Contents. Chapter 1. Complex Numbers 3 1. Basic properties 3 2. The Argand Plane 6 3. Powers of z Loci and Regions in C 13

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1 Contents Chapter. Complex Numbers 3. Basic properties 3. The Argand Plane 6 3. Powers of z 4. Loci and Regions in C 3 Chapter. Complex Functions 7. Continuity of complex functions 7. The Complex Derivative 3. Harmonic Functions 4 Chapter 3. Transcendental Functions 9. The Exponential Function 9. Trigonometric and Hyperbolic Functions 3 3. The Complex Logarithm Complex Exponents 36 Chapter 4. The Complex Integral 39. Complex Path Integration 39. Cauchy s Theorem The Fundamental Theorem of Calculus The Cauchy Integral Formula 5 5. Applications of the Cauchy Formula 54 Chapter 5. Complex Power Series 57. Convergence of Complex Series 57. Uniform Convergence 6 3. Taylor Series 6 4. Laurent series 64 i

2 ii CONTENTS Chapter 6. Residue Theory and Applications 69. Complex Residues 69. Isolated Singularities 7 3. Applying Residues to Real Integration 76

3 CHAPTER Complex Numbers. Basic properties x + solved by x ±i; say i i (cf. x x ± ) Definition.. Complex number z a+ib st a, b R say z C. If w c+id C, then define z + w (a + c) + i(b + d) (z + w w + z) z w (a + ib)(c + id) (ac bd) + i(ad + bc) (Check: zw wz) Write a Re(z) real part of z b Im(z) imaginary part of z Note: if e R, then e + z (e + a) + ib w e c e, d w ie c, d e ) ie + z a + i(b + e) e z ea + ieb ie z eb + iea a + i a + ib ib Powers of i: i i 3 i i i i 4 i i 3 (i i) ( ) i 5 ii 4 i

4 . COMPLEX NUMBERS Example: Solution: Find Im [(3 + 4i)( i)] (3 + 4i)( i) 3. 4.(i i) + 4.i 3.i i(4 6) i Im[(3 + 4i)( )i] Definition.. Given z a + ib C, define the conjugate of z: z a ib z + z (a + ib) + (a ib) (a + a) + i(b b) a + i a Re(z) ( Re( z)) z z (a + ib) (a ib) (a + ib) a + ib (a a) + i(b + b) + ib ib iim(z) iim( z) z z (a + ib)(a ib) (a a b( b)) + i(a ( b) + ba) a + b + i( ab + ab) a + b + i a + b. Division of Complex Nos: z a + ib, w c + id z w z (a + ib)(c id) (ac + bd) + i(bc ad) w w w (c + id)(c id) c + d ( ) ( ) ac + bd bc ad + i c + d c + d

5 Special Case: Conjugation: (iii) (iv) ( z) z z a, b w w (i) (ii) ( ) w Corollary: e R, e z ē z e z. BASIC PROPERTIES 3 z + w z + w z w z w ( ) ( ) c d i c + d c + d } ( ) ( c d + i c + d c + d ( c c + ( d) i c + i( d) w Exercise! ) d c + ( d) ) z w z + ( w) z + ( w) z w (using the above, with e ) ( z w ) z ( ) z w w z w Example: Write [ +i i 3+4i 3 4i] in the form a + ib. Solution: (i) (ii) (iii) + i 3 + 4i i 3 4i ( + i)(3 4i) (3 + 4i)(3 4i) (i)(3 + 4i) (3 4i)(3 + 4i) + i 3 + 4i i 3 4i (6 + 4) + i(3 8) i5 ( ) 5 5 i i i i i 5 ( i 6 5 )

6 4. COMPLEX NUMBERS (iv) Therefore [ + i 3 + 4i i ] 3 4i ( i ) ( i ) 5 5 (8 + i) (8 + i) 5 ( 8 + i(8 + 8) ) i The Argand Plane z x + iy (x, y) R, x Re(z) y Im(z) Definition.. Modulus (or length) of z : z x + y z z Properties: () zw (zw)(zw) (z z)(w w) ( z z ) ( w w ) z w

