) Log(z). L (z) for some L(z). We can take To prove this, assume z >, I(z) > (not essential - slightly different diagrams needed for other cases). I claim L(z) (centred at ). dw w where consists of the line segment from to z and the circular arc 2 from z to z followed by 2 followed by the reverse of the line segment [, z] is a closed contour, so as U is a star domain, the integral over this contour is zero, i.e. dw w + dw 2 dw i.e. L(z) w + z w 2 dw w. dw w 43

44 where dw w z dt t log z. 2 is parametrised by (t) z e it, for t θ Arg(z), (t) i z e it. Then Therefore 2 dw w Arg(z) iarg(z) i z e it dt z eit L(z) log z + iarg(z) Log(z). This constitutes another proof that Log(z) is holomorphic on the slit plane, with L (z) /z. 2. Cauchy s Integral Formula Theorem (Cauchy s Integral Formula for a Circle) Let Ube a domain. Suppose D(a, r) U. Let be the circle centre a and radius r. If f is holomorphic on U then f(z) f(w) 2πi w z dw for z D(a, r). Note that this formula expresses the values of f in the interior of the disc in terms of its values on the boundary of the disc. Proof (technical claim) document on the website. There exists r > r such that D(a, r ) U. The proof of this initial remark is in a Assume that f is holomorphic on D(a, r ). Let z D(a, r). Let be a small circle, centre z radius s > contained inside D(a, r). We claim that f(w) w z dw f(w) w z dw. Draw a line horizontally through z. Use this to define contours 2 and 3 as shown. 44

45 We claim that that 2 f(w) w z dw + 3 f(w) w z dw In the first sum, the integrals over the horizontal edges cancel. The integrals over over arcs of the big circle add to Consider 2 f(w) w z contour over which f(w) w z dw. The function f(w) w z f(w) w z f(w) w z. f(w) f(w) w z dw, and those of the small circle add to w z dw. is not holomorphic on w z, so what we want to do now is consider a is holomorphic and dodges this singularity. Now consider U which is obtained by deleting z and all points vertically below it from D(a, r ); U is a star domain. By Cauchy s theorem, dw and similarly, 2 f(w) w z 3 f(w) dw w z f(w) w z dw f(w) w z dw. Parametrise by (t) z + δe it with t [, 2π], and (t) iδeit. f(w) 2π w z dw i 2π f(z + δe it ) δe it iδe it dt f(z + δe it )dt f(w) w z dw. Now let δ. We claim that 2π lim f(z + δe it )dt 2πf(z). δ So we need to justify this claim. Note that 2π f(z + δe it )dt 2πf(z) 2π [ f(z + δe it ) f(z) ] dt and as f is continuous at z, given ε > there is some η > such that f(u) f(z) < ε if u D(z, η). So if δ < η, 2π [ f(z + δe it ) f(z) ] 2π dt f(z + δe it ) f(z) dt 2πε. So this shows that lim δ 2π f(z + δe it )dt 2πf(z). Therefore 45

46 f(w) 2π dw lim w z i f(z + δe it )dt δ 2πif(z) as required. Example e z2 z + dz where is the circle centre, radius 2. Start off by parametrising (t) 2e it with t [, 2π]. Also, note that z + z ( ). By Cauchy s integral formula, as f(z) e z2 is holomorphic on C, we have e z2 dz z + 2πif( ) 2πie. Example e 2z z 2 dz where is the circle centre, radius. Note that where f(z) e2z z+. f(z) is holomorphic on and inside, so the integral is e 2z z 2 e 2z (z + )(z ) f(z) z 2πif() 2πi e2 2 πie 2. 46

47 Cauchy s Integral Formula for other closed contours The formula holds for f holomorphic on and inside a rectangular contour. The idea is the same, i.e. that integrating over the rectangle is the same as integrating over a small circle, and this reduces to proving that integrals along the two contours are zero etc. We claim that the formula is true for all simple closed contours, if a contour is traversed anticlockwise. The idea is to reduce to an integral over a small circle, and break up the different into contours each contained in a star domain. There is a well-defined notion of anticlockwise on. then if f is holomorphic on a domain containing and its f(w) interior, f(z)dz and for z in the interior of, f(z) 2πi w z dw. Cauchy s Integral Formula for the Derivative Theorem Let be a simple close contour. Let f be holomorphic on and its interior (we always assume is taken in an anticlockwise direction). If z is in the interior of, then f (z) f(w) 2πi (w z) 2 dw. Proof The idea is to consider Cauchy s integral formula and differentiate it with respect to z. But we shall proceed carefully! Define F (z) f(w) 2πi (w z) 2 dw. To show that F (z) f (z) we need to prove that ( ) f(z + h) f(z) lim F (z). h h Let z be in the interior of. Pick δ > such that D(z, δ) is contained in the interior of. We claim that f(w) is bounded on. If I have some parametrisation of, then : [a, b] C then t f ((t)) is a continuous real-valued function on this interval, so it is bounded. There is M > with f(w) M for all w on. There is M > with f(w) M for all w on. 47

48 Let < h < δ. Then f(z + h) f(z) F (z) h 2πi 2πi 2πi h 2πi h (w z h) (w z) (w z) }{{}}{{} 2 f(w)dw f(z+h) f(z) [ (w z h)(w z) (w z) 2 (w z) (w z h) (w z) 2 (w z h) f(w)dw f(w)dw (w z) 2 (w z h) ] f(w)dw f(w) M, w z δ for w on and D(z, δ) doesn t meet. By the alternate triangle inequality, we have w z h δ h f(z + h) f(z) F (z) h h M 2π δ 2 (δ h ) length() as h. So as h, whence F (z) f (z). f(z + h) f(z) h F (z) Using this for z in the interior of we get f (z + h) f (z) lim 2 f(w) h h 2πi (w z) 3 dw The proof is very similar to the above but the algebra is slightly fiddlier (if there was a L A T E Xsymbol for sad face, I d be obliged to put one here). Theorem Let U be a domain, with f holomorphic on U. Then f is also holomorphic on U. Consequently, if a complex function is differentiable once, then it is infinitely differentiable! Wow! Proof Let z U, be a circle with centre z with with and its interior contained in U (this is okay because U is open). Then f (z + h) f (z) lim h h exists (derived above), and then f (z) exists and so on! 48