7 . THE ARGAND PLANE 5 () Example: Solution: z w z ( w z w z w z Compute (3 + 4i)( + i) 3 4i ) ( ) w w w z w z w 3 + 4i (3 + 4i)(3 4i) i + (3 + 4i)( + i) 3 4i 3 4i i + i 3 4i 5 5 Note: z z z z z x + iy (x, y ) w x + iy (x, y ) z + w (x + x, y + y )

8 6. COMPLEX NUMBERS Triangle Inequality (i) z + w z + w (ii) z w z w Proof of (ii) z w (z w)(z w) z z z w w z + w w (w z (z w)) z Re(z w) + w ζ α + iβ z Re(z w) + w Re(ζ) ζ z z w + w ( z w ) ( z w z w ) z w z w Induction: z + z + z 3 z + z + z 3 z + z + z 3 z + + z n z + z + + z n Polar Representation of z z (x, y) (r, θ) r z tan ( y x + y, θ x) if x > tan ( y x) + π if x <

9 . THE ARGAND PLANE 7 Principal Argument Arg(z) θ + kπ ( π, π] (for a suitable integer k). Now x r cos θ y r sin θ ] z r(cos θ + i sin θ) : rcis(θ) z r cos θ + ir sin θ, w r cos θ + ir sin θ z w r r (cos θ cos θ sin θ sin θ ) + ir r (cos θ sin θ + sin θ cos θ ) But cos θ cos θ sin θ sin θ cos(θ + θ ) cos θ sin θ + sin θ cos θ sin(θ + θ ) z w r r (cos(θ + θ ) + i sin(θ + θ )) r r cis(θ + θ ) Example: Write ( + i)( + 3i) in polar form. (Answer : rcis(θ) cis( π ))

10 8. COMPLEX NUMBERS z r cos θ + ir sin θ ( ) r cos θ r sin θ + r cos θ cos θ ( ) sin θ i r r ( i r sin θ r sin θ + r cos θ ) r (cos θ i sin θ) cos( θ) + i sin( θ) r r cis( θ) Example: Express ( + i)( i 3) 3cis(π/8) as rcisθ r r r 3 3 θ π 4 θ tan ( 3) θ 3 π/8 π 3 ( + i)( i 3) 3cis(π/8) r r r 3 cis(θ + θ θ 3 ) 3 cis( 3π 4 )

11 3. POWERS OF z 9 3. Powers of z Summary: z r (cos θ + i sin θ ) w r (cos θ + i sin θ ) z w r r (cos(θ + θ ) + i sin(θ + θ )) z w r (cos(θ θ ) + i sin(θ θ )) r Induction: (r ) z n r n (cos(nθ) + i sin(nθ)) ( ) n w n (cos(nθ) i sin(nθ)) w rn De Moivre: (cos θ + i sin θ) n cos(nθ) + i sin(nθ) Fractional Powers: z r(cos θ + i sin θ) w ρ(cos φ + i sin φ) ρ m (cos(mφ) + i sin(mφ)) r(cos(θ) + i sin(θ)) } Suppose w m z Note w is not unique! ρ m r r m But cos(mφ) cos(θ) sin(mφ) sin(θ) ] φ θ m + πk m k, ±, ±, All distinct values of φ are covered if k,,, m m distinct values of w. ( ( Write (w )z m r m cos θ + ) ( πk m m + i sin θ + )) πk m m k,, m Example: Find ( i) /3 z i r θ π 4

12 . COMPLEX NUMBERS ( i) /3 ( /6 cos ( π + ) ( πk 3 + i sin π + )) πk 3 k,, ( /6 cos ( ( π ) i sin π )) (k ) ( /6 cos ( ) ( 7π + i sin 7π )) (k ) ( /6 cos ( ) ( 5π + i sin 5π )) (k ) What about ( i) /3? w ( i) + i r, θ π w /3 ( /6 cos ( π + πk 4 3 Note ( π ) cos 6 ( π cos ) ( π sin ) ( + i) )) ) + i sin ( π + πk 3 ( π ) ( π ) ( π ) cos sin cos ( i) /3 4/3 ( + 3) i( 3 )) (k ) 4/3 (( 3) i( + 3)) (k ) /3 ( + i) (k )

13 4. LOCI AND REGIONS IN C 4. Loci and Regions in C Example: () Sketch {z C z z > Re(z)} Ω x + y x (x ) + y () z 3 + i < (3) Im(z) Re(z ) z x + ixy y Re(z ) x y y x y y + y x (y + ) x ( ) Im(z) Re(z ) (y + ) x 4

14 . COMPLEX NUMBERS (4) Σ {z C Re(z) 3 } (5) Ω Σ {z C z > Re(z) 3 }

15 4. LOCI AND REGIONS IN C 3 Some important terms: Definition 4.. () A neighbourhood of z C is a set D r (z ) {z z z < r} () A neighbourhood of z is punctured or deleted when we write D r (ẑ ) { < z z < r} for some r >. (3) A set S is open if every z S has a nbd D r (z ) S for some r >. (4) z C is a boundary point (write z S or z Bd(S)) for S if for all r > D r (z ) S, and D r (z ) S (where S denotes the complement of S in C). (5) z C is an interior (resp exterior ) point of S if there exists r > such that D r (z ) S (resp. D r (z ) S). (6) S is connected if any two pts z, z S can be joined by a continuous path inside S. (7) Ω C is a domain if open and connected. (8) Ω C is a simply connected domain if every loop (i.e. every path from z Ω to itself) inside Ω can be shrunk down to z without going outside Ω. (cf. Examples 4,5)

16 4. COMPLEX NUMBERS (9) Ω is multiply connected if some path(s) cannot be shrunk (i.e., if Ω has one or more holes or punctures; cf. Ex.,) () z C is an accumulation point of S if for every r >, D r (z ) S ( every boundary point of S is an accumulation point) () S is a region if S Ω Σ (Σ Ω cf. Ex.,5), and a closed region if S Ω Ω for some domain Ω. () S is bounded if there exists R > st S D R ()

17 CHAPTER Complex Functions. Continuity of complex functions f(z) w C Dom(f) C, Range(f) C z x + i y w u + i v Example: f(z) z + iz } f(z) u(x, y) + i v(x, y) w z + iz x + y + i(x + iy) x + y y + ix u(x, y) x + y y v(x, y) x ] Conversely z x + i y z + z x z x i y z z i y Example: Consider (u, v) : R R with ] x (z + z) y i (z z) u(x, y) xy v(x, y) x + y. Find f : C C such that f(z) u(x, y) + i v(x, y). Solution: xy i 4 (z + z)(z z) i 4 (z z ) 5

18 6. COMPLEX FUNCTIONS { (z + z) i (x + y ) i } 4 (z z) i { z + z + z (z z + z ) } i { z + z + z (z z + z ) } 4 i { z + 6 z z } 4 f(z) u + iv i z + 3i z i (3 z z ) Definition.. Let f : C C be a complex function, and L C. We say lim z z f(z) L if for every ε > there is a δ > such that < z z < δ f(z) L < ε. Example: f(z) sin( z ) z + i Re(z) Show lim z f(z) Proof Note θ < sin(θ) 6 θ sin(θ) θ θ3 3! + θ5 5! θ7 7! + sin(θ) θ < + θ 6 sin z θ 3! + θ4 5! θ6 7! + z < z 6 Now f(z) sin z z sin z z + Re(z) ( inequality) + z ( Re(z) z ) < z 6 + z 7 6 z z z < 6 6 ε f(z) < ε for any ε > choose δ 7 7 ε Example: f(z) z + i Arg(z). Show lim f(z) does not exist z

19 Remark lim f(z) exists if and only if z z Recall Arg(z) ( π, π]. CONTINUITY OF COMPLEX FUNCTIONS 7 lim (x,y) (x,y ) lim arctan(y (x,y) (,) x ) u(x, y) exists and lim (x,y) (x,y ) π if (x, y) 3rd Quadrant (x <, y < ) π if (x, y) nd Quadrant lim Arg(z) does not exist. z Theorem: If lim f(z) L and lim g(z) L, then z z z z (i) lim (f + g)(z) L + L z z (ii) lim f(z)g(z) L L z z (x <, y > ) f(z) (iii) lim z z g(z) L if L. L Definition.. f(z) is continuous at z if all three conditions hold : (a) f(z ) is defined (b) lim z z f(z) exists (c) f(z ) lim z z f(z) Examples v(x, y) exists. () f(z) z + i Arg(z) continuous at z for all z C \ {Im(z), Re(z) } sin z () g(z) + i Re(z). z Note g() not defined, but if we consider g(z) z ĝ(z) z then ĝ cts at