49 2.. Corollary (Cauchy s Formula for the nth Derivative) If f is holomorphic, the nth derivative of f exists for all n N. We also get Cauchy s integral formula for the nth derivative, under the hypotheses of Cauchy s integral formula. f (n) (z) n! f(w) dw 2πi (w z) n+ Outline proofs. Follow proof for f (z) making appropriate modifications. 2. Integrating f (n) (z) CIF 2πi f (n) (w) dw by parts n times. w z Example sin z z 4 dz where is the unit circle. sin z z 4 dz f(z) dz where f(z) sin z (holomorphic on C) z3+ By Cauchy s integral formula, However, f () cos, so f () 3! 2πi 6 2πi 6 2πi f(z) dz (z ) 3+ f(z) z 4 dz sin z z 4 dz sin z z 4 dz πi 3. Example e z z 2 (z + ) dz where is the circle with centre and radius 2. The denominator of the integrand vanishes at z and z, both of which are inside, so we can t apply Cauchy s integral formula directly. 49

50 Method : Break up into two contours each encircling one of the values z or z. So f(z)dz f(z)dz + f(z)dz 2 e z z 2 (z + ) dz g(z) z 2 dz + h(z) z + dz 2 where g(z) ez ez z+ and h(z), holomorphic on and inside z 2 and 2 respectively. By Cauchy s integral formula, e z z 2 (z + ) dz 2πig () + 2πih( ) [ e z 2πi z + e z ] [ ] e z (z + ) z 2 + 2πi z z 2 + 2πie 2πi e. Method 2: expand into partial fractions. z 2 (z+) so z 2 (z + ) A z + B z 2 + C z + e z z 2 (z + ) dz A e z z dz + B e z z 2 dz + C e z z + dz Method 3: Calculus of residues (later). Cauchy Estimates Theorem Let f be holomorphic on a domain containing the closed disc D(a, r). Let M be an upper bound for f(z) for z on the circle centre a, radius r. That is, f(z) M for all z with z a r. Then f (n) (a) n!m r n. Proof Let be the circle centre a and radius r. Then by Cauchy s integral formula, On, f(w) M by hypothesis, where w a r. The length of is 2πr. So by the ML-bound, f (n) (a) n! f(w) dw. 2πi (w a) n+ 5

51 f (n) (a) n! 2π n!m r n. M 2πr rn+ 2. Liouville s Theorem Corollary (Liouville s Theorem) Let f be holomorphic on C (such a function is entire). Then if f is bounded on C, i.e. there exists M with f(z) M z C, then f is constant. Proof We can apply Cauchy s estimates with this bound M, and by any a and any r. f (a) M r ( a Cand r > ). We can let r, and we deduce that f (a) a C, i.e. f. If z C, f(z) f() z f (w)dw f(z) f() z C so f is constant. Corollary (Fundamental Theorem of Algebra) Let P (z) z n + a z n a n z + a n be a polynomials of degree n. Then P (z) has at least solution in C. Wow! Proof Let f(z). If P (z) has no root, then f(z) is an entire function. We claim that if this is the case P (z) then f(z) is bounded. f(z) z n + a an with z. We see that f(z) as z. 5 z z n

52 There is R > such that f(z) < if z > R. But f(z) is continuous on D(, R) so f(z) is bounded on D(, R) (a continuous real-valued function on a closed bounded set is bounded there). So f is bounded by Liouville s theorem. Thus f is constant and so P /f is constant. But then P (z) z n + a z n a n z + a n, so P (z) so P isn t constant! This is our contradiction. Therefore we must have that P (z) has a zero somewhere. Corollary If P (z) z n + a z n a n z + a n then P (z) (z z )(z z 2 )... (z z n ) for some z,..., z n C. Proof By the Fundamental Theorem of Algebra, where Q(z) is a polynomial of degree n. Keep extracting factors to get the desired result. P (z ) for some z P (z) (z z )Q(z) 2.. Corollary (Generalised Liouville Theorem) Let f be entire. Suppose there are n, C, R such that f(z) C z n for sufficiently large z with z > R. Then f is a polynomial of degree at most n. Proof Let a C. Then for each r >, f (n+) (a) (n + )! r n+ M(r) where M(r) is some upper bound for f(z) when z a < r (by Cauchy estimates). If we take r large enough, then when z a r. Take M(r) C (r + a ) n. Then f(z) C z n C ( z a + a ) n (by the triangle inequality) C (r + a ) n f (n+) (z) (n + )! (r + a ) n r n+ as r. 52

53 f (n+) (z) f (n+) (z) f(z) A + A z A n z n (integrating n + times). Example Suppose f(z) z + for all z C where f is holomorphic. Prove that f(z) Az + B where A and B. Proof If z then f(z) z + 2 z so by the Generalised Liouville theorem, f(z) Az + B where A, B C. Now By Cauchy s estimates, with z r >. On z r, B f() + by the condition f(z) z +. A f () r M(r) f(z) z + r +. Take M(r) r +.Then A r+ r as r, so A. 2.2 Power Series 2.2. Theorem (Taylor s Theorem) Let f be holomorphic on a domain U, and suppose D(a, r) U. Then f(z) if z D(a, r) (the series converges for such z). f (n) (a) n (z a)n n! 53

54 Proof By considering f(z + a) in place of f(z), we can assume a. So D(, r) U. Let < R < r. Then, the circle centre radius R is contained in D(, r). If z < R, f(z) f(w)dw 2πi w z by Cauchy s integral formula. Now Therefore N n w z w w z w ( z ) n w n ) N+ ( z ) n ( z w w w w ( ) z (geometric series formula) w ( z ) N+ w z w w z w z ( z ) N+ N w w z + n w ( z w ) n f(z) (( z ) N+ w 2πi w z ( z ) N+ w 2πi I n + + N n N w z f(w)dw + N a n z n ( ) n ) ( z ) n f(w)dw w w n z n f(w) w n+ dw where a n f(w)dw 2πi f (n) () w n+ n! by Cauchy s integral formula for the nth derivative. Now we apply the ML-bound to I N. I N ( z /R) N+ M 2π R z f(w) is bounded by M, say on, so that w R on. ( ) N+ Since z < R, z R as N, IN as N. By putting this limit into ( ), we have f(z) a n z n n n 2πR f (n) () z n if z D(, R). n! 54