20 8. COMPLEX FUNCTIONS Theorem () If f(z) cts at z and g(z) cts at z, then (a) (f ± g)(z) cts at z (b) f(z)g(z) cts at z (c) f(z)/g (z) cts at z if g(z ) () If f(z) cts at z and g(z) cts at f(z ) w, then the composition g f(z) g(f(z)) is cts at z. (3) f(z) u(x, y) + i v(x, y) ; f(z) cts at z if and only if u(x, y) cts at (x, y ) and v(x, y) cts at (x, y ). (4) If f cts at z for all z R (a region in C), then f(z) also cts on R. If R closed and bounded, then M R such that f(z) M for all z R, and z R st f(z ) M. Example f(z) z + i Arg(z). Find M st f(z) M on R {z C z 3 } f(z) z + (Arg(z)) 3 z 3 + ; π 6 Arg (z) π 6 Claim: M ( + 3). The Complex Derivative Definition.. Given f(z) a complex function, we define the derivative at z C by f f(z + z) f(z ) (z ) lim z z

21 . THE COMPLEX DERIVATIVE 9 Note: z x + i y Special Case: y y y f(z) u(x, y) + iv(x, y). f (z ) lim x lim x lim x x [u(x + x, y ) + i v(x + x, y ) u(x, y ) i v(x, y )] x [u(x + x, y ) u(x, y ) + i(v(x + x, y ) v(x, y ))] x [u(x + x, y ) u(x, y )] + i lim x x [v(x + x, y ) v(x, y )] u x (x, y ) + i v x (x, y ) Similarly: (x x, x ) f (z ) lim y i y [u(x, y + y) u(x, y ) + i(v(x, y + y) v(x, y ))] i u y + v y (x, y ) Conclusion: if f (z ) exists, then u (x x, y ) v (x } y, y ) u (x y, y ) v (x Cauchy-Riemann Equations x, y ) Remark: If u x v y, u y v x at z C, and u x, u y, v x, v y, u, v are all cts on D r(z ) for some r >, then f (z ) exists. Example: () f(z) z + i Arg(z) u(x, y) x + y v(x, y) arctan( y x )

22 . COMPLEX FUNCTIONS u x x x + y u y y x + y v x v y u x v y x + y v x u y x + y y x + y x x + y f (z ) exists for all z {z C z, Arg(z) π} but nowhere else. () f(z) c (constant C) f (z ) for allz C (exercise) f(z) z n f (z ) lim z z [(z + z) n z n ] lim z [ z n + z ( ) ( ) ] n n z n z + + z ( z) n + ( z) n z n n nz n f (z ) exists for all z C. Properties of the Derivative: Suppose f (z ) and g (z ) exist. Theorem () (f ± g) (z ) f (z ) ± g (z ) () (f g) (z ) f (z )g(z ) + f(z )g (z ) Product Rule (3) ( f g ) (z ) f (z )g(z ) f(z )g (z ) (g(z )) (g(z ) ) Quotient Rule

23 (4) (fog) (z ) f (g(z ))g (z ) Chain Rule.. THE COMPLEX DERIVATIVE Pf: -4 Same as for functions of a real variable. Definition.. f(z) is analytic at z C if f (z) exists for all z D r (z ), some r >. Example: () f(z) z + iarg(z) f (z ) exists for z, Arg(z ) π only nowhere analytic. () f(z) z n (n ).f (z ) nz n for all z C. Example: analytic for all z C (Such f are said to be entire functions.) if f(z) is analytic for all z Ω C, we denote the derivative function df dz, such that df dz (z ) f (z ) for all z Ω. () f(z) c constant df dz () f(z) z n df dz nzn (3) f(z) a + a z + + a n z n + a n z n df dz a + a z + + (n )a n z n + na n z n (4) f(z) z n df dz nz n n z n+

24 . COMPLEX FUNCTIONS Remark (5) Product, Quotient, Chain Rules for d dz (cf. previous Theorem) () If f(z) analytic on Ω, then f(z) continuous on Ω (Proof easy). () If f(z) analytic on Ω, then df dz analytic on Ω (Proof discussed later). L Hôpital s Rule: If f, g analytic at z C, f(z ) g(z ), then ( ) f(z) df lim z z g(z) lim z z dz /dg f (z ) dz g (z ) Pf: f(z) lim z z g(z) lim z z f(z) f(z ) z z g(z) g(z ) z z Now df dz, dg dz continuous near z lim Example: lim z i z 7 + z + i z 4 lim z z 3i 4 f(z + z) f(z ) lim z z z z df lim z dg dz g (z ) Theorem: If f, g analytic at z C, then g(z + z) g(z ) z dz f (z ) () f ± g analytic at z () fg, f/g (g(z ) ), f g all analytic at z. Example: f(z) z 3 + z g(z) z + f, g are entire functions f g (z) z3 + z z + ie z ± i. f (z ) g (z ) is analytic for all z C st z +, Definition.3. If f(z) not analytic at z C, but for all r > there exists ξ D r (ẑ ) such that f analytic at ξ then we say z is a singularity of f. (Ex. f(z) z3 +z z + has singularities at z ± i.) 3. Harmonic Functions Consider f(z) u(x, y) + i v(x, y) analytic. Then u x v y ; u y v x

25 3. HARMONIC FUNCTIONS 3 Since df dz f (z) also analytic, it follows that u & v are twice differentiable, hence (Theorem in R : If on Ω R.) u x v x y & u y v y x. ( ) v x y, v y x, v x, v y all continuous fns on Ω R, then v x y v y x ( ) u + u. x y Write +, then f analytic on Ω u, v satisfy Laplace s eqn : x y u, v, on Ω. Definition 3.. ϕ(x, y) : R R is harmonic on Ω R if ϕ on Ω. When we say that ϕ is harmonic, it will also be assumed that all the second partial derivatives of ϕ are continuous on Ω. (Hence f u + i v analytic u, v harmonic) Example: ϕ(x, y) x 3 3y x ϕ. ϕ x 6x ϕ y 6x Theorem: If ϕ(x, y) harmonic on a simply connected domain Ω R, there exists an analytic fn f(z) on Ω C such that ϕ(x, y) Re(f). Also there exists analytic g(z) on Ω C such that ϕ(x, y) Im(g). Proof: We have to use a Lemma which we ll prove later: Lemma: If Ω simply connected, and F (z) analytic on Ω, then there exists f(z) analytic on Ω such that ( Idea: f(z) z z df dz F (z). F (ξ)dξ, well defined if all paths from z to z inside Ω are continuously equivalent, i.e.,ω is simply connected). Now consider F (z) ϕ x i ϕ : µ + iν y

26 4. COMPLEX FUNCTIONS F analytic since the partial derivatives satisfy µ x ν y & µ y ν x and are all continuous, there exists such that f(z) ϑ(x, y) + i λ(x, y) analytic ϕ x i ϕ y df dz ϑ x + i λ x ϑ x i ϑ y ϕ x ϑ x, ϕ y ϑ y ϕ(x, y) ϑ(x, y) + c ie ϕ Re(f) c. Alternatively: Consider G(z) ϕ y + i ϕ x ν + iµ ( G analytic, since ν x µ ) y, ν y µ x have g(z) u(x, y) + i v(x, y) analytic such that ϕ y + i ϕ x dg dz u x + i v x v y + i v x ϕ(x, y) v(x, y) + c Im(g) + c Example: ϕ(x, y) x 3 3y x. Find f(z)st ϕ Re(f) ± c f(z) ϑ(x, y) + i λ(x, y) st ϑ x λ y ϕ x 3x 3y ϑ y λ x ϕ y 6xy