55 If z D(, r), z < r so I can choose R with z < R < r. Using this particular r, we can get f(z) n f (n) () z n. n! Taylor s theorem says that holomorphic functions are locally power series. Remark Power series expansions may not hold for all points of the domain, but certainly in any disc surrounding a given point. For example, f(z) z and a. f is holomorphic on C\{}. f is also holomorphic on D(, ). In this disc, f(z) z + ( (z )) ( ) n (z ) n. However, outside of this disc, the series is divergent. f isn t given by this series outside of the disc. Again, as in real analysis, there are 3 possibilities.. f(z) n b n(z a) n converges z C. n 2. f(z) n b n(z a) n converges only for z. 3. There exists R (, ) (the radius of convergence) such that Proof: Same as in real analysis. f(z) { converges absolutely z a < R diverges z a > R. The radius of convergence makes more sense over C than R, since the power series converges on a disc of radius R. Theorem If the power series n b n(z a) n is convergent on D(a, R), then it defines a holomorphic function f there and moreover f (z) nb n (z a) n on D(a, R). n 55

56 Proof Assume a. otherwise replace z a with z. Let f (z) n nb nz n. Robin claims that f converges on D(, R). Let z D(, R). One can choose r with z < r < R. Then n b nr n converges, so b n r n, so (b n r n ) is a bounded sequence. There is M R with b n r n M for all n. nb n z n z n r n so nbn r n ( ) z n n r r nb nr n M r n ( z r ) n and so nb n r n M r n n ( ) z n r which is convergent by the ratio test. Therefore f (z) n nb nz n is absolutely convergent for z D(a, R). We repeat the argument for f 2 (z) n(n )b n z n 2 n2 which also converges on D(a, R). To show that f is differentiable and f f, consider f(z + h) f(z) h f (z). We need to prove that this tends to zero as h. 56

57 Therefore f(z + h) f(z) [ (z + h) n z n ] f (z) b n nz n (using the fact that f(z) h h n Now (z + h)n z n n ( ) n z n k h k (by the binomial theorem) h k k (z + h)n z n n ( ) n nz n h z n k h k 2 h k k2 (z + h) n z n nz n n ( ) n h h z n k h k 2 k k2 n ( ) n h z n k h k 2 (by the triangle inequality) k k2 n n! h k!(n k)! z n k h k 2 k2 n n(n )(n 2)! h k(k ) k2 n h n(n ) h n(n ) n(n ) 2 f(z + h) f(z) h k2 k z n k h k 2 (n k)! (k 2)! b n z n ) n (n 2)! k(k ) (k 2)!(n k)! z n k h k 2 n ( ) n 2 z n k h k 2 (as k 2,k(k ) 2 2) k 2 ( z + h ) n 2 f (z) h 2 n(n ) b n ( z + h ) n 2. If h < R z then z + h < R so n2 n(n ) b n ( z + h ) n 2 is absolutely convergent. Let z < r < R. If h < r then f(z+h) f(z) h f (z) h M 2. n2 Indeed, lim h f(z+h) f(z) h f (z), so f is differentiable with f (z) nb n z n. n Inside their disc of convergence, power series define functions that are holomorphic. Remark: Just as with power series over R, the radius of convergence can be computed by the ratio test, that is, 57

58 if lim n b n b n+ then b n (z a) n has radius of convergence equal to R. R 2.3 Zeros of holomorphic functions If f is holomorphic, then a zero of f is some z C such that f(z ). How does f behave near a zero? Let f be holomorphic on U, and f(a) (a U). By Taylor s theorem, f(z) n b n(z a) n (starting at n as f() ). This is valid in some disc centre a. Case (trivial): All the b n f(z) on a disc D(a, r), r >. Case 2: Some b n. Let N be the smallest integer with b N. Then f(z) nn b n(z a) n on some D(a, r). In this case we say f has a zero of order N (this integer is unique) at z a. A zero of order is simple A zero of order 2 is double A zero of order 3 is triple, and so on. The order of a zero of f is m at z a if and only if f(a)... f (m ) (a) f (m) (a). Remarked that we can have f(z) in some D(a, r) if not then f(a) or f has a zero of order m at a. Then as f(z) (z a) m b m as z a, f(z) in some punctured open disc D (a, r) {z C < z a < r}. Zeros of order m are isolated, surrounded by a disc where f(z) (except from at a). Theorem If f is holomorphic on a domain U, and f is identically zero on some disc D(a, r) with D(a, r) U, then f is identically zero on U. Proof (vague outline) We need the fact that U is connected. Let V {a U f is identically zero on some D(a, r)}. Then one proves that. V is open 2. U\V is open 3. In a connected open set U, the only V satisfying the first two rules are V or V U. 58

59 We omit proofs of these three properties, and say that V by the hypothesis so V U f is identically zero on U Corollary (Identity Theorem) Let U be a domain. Let (z n ) be a sequence of points in U converging to w U with z n w n. Then if f and g are holomorphic on U and f (z n ) g (z n ) for all n, then f(z) g(z) for all z U. Proof Let h(z) f(z) g(z). h (z n ) as h is holomorphic and continuous. So also z n w h (z n ) h (w) h(w). By the Theorem above, either h is identically zero on U, i.e. f g on all of U, or h has a zero of order m at w. Then w is an isolated zero of h, so that h(w) on some D(w, r). Let z n w but z n w for large enough n. Then z n D (w, r), the punctured disc. But this means that h (z n ) which is a contradiction. Examples. f(z) z n has a zero of order n at z. 2. f(z) sin z z z3 3! + z5 5!... has a simple zero at z and at z π (and other integer multiples of π); f(z + π) sin(z + π) sin z ( ) n z 2n+ (2n + )! n ( ) n (z π) 2n+ f(z) (2n + )! n (z π)3 (z π) so sin zdoes have a simple zero at z π. ( ) 3. f(z) cos z z2 2! + z4 4!... z2 2!... so cos z has a double zero at z. 59

60 2.4 Laurent Series 2.4. Theorem (Laurent s Theorem) Let f be holomorphic on the annulus A {z C r < z a < R} where r < R ( ). Then f(z) where b n C for each n Z. This series is called a Laurent series and equals n b n (z a) n... + b 2 (z a) 2 + b (z a) + b + b (z a) + b 2 (z a) Proof Let z A. Pick r, r 2 with r < r < z a < r 2 < R. Assume without loss of generality that a. Create contours and 2, defined to be circles (both centre ) with radii r and r 2 respectively. I claim that So f(z) [ 2πi f(w) w z dw 2 I( 2 ) I( ) I( 3 ) + I( 4 ) ] f(w) w z dw. f(w) (where in this case, I() is shorthand for 2πi w z dz). Applying Cauchy s integral formula to I( 3 ), we get I( 3 ) 3 f(w) dw 2πif(z). w z Also, by Cauchy s theorem, I( 4 ). This means that f(z) I( 2 ) I( ). Now I( 2 ) where b n f(w) 2πi 2 dw (from Taylor s theorem). w n+ 6 b n z n n

61 2πi f(w) w z dw ( ) 2πi z ) 2πi ( z 2πi ( z ) z m m f(w) w dw z w m f(w) z m dw m ( ) w m f(w)dw 2πi } {{} b m m b m z m. Technical point: The formulae for b n involve r or r 2. But a similar argument to Robin s circle chopping contour argument shows that b n is independent of the radii r, r 2. So the formula f(z) b n z n n is true on the annulus. The Laurent series can be split up as follows. n b n (z a) n b n (z a) n + n n b n (z a) n The first term is a power series in (z a) which converges absolutely for z a < R. The second term is a power series in (z a) which converges absolutely for < R. If R < R then the Laurent series converges absolutely for R < z a < R. z a Example f(z) (z )(z 2) z 2 z is holomorphic everywhere except for z, 2. In particular, f is holomorphic on A {z C < z < 2}. 6

62 and z 2 ( ) z 2 2 ( ) 2 z 2 ( z ) n z (as < ) n z n 2 n+ n Therefore ( z z z z n n m ) z n (as z n+ z < ) z m (m (n + )) This means that z 2 z z n 2 n+ + n has Laurent series on the annuli (z 2)(z ) which are different to the ones we calculated! m {z C < z < } (Taylor series; no terms z n with n < ) {z C 2 < z } (Laurent series only with terms z n with n < ) z m 2.5 Singularities Definition Let U be a domain on which f is holomorphic. Let a / U but suppose D (a, r) {z C < z a < r} U for some r >. We say that f has an isolated singularity at z a. For example, z, sin z z, e/z all have isolated singularities at z. If f is holomorphic on D (a, r), then Laurent s theorem states that f(z) b n(z a) n on D (a, r). 62

63 is convergent for all z a. f(z) f (z) + f 2 (z) b n (z a) n + n b n (z a) n + n n m b n (z a) n b m (z a) m }{{} principal part of a Definition Case (i): If b m for m >, so f 2 (z), then f has a removable singularity at a. Then { f(z) z a f (z) z a is holomorphic on D(a, r) and f (z) n b n(z a) n. Case (ii): f 2 (z) but the series has finitely many terms. b f 2 (z) n N b n (z a) n b N (z a) N b z a where b N. Then f(z) n N b n(z a) n. We say f has a pole of order N at z a (for example, f(z) /z has a pole of order at z ). Poles of order are simple poles Poles of order 2 are double poles, and so on. If f has a pole of order N are a, then (z a) N f(z) has a removable singularity at a, and therefore (z a) N f(z) converges towards a finite limit as z a. This implies that f(z) as z a. Case (iii): f 2 (z) has infinitely many terms. We say f has an essential singularity at a. For example, f(z) e /z n z n n! m z m ( m)! has an essential singularity at z. Essential singularities are complicated! 63

64 Example f(z) sin z z has a removable singularity at z. f(z) z z3 3! + z5 5!... z z2 3! + z4 5!... There are no terms of negative exponent in this series. f has a removable singularity. Theorem (Picard s Great Theorem) Suppose f has an essential singularity at a. If f is defined on D (a, r) then f takes all complex values with at most one exception on D (a, r). At an isolated singularity a, f(z) n b n(z a) n. Here is an example to illustrate this. The complex function that has been plotted is f(z) e /z I R f(z) e /z has an essential singularity at z. We can see that every possible complex value is reached (with all brightnesses and colours) in an arbitrarily large disc about z. This illustrates the fascinating (and rather nasty) behaviour of essential singularities. The principal part of f at a is n n b n (z a) n b n (z a) n 64

65 which converges for all z a. The residue of f at a is Res za f(z) b. If f has a removable singularity at a, then Resf(z). za If f has a simple pole then Resf(z). za 2.6 Calculus of Residues Theorem (Calculus of Residues) Suppose is a closed, simple contour traversed anticlockwise. Let f be a function which is holomorphic on and inside except for a finite number of isolated singularities inside (call them a,..., a k ). Then f(z)dz 2πi k Resf(z). za j This theorem subsumes Cauchy s theorem, Cauchy s integral formula and Cauchy s integral formula for the nth derivative. j Proof At each singularity a j, let g j (z) be its principal part. F (z) f(z) k j g j(z) will have only removable singularities and so can be considered holomorphic. F (z)dz by Cauchy s theorem (as the contour is closed) so f(z)dz k j g j (z)dz. b g j (z) + h j (z) z a j g j (z)dz b + h j (z)dz z a j where h j (z) b n n2 (z a j ) n. But h j (z) d dz n2 Putting it all together, Res f(z) 2πi + h j (z)dz by Cauchy s integral formula za b n so (n )(z a j ) n h j(z)dz by the Fundamental Theorem of Calculus. f(z)dz k j 65 2πi Res za j f(z)