27 3. HARMONIC FUNCTIONS 5 ϑ ϕ λ 3(x y )dy 6 xy dx 3x y y 3 + C(x) 3x y + D(y) C(x) c, D(y) y 3 + c λ(x, y) 3x y y 3 + c f(z) x 3 3y x + i(3x y y 3 ) + ic ie., f(z) (x + iy) 3 z 3 + ic Definition 3.. Given ϕ st ϕ, say ψ is the harmonic conjugate of ϕ if ϕ(x, y) + iψ(x, y) : f(z) analytic Example: ϕ x 3 3y x ψ 3x y 3 (+c). Remark: If ϕ harmonic Re(f) for some analytic f(z), then ϕ Im(if). Example: ϕ(x, y) x x + y, ϕ x x(x 3y ) ϕ (x + y ) 3 y ϕ on R \{(, )} F ϕ x i ϕ y y x (x + y ) + i xy (x i y) (z) (x + y ) (x + y ) (zz) z d dz ( z ) ϕ(x, y) Re( z ) Im( i z ) ψ(x, y) y x + y Im( z ) Re( i z )

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29 CHAPTER 3 Transcendental Functions. The Exponential Function Definition.. e z e x+iy e x (cos y + i sin y) e iy : cos y + i sin y for all y R Properties: (i) e x e Re(z) e z, y Im(z) Arg(e z ) ± πk k,,, (ii) e, e iπ (iii) u(x, y) e x cos(y), v(x, y) e x sin(y) u x ex cos(y) v y u e x sin(y) v y x e z analytic on C (i.e. e z entire). d (Exercise: show dz [ez ] e z (iv) e z+w e Re(z+w)+iIm(z+w) and e Re(z+w) (cos(im(z + w)) + i sin(im(z + w))) d dz [ef(z) ] f (z)e f(z).) e z e w e Re(z) e Re(w) (cos(im(z))) + i sin(im(z)))(cos(im(w)) + i sin Im(w)) e Re(z+w) {cos(im(z)) cos(im(w)) sin(im(z)) sin(im(w)) +i(sin(im(w)) cos(im(z)) + cos(im(w)) sin(im(x)))} e Re(z+w) (cos(im(z) + Im(w)) + i sin(im(z) + Im(w))) e z+w ( (e z ) m e mz ) (v) e z w e z /e w (Proof similar) 7

30 8 3. TRANSCENDENTAL FUNCTIONS Special Case: z e w e w x : (e iy ) m e imy cos(my) + i sin(my) (cf. De Moivre) Example: (a) f( + i) (b) f ( + i) f(z) e (z z )π. Find (c) {z C f(z) } (a) z + i i z + 3i z z (b) (c) ( ( f( + i) e ( + 3i )π e π π ) ( )) 3π cos + i sin ie π/ ( f (z) e (z z )π d z ) π dz z ( πe π(z z ) + ) π ( + z ) f(z) z ) f ( + i) πie ( π/ + ( + i) ( ) πie π/ + ( i) 4 πie π/ (4 i) 4 f(z) e (z z )π e Re(z z )π e π(x e π x x x +y x x +y ) ( ) x x (x + y ) or x. x +y

31 . TRIGONOMETRIC AND HYPERBOLIC FUNCTIONS 9. Trigonometric and Hyperbolic Functions Recall e iy cos y + i sin(y) e iy cos( y) + i sin( y) cos(y) i sin(y) cos(y) ( e iy + e iy) sin(y) ( e iy e iy) i ( e iy e iy) i Definition.. cos(z) ( e iz + e iz) sin(z) i ( e iz e iz) Properties: () e iz, e iz entire sin(z) and cos(z) entire. () d dz sin(z) d [ ( e iz e iz)] dz i ( ie iz + ie iz) ( e iz + e iz) i cos(z) d cos(z) dz sin(z) (Exercise) (3) sin (z) + cos (z) (Exercise) (4) sin(z ± w) [ e i(z±w) e i(z±w)] i [ e iz e ±iw e iz e ±iw] i [( sin(z) cos(w)±cos(z) sin(w) 4i e iz e iz) ( e iw + e iw) ± ( e iz + e iz) ( e iw e iw)] [ 4i e i(z+w) + e i(z w) e i(z w) e i(z+w) ± e i(z+w) ± e i(z w) ± e i(z w) ± e i(z+w)] [ i e i(z+w) e i(z+w)] [ or i e i(z w) e i(z w)] sin(z ± w). (5) Similarly, cos(z ± w) cos(z) cos(w) ± sin(z) sin(w).

32 3 3. TRANSCENDENTAL FUNCTIONS Example: Where does fail to be analytic? cos(iz) Note: f(z) cos(iz) is entire analytic if f(z). f(z) But cos(iz) [ e z + e z] [ e x (cos y i sin y) + e x (cos y + i sin y) ] cos(y) ( e x + e x) + i sin(y) ( e x e x) cos(y) ; e x e x, i.e. y π ± πk ; x Now define tan(z) sin(z) cos(z), sec(z) cos(z), etc Recall Hyperbolic sine and cosine: sinh(x) ex e x ; cosh(x) ex + e x Now define sinh(z) ez e z ; cosh(z) ez +e z Properties: () sinh(z), cosh(z) entire. () d sinh(z) cosh(z); d cosh(z) sinh(z) dz dz (3) sinh(iz) eiz e iz i sin(z) cosh(iz) eiz + e iz cos(z) (4) cosh (z) sinh (z) (cosh(z) + sinh(z))(cosh(z) sinh(z)) e z e z Example: Note: Now define tanh(z) sinh(z) cosh(z) ; sech(z) cosh(z) ; etc. Give the domain of C in which sech(z) is analytic. cosh(z) is entire sech(z) fails to be analytic when cosh(z), i.e., e z + e z e x (cos y + i sin y) + e x (cos y i sin y)

33 i.e., cos y(e x + e x ) and sin y(e x e x ) 3. THE COMPLEX LOGARITHM 3 i.e., y π ± πk and x. sech(z) analytic on Ω C \ { + i ( π ± πk) k,,, } 3. The Complex Logarithm Recall e z e x (cos(y) + i sin(y)) e z e x Arg(e z ) y Im(z) ± πk e log z z e Re(log(z)) z Re(log(z)) ln z Arg(e log(z) ) Arg(z) Im(log(z)) ± πk log(z) ln z + iarg(z) ± πki ; k,,, Definition 3.. Log(z) ln z + iarg(z) Example: z ( + 3i); z ; Arg(z ) π 3 Log(z ) ln() + πi πi 3 3 Log(z) ln() + iarg(z) πi 3 } Log(z ) Log(z ) πi In general, log(z z ) ln z z + iarg (z z ) ± πki ln z + ln z + iarg(z ) + iarg(z ) ± πki log(z z ) log(z ) + log(z ) ± πki log(z n ) n log(z) ± πki. Example: Solve (e z ) e z Now e z e z + e z iff e z i.e. z ln ( ) ± πki

34 3 3. TRANSCENDENTAL FUNCTIONS Note log(e z ) ln e z + iarg (e z ) ± πki log e Re(z) + iim(z) ± πki Re(z) + iim(z) ± πki z ± πki Log(e z ) z if Im(z) ( π, π] Consider Log(z) ln z + iarg(z) u + iv u x (x + y ) / (x + y ) x x / x + y u y (x + y ) / (x + y ) y y / x + y arctan( y ) x x π v Arg(z) x, y > π x, y < cts on C\{y, x } v x + ( ) y y x y x + y ; v y + ( y u x v y x and u y v x on C\{y, x } x ) x x x + y Now d dz Log(z) u log(z) x x + i v x x x + y i y x + y z z z z d (cf. log x ) Log(z) analytic on Ω C\{y ; x } dx x Definition 3.. : The analytic function f(z) Log(z) on the domain Ω C\{Im(z) ; Re(z) } is called the Principal Branch of log(z).