66 as required. Calculating Residues Case (i): f has a simple pole at z a. f(z) (z a)f(z) b lim z a f(z)(z a) b z a + b + b (z a) +... near z a. Case (ii): f has multiplie poles (order N 2). f(z) Method (NOT recommended) b N (z a) N b z a + b +... near z a. (z a) N f(z) b N b (z a) N + b (z a) N +... ( ) d N ( (z a) N f(z) ) (N )!b + N!b (z a) +... dz The residue is then ( ) d N ( b lim (z a) N f(z) ). z a (N )! dz Method 2 (Robin s favourite) w z a or z w + a so f(w + a) b N w N b w + b +... Expand f(w + a) as a Laurent series as far as the w term. Case (iii): Essential singularity Have to compare Laurent series. These are usually rather nasty, and are rare on exercises and problem sheets. 66

67 Example Res zi z 2 + This is where f(z) z 2 has two simple poles at z ±i. + Note that z i z 2 + z i (z + i) (z i) provided z ±i. z + i So Res zi z 2 + lim z i z i z 2 + 2i i 2 In practice we are usually given f(z) g(z) h(z) z a where g, h are holomorphic, g(a) and h has a simple pole at g(z) Res za h(z) lim z a So for the example above, we would have (z a)g(z) h(z) lim z a g(z) lim z a g(a) h (a). z a as h(a) h(z) h(a) so g(z) and h (z) 2i. Res zi z 2 + g(a) h (a) 2i Example Res tan z sin π 2 z π 2 cos ( ) π 2 sin π 2 sin π 2. Example Res zi (z 2 + ) 2 67

68 f(z) (z 2 has double poles at z ±i. + ) 2 f(w + i) (w 2 + 2iw) 2 (2iw) ( 2 iw ) 2 2 ( ) 4w 2 iw w2 4 ( 4 w 2 c + c w + c 2 w ) ) ( iw w2 (c + c w + c 2 w ) 4 The residue is the w coefficient, which is c 4. By comparing constants, we have c. By comparing w coefficients, we have c ic c ic i. So Res z i (z 2 + ) 2 i 4. Example e z + z 2 dz where is the circle centre radius 2. f(z) ez has singularities at z ±i, both of which are inside (and are simple poles). Then + z2 e z ( ) dz 2πi Resf(z) + Res + z2 f(z) zi z i Both residues can be calculated by method of writing f(z) g(z) h(z) point. where h has a simple zero at the singular and similarly Res f(z) zi ei 2i e z ( + z 2 ) ez zi 2z zi 68

69 So e i Resf(z) z i 2i e z ( ) e i dz 2πi + z2 2i + e i 2i π ( e i e i) 2πi sin. Example dz z 2 (e z ) where is the circle centre radius 2i radius 3. Let f(z) z 2 (e z. This function has singularities at z (triple pole) and not forgetting z 2nπi, n Z ) (as e 2nπi ). However, only z is inside. f(z) Comparing z 3 coefficients gives. z 2 z + z2 2! + z3 3! +... z 3 + z z 3 + az 2 + bz ! + z2 3! +... ) z 3 z2 (z ( + + z2! 3! az 2 + bz +... ) Comparing z 2 coefficients gives 2 + a a 2. Comparing z coefficients gives a + b so b 6 2 a Hence dz z 2 (e z ) 2πi 2 πi Integrals of the form 2π F (cos t, sin t)dt (where F is a nice 2-variable function) When is the unit circle centre, with parametrisation (t) e it, this integral becomes 69 ( ) F z+z 2, z z dz 2i iz.

70 Example 2π dt a + cos t where a >. This equals Let f(z) I 2 i a + z + z 2 dz iz dz z 2 where is the unit circle. + 2az + z 2 + 2az +. This has poles at the roots of z2 + 2az +, i.e. at z a ± a 2 R. Clearly because a >, a 2 > so a a 2 < a < so the negative root isn t in the unit circle. The product of the two roots is so a + a 2 unit circle. Hence a a 2 (, ) so the positive root is inside the I 2 i 2πi Res z a+ a 2 f(z) 4π h ( a + ) a 2 where h(z) z 2 + 2az + as f(z) h (z) has a simple pole at z a. h (z) 2z + 2a 2(z + a) h (b) 2(a + b) 2 a 2 So I 2π a 2. Example 2π cos 2t 5 3 cos t dt Now 7

71 So cos 2t ei(2t) + e i(2t) 2 z2 + z 2 where is the unit circle. 2 z 2 +z 2 2 dz I 5 3 z+z zi 2 z 2 + z 2 i 3z 2 + z 3 dz + z 4 i z 2 (3z 2 z + 3) dz f(z) + z 4 z 2 (3z 2 z + 3) + z 4 z 2 (3z )(z 3) This function has a double pole at z and also simple poles at z 3 and z 3. Only z and z 3 are inside the unit circle. Therefore z is a double pole, so write I (Res i 2πi f(z) + Resf(z) z z 3 ) f(z) z2 + z 2 3z 2 z + 3 z az +... z 2 ( + z 4) ( 3 z + 3z 2) ( z 2 Comparing z coefficients gives 3a 3 a 9. So Resf(z) z 3 + z 4 (3z 3 2 z + 3) (6z ) z (2 ) ) 3 + az +... z 3 7

72 and Resf(z) z 9 so ( I 2π 9 4 ) 36 π The Semicircle Method Take a contour of the form as shown below. is the interval [ R, R] and 2 is the semicircular arc of radius R going from R to R. INSERT DIAGRAM Let R. Then f(z)dz R R f(x)dx f(x)dx. We hope that f(z) sufficiently quickly as z to ensure that 2 f(z)dz. If these conditions are satisfied, then f(x)dx lim f(z)dz. R Example I dx + x 2 Let f(z) + z 2. On 2, z R and so z 2 + R 2 (by the alternate triangle inequality, z + w z w z, w C) z 2 + R 2. Then if R >, + z 2 R 2. The arc 2 has length πr, so by the ML-bound, 72