35 Remark: 3. THE COMPLEX LOGARITHM 33 f(z) log z + iarg(z) + πik, for any fixed k ±, ±, defines a secondary (or non Principal) branch of the complex logarithm on Ω. (Note: Ω simply connected). The half-line {Im(z) ; Re(z) } ( is ) called a branch cut ( ) for log(z). Example: Consider φ(x, y) arctan where π < arctan π Find (a) The harmonic conjugate ψ(x, y) (b) The domain Ω on which φ + iψ is analytic. (a) F (x, y) φ x i φ y x y ; φ x ( + φ y F (x, y) y + ix x + y i z z z ( + x y x y i z df dz ) y x y y x + y ) x y x x + y x (y + ix i z) f(z) i log(z) Arg(z) + C + i log z ( ψ log(x + y ) ( ) ) (b) But φ(x, y) Re(f) arctan arg(i z) π arg(z) x y where π < arctan ( ) x π π < π y arg(z) π π arg(z) < 3π Define f(z) π arg(z) + i log z arg(z) + i log z where π arg(z) < 3π f φ + iψ has an analytic branch on Ω C\{Re(z) ; Im(z) }

36 34 3. TRANSCENDENTAL FUNCTIONS Example: Find the maximum domain Ω C on which f(z) Log ( ) z z is analytic. Note: Log ( ) ( z z log z z + iarg z ) z Clearly z,, Arg ( ) z z ±π arg(z ) arg(z) ± π But arg(z ) arg(z) ± π z (, ) Log ( ) z z analytic on C\[, ] Ω 4. Complex Exponents Recall x R x c e c log(x) for any c R Define z z c e c log(z) for any c C Example: () z i, c 3 c log(z) (log i + iarg(i) ± πik) 3 πi 6 ± πki 3 k,, πi 6 ; 5πi 6 ; 9πi 6 3πi πi

37 () 4. COMPLEX EXPONENTS 35 e π 6 i 3 + i (k ) i 3 e 5π 6 i 3 + i (k ) e π i i (k ) z + i, c i c log(z) i(log + i + iarg( + i) ± πik) i(log + i π 4 ± πki) π 4 ± πk + i log ( + i) i ±8k e ( 4 )π e i log ±8k e ( 4 )π ( cos(log ) + i sin(log Now f(z) z c e clog(z) f(z) is analytic on the same domain Ω C\{Im(z) ; Re(z) } as Log(z) (note e z is entire) we may define the Principal Branch of z c as e clog(z) on Ω. Moreover d dz zc d dz eclog(z) e clog(z) d dz (clog(z)) c z eclog(z) c z zc cz c ) Example: analytic? Using the principal branch of f(z) z cosh(z), find f ( ) πi. Where is f(z) d dz zcosh(z) cosh(z)z cosh(z) d dz cosh(z) cosh(z) z cosh(z) + sinh(z)z cosh(z) Log(z) ( ) πi Now cosh ( ) πi sinh ( ) πi f eπi/ + e πi/ eπi/ e πi/ i ( ) πi + ilog ( ) πi

38 36 3. TRANSCENDENTAL FUNCTIONS Note cosh(z) is entire z cosh(z) e cosh(z) log(z) is analytic on Ω {Im(z) ; Re(z) }. (Exercise: Do the same for f(z) z tanh(z) )

39 CHAPTER 4 The Complex Integral. Complex Path Integration Definition.. : A smooth path between two points z and z in C corresponds to a map γ : [, ] C, γ(t) x(t) + iy(t), such that (i) γ() z ; γ() z (ii) x(t) and y(t) are differentiable functions for all t (, ) (iii) For each z γ([, ]), there is at most one t [, ] such that γ(t) z. Remark: γ may also be represented in implicit form as a graph y f(x) or x g(y) in R. Examples: () z 3, z + i, γ(t) (3 + t) + it ; implicit form: y x 3 between 3 and + i () z i, z, γ(t) e i( t) π ; implicit form: y x between i and. (3) z, z + i, γ(t) t + it ; implicit form: y x between and + i 37

40 38 4. THE COMPLEX INTEGRAL Definition.. : (i) Complex line element z x + i y Infinitesimal line element dz dx + idy Hence if γ(t) x(t) + iy(t) smooth path in C, can define path element dz dx dy dt + i dt. Now dt dt (ii) for any complex function f(z) u(x, y) + iv(x, y), define the Path integral z f(z)dz : z z x x (u(x, y) + iv(x, y))(dx + idy) udx y y (ux vy )dt + i { x vdy + i vdx + x (vx