73 f(z)dz 2 πr R 2 (if R > ) as R 2 f(z)dz as R. Therefore we can use the semicircle method (therefore calculus of residues) to evaluate the integral. f(z) is holomorphic except at z ±i (two simple poles). Only one of these lies inside for large R, which is z i. Therefore f(z)dz 2πiRes zi + z 2 2πi d dz ( + z2 ) 2πi 2i π. zi Confirming that this is true, we could use traditional integration methods: dx + x 2 [arctan x] π ( 2 π ) 2 π. Example I dx ( + x 2 ) 2 dx 2 ( + x 2 ) 2 since the integrand is an even function of x and is therefore symmetrical about x. Let f(z) ( + x 2 ) 2. Then f(z) (R 2 ) 2 if z is on 2 from the previous example. The same inequality holds because the integrand and limit are the square of those in the previous example, and squaring both sides of the inequality preserves the sign. Then by the ML-bound again, 73

74 f(z)dz 2 Therefore the condition to use the semicircle method is met. πr (R 2 ) 2 as R. f(z) has the same singularities as in the previous example (z ±i, double poles this time!) and once again z i is inside the contour for large R. From a previous lecture, Res zi ( + z 2 ) 2 4i, so dz ( + z 2 ) 2 2πi 4i dx ( + x 2 ) 2 π 4. Again this can be verified using traditional integration: π 2 dx ( + x 2 ) 2 dx π/2 ( + x 2 ) 2 sec 2 θdθ ( + tan 2 θ) 2 (setting x tan θso dx sec 2 θdθ) π/2 π/2 π/2 sec 2 θdθ sec 4 θ cos 2 θdθ ( + cos 2θ) dθ 2 [θ + 2 ] π/2 sin 2θ 2 2 π 2 π 4. Example I dx (x 2 + a 2 )(x 2 + b 2 with < a < b. ) Let f(z) (z 2 + a 2 )(z 2 + b 2 ) On 2 (the semicircular arc) where z R, we have f(z) (R 2 a 2 )(R 2 b 2 ) 74

75 and using similar reasoning to the previous examples and the ML-bound, we have f(z)dz πr (R 2 a 2 )(R 2 b 2 ) as R. f(z) has 4 singularities, at z ±ai and z ±bi (each simple poles). For large R > b, the contour encloses 2 of these poles (z ai and z bi). Therefore [ f(z)dz 2πi Res ] f(z) zbi f(z) + Res zai [ 2πi d dz ((z2 + a 2 )(z 2 + b 2 + )) zai d dz ((z2 + a 2 )(z 2 + b 2 )) zbi ] and d dz [ (z 2 + a 2 )(z 2 + b 2 ) ] 2z(z 2 + a 2 ) + 2z(z 2 + b 2 ) 2z ( 2z 2 + a 2 + b 2) so the integral becomes [ ] f(z)dz 2πi 2ai ( 2a 2 + a 2 + b 2 ) + 2bi ( 2b 2 + a 2 + b 2 ) [ ] 2πi 2ai (b 2 a 2 ) + 2bi (a 2 b 2 ) [ ] π a (b 2 a 2 ) b (b 2 a 2 ) [ ] b a π ab (b 2 a 2 ) π ab(a + b) Example I cos x + x 2 dx This is a wolf in sheep s clothing! The snag is that cos x +x as R, so we need to use a more strict 2 bound. We could try using g(z) eiz and then apply the semicircle method. + z2 For z in the upper half-plane, e iz since e iz e R(iz) e I(z) if y I(z). On 2 (the semi-circular arc), 75

76 e iz + z 2 z 2 + R 2 (R > ) so by the ML-bound, we have g(z)dz πr R 2 as R. So the condition to apply the semicircle method holds. However, g(x)dx cos x + x 2 dx R g(z) has a simple pole, z i, which lies inside. Therefore e ix + x 2 dx cos x + i sin x + x 2 dx [ ] g(x)dx. [ cos x + x 2 dx R ] g(x)dx ] [ R 2πiRes g(z) zi ] R [2πi e (using an adaptation of a previous example) 2i [ π ] R e π e Slice of pie Method I dx + x To avoid adding 5 residues I use a contour enclosing just one pole. The poles of f(z) hundredth roots of unity, i.e. e i[ π + 2πk ]. are the + z Consider α e πi. Integrate over a slice of pie contour, the boundary of a sector of a circle of centre and radius R, and angle 2π. 76

77 INCLUDE DIAGRAM only contains one pole in its interior. Poles of f are located at α, α 3, α 5,..., α 99 α with only α enclosed in. Fortunately though, z α is a simple pole, so by the calculus of residues, f(z)dz 2πiRes f(z) za 2πi d dz ( + z ) zα 2πi α 99 2πiα α 2πiα (as α ). Clearly f(z)dz f(x)dx as R. On 2 (the arc), z R, so z + R. If R >, Also, length( 2 ) 2πR. Hence f(z) R. f(z)dz 2 2πR R as R 2 f(z)dz as R. Now we need to integrate over 3 (the line segment from to α 2 R). Parametrise this by z α 2 t for t R. Then f(z)dz 3 3 R α 2 R f(z)dz f(α 2 t)α 2 dt α 2 R + α 2 dt t + t dt (as α2 ( α ) 2 ( ) 2 ) α 2 I as R, where Iis the original integral that we wanted to find. 77

78 Therefore as R, f(z)dz ( α 2 )I 2πiα for R > 2πiα I ( α 2 ) 2πi (α α ) 2πi ( ) e iπ e iπ π sin π π cosec π. The slice of pie method is often useful if the integrand depends only on x n where n is a large integer. This cannot be verified using traditional integration! Example I sin x x dx Now since sin x x However, sin x x as x we have no problems with any sort of poles. is not absolutely integrable on [, ), i.e. I as R lim R sin x x sin x x dx dx diverges! So if we interpret then it will emerge that this limit exists. We would like to attack this integral using the semicircle method... Note that 2I? I sin x x dx [ e ix ] x dx although this is somewhat dubious as the integral goes right through the singularity at z. So we have to use a modified version of the semicircle method where the contour omits z somehow. We create a new contour that circles around z with radius ε, and then consider the limit as R and ε. f(z) e iz /z only has the one singularity, at z, which is outside, so 78

79 by Cauchy s theorem. f(z)dz Now let R and ε. Then f(z)dz and f(z)dz 3 R ε ε f(z)dz + f(z)dz 3 e ix x dx e ix x dx R R e ix ε R 2i ε R ε x dx e ix e ix x sin x x 2iI as R and ε. Consider 4 f(z)dz f(z)dz, and the parametrisation z εe it on t [, π]. 4 Now by using Laurent series, f(z) z + i z 2! iz2 3! + z3 4! g(z) where gis holomorphic on C z z + G (z) where Gis holomorphic on C. Hence Therefore dz g(z)dz 4 4 z + G (z)dz 4 dz + G(ε) G( ε) by the Fundamental Theorem of Calculus 4 z dz Now 4 z dz z πi 4 π εie it dt εeit 4 f(z)dz πi + G() G() as ε. 79

80 We need to prove that 2 f(z)dz as R. The ML-bound isn t enough to prove this, as e iz z the large arc, but 2 has length πr so this bound implies f(z)dz π 2 which is not good enough! So we need to work harder and use a different bound. Jordan s Inequality R on for t [, π 2 ]. sin t 2t π So we can use Jordan s inequality on a parametrisation of 2 f(z)dz. Let z Re it with t [, π]. Then 2 f(z)dz i π e ireit π π Re it ireit dt e ireit dt i e R sin t+ir cos t dt π f(z)dz e R sin t dt π/2 π/2 e R sin t dt (as the integrand is symmetrical about the line t π 2 ) e 2R π t dt (by Jordan s inequality and the fact that expis an increasing function, so the inequality is preserved) < 2 e 2R π t dt π R as R. f(z)dz as R. 2 So f(z)dz + f(z)dz + 2 f(z)dz + 3 f(z)dz 4 2πI πi + (letting R and ε ) Therefore 8

81 sin x x dx π 2 which is a very important integral in Fourier analysis Rectangle Method Example (Rectangle Method) I dx cosh x The rectangle method is useful for f(z) periodic with period iλ for λ R. Remember that cosh z is periodic over C but not R. Let f(z). Integrate this over the contour, which traverses the rectangle with vertices ±R and ±R+πi cosh z anticlockwise. is the path from R to R. The poles of cosh z cos iz occur when iz (2n + ) π 2 for n Z, therefore when z i (2n + ) π 2. Inside there is only one simple pole, which is z πi Therefore 2. f(z)dz 2πiRes f(z) z πi 2 2πi d dz (cosh z) 2πi sinh z 2πi i sin π 2 2π z πi 2 z πi 2 Also, note that and R f(z)dz f(z)dz 3 R 3 R R R R dx cosh x f(z)dz dx cosh(x + πi) dx cosh x because cosh(x + πi) cos(ix π) cos ix cosh x. 8

82 Therefore f(z)dz + f(z)dz 2I as R 3 2 dx cosh x. Consider 2 f(z)dz. On 2, z R + it with t [, π]. cosh 2 z cos 2 iz cos 2 (ir t) cos 2 t + sinh 2 R sinh 2 R (as cos 2 t > t). So on 2, cosh z sinh R. Therefore by the ML-bound, with length( 2 ) π, we have f(z)dz π 2 sinh R as R. f(z)dz as R. 2 And by similar reasoning, 4 f(z)dz as R. Letting R in f(z)dz 2π, we get 2 dx cosh x 2π dx cosh x π. Example (Summing a series using contour integration) S n n 2 Let f(z) cot z z 2 82

83 where z has poles where tan z, i.e. at z nπ for n Z. All these poles are simple, because tan (nπ) sec 2 nπ so all these poles are simple. So f(z) has simple poles at z nπ for n Z\{}, but z is a triple pole. Let N N. Let be the contour which is the perimeter of a square with vertices ± ( N + 2) π and ± ( N + 2) πi. This means that encloses all poles Nπ,..., π,, π,..., Nπ. Then where a n Res znπ f(z) When n, nπ is a simple pole, so is a triple pole of f(z): f(z)dz 2πi [ a + ] N (a n + a n ) n a n Res f(z) znπ ( cos z ) Res znπ z 2 sin z cos z d dz (z2 sin z) znπ cos z 2z sin z + z 2 cos z znπ cos nπ 2nπ sin nπ + (nπ) 2 cos nπ ( ) n (nπ) 2 ( ) n n 2 π 2. f(z) z2 2! + z4 ( 4!... ) z 3 z2 3! + z4 5!... z 3 ( + c 2 z 2 + c 4 z ) ( + c 2 z ) ( z2 ) ( z2 3! (c 2 z 2 c 2z 4 ) 3! +... z2 2! +... ) +... z2 3! 2! +... so by equating z 2 coefficients, we obtain 3! + c 2 2! the Laurent series of f(z) and therefore Therefore a Res z f(z) 3. c This is the z coefficient in 83

84 f(z)dz 2πi [ a + ] N (a n + a n ) n [ 2πi N ] n 2 π 2 n [ 4πi π 2 π2 N ] 6 + n 2. n To prove that n n 2 π2 6 Note that we can show that f(z)dz as N. cot 2 (x + iy) cos(x + iy) 2 sin(x + iy) 2 cos2 x + sinh 2 y sin 2 x + sinh 2 y. Suppose that z x + iy lies on. Case (i): z is on a horizontal side of, i.e. y ± ( N + [ ( ) ( ] 2) π (with x N + 2 π, N + 2) π ). Then cot z 2 + ( sinh2 N + ) 2 π sinh 2 ( N + 2) π ( cosh2 N + ) 2 π sinh 2 ( N + 2) π ( coth 2 N + 2 ) π Case (ii): z is on a vertical side of, i.e. x ± ( N + [ ( ) ( ] 2) π (with y N + 2 π, N + 2) π ). Then cos x cos 2 x sin 2 x So cot z 2 sinh2 y + sinh 2 y tanh 2 y as < coth ( N + 2) π. 84

85 On, cot z coth ( N + 2) π, and z ( N + 2) π so has length 4(2N + )π. Therefore by the ML-bound, f(z) coth ( N + ) 2 π ( ). N π 2 Therefore cot z z 2 dz 4(2N + ) coth ( N + ) 2 π ( ) N π 2 8 ( N + ) ( 2 coth N + 2) π ( ) N π 2 8 coth ( N + ) 2 π ( ) N + 2 π 2 as N. cot z dz as N. z2 as required! 4πi π 2 [ π2 N ] 6 + n 2 as N n N n n 2 π Rouché s Theorem Recall that f is holomorphic on a domain U if f is differentiable there. We call f meromorphic on a domain U if f is holomorphic on U except possibly for poles. For example: All rational functions are meromorphic on C. f(z) e /z is not meromorphic on C (because is not a pole of f(z)) but it is on C\{}. Rouché s theorem is about containing zeros inside a simple closed contour. Let f(z) be a meromorphic function and consider f (z)/f(z). If a is in the domain of f (assuming f is not constant) its Laurent series is 85

86 If N, f(z) a n (z a) n (a N ) nn with N ord f(z) za f (z) na n (z a) n nn f (z) f(z) Na N(z a) N + (N + )a N+ (z a) N +... a N (z a) N + a N+ (z a) N ( ) NaN + (N + )a N+ (z a) +... z a a N + a N+ (z a) +... So f (z)/f(z) has a simple pole of residue N at z a. If N, then f(a), and so f (z) f(z) is holomorphic at z a. f (z) f(z) simple poles precisely at the poles and zeros of f. Moreover, f (z) Res za f(z) ordf(z). za is also meromorphic, as it has at most Now we apply the residue theorem. If f is meromorphic on and inside (a simple closed contour) and f has no zeros or poles on, then f (z) dz 2πi (residues of f ) f(z) f in 2πi (orders of fat and poles in ) f (z) 2πi f(z) dz Z P where Z (order of zeros of fin ) and P (orders of poles of fin ). If f is holomorphic inside, then P and f (z) 2πi f(z) dz Z #(zeros of f in countered according to multiplicity) Theorem (Rouché s Theorem). If Let be a simple closed contour. Let f and g be holomorphic on and inside f(z) g(z) < f(z) + g(z) for all z on, then Z f Z g, where Z f total order of zeros of finside and similarly for Z g. (A more usual statement and stronger hypothesis is that f(z) g(z) < f(z) on ) 86

87 Proof The hypothesis implies that f and g are nonzero on. If f(z) then g(z) < + g(z) i.e. g(z) < g(z) is false so f(z) on and similarly for g(z). We claim also that f(z) g(z) If f(z) g(z) t, for t R+ then can t be a negative real number. f(z) tg(z) f(z) g(z) ( + t)g(z) f(z) g(z) ( + t) g(z) and f(z) + g(z) g(z) + t g(z) ( + t) g(z) which contradicts that f(z) g(z) < f(z) + g(z). For z on, f(z) U C\{ t t R, t } (the slit plane). On U, Log is holomorphic. So Logf(z) g(z) g(z) is holomorphic on a domain containing. As is a closed contour, h (z)dz, i.e. h(z) Log f(z) g(z) h (z) f (z) f(z) g (z) g(z). f (z) f(z) dz g (z) g(z) dz Z f Z g as required. Example Prove that all roots of z 5 + 7z + 2 lie in the annulus {z C < z < 2}. We consider the numbers of possible roots inside each disc and then subtract one from the other. Large disc: let be the circle with z 2., and f(z) z 5 + 7z + 2. Then let g(z) z 5 (the term in f with greatest modulus). Then on, 87

88 On, g(z) z 5 z Now since f(z) g(z) 7z z + 2 (by the triangle inequality) 26 (as z 2on ) f(z) g(z) < g(z) f(z) + g(z) on, then by Rouché s theorem, we conclude that the number of solutions of f(z) in the annulus is equal to the number of solutions of g(z) in the disc with z < 2, which is 5 (counted with multiplicity) and z 5 has a quintuple zero at z and no others. Small disc: let be the unit circle, and this time we take g(z) 2, because z 5 < z < 2 when z <. On, we have z and so f(z) g(z) z 5 + 7z z z 8 < g(z). By Rouché s theorem, f and g have the same number of zeros inside, i.e. none! Therefore all zeros of f lie in the annulus {z C < z < 2}. To visualise this, see the diagram below. 3 2 I R 88

89 Example (From 29 exam) How many zeros (counted with multiplicity) has f(z) z 5 5z 4 + z 2 + 9z in the annulus {z C < z < 3}? Large disc: let be the circle centre radius 3. Let g(z) 5z 4 (not z 5, as 5z 4 > z 5 when z 3, as on ). Hence on, we have f(z) g(z) z 5 + z 2 + 9z z 5 + z z + (by the triangle inequality) and g(z) 5 z 4 45 on, so f(z) g(z) < g(z). Rouché s theorem applies. f and g have the same number of zeros inside, i.e. 4 (counted with multiplicity). Small disc: let be the unit circle, i.e. where z. This time, let g(z) 9z (as all the other terms have smaller moduli than 9z when z ). Then f(z) g(z) z 5 5z 4 + z 2 z z 4 + z 2 + (by the triangle inequality) on. Also, g(z) 9 > 8 on. Rouché s theorem strikes again! f(z) has the same number of zeros as g(z) inside, i.e.. Therefore in the annulus {z C < z < 3}, f(z) has 4 3 zeros (counted with multiplicity). You can see this below. 89

90 4 3 2 I R Proof of Fundamental Theorem of Calculus (Using Rouché s Theorem) Rouché s theorem implies the Fundamental Theorem of Calculus! An important thing to note is that Let f(z) z n + a z n a n z + a and g(z) z n, which is term in f(z) of greatest modulus. We consider these functions on the contour, the circle centre radius R. Then Then f(z) g(z) a z n a n z + a a z n a n z + a (by the triangle inequality) f(z) g(z) g(z) a R n + a 2 R n a n R n a R + a 2 R a n R n There is R such that f(z) g(z) g(z) Rouche s theorem applies. < if z is on and R > R. For such R, f(z) g(z) < g(z) holds on so So f and g have the same number of zeros (counted with multiplicity) inside, i.e. n zeros. So f(z) has exactly n zeros on C. 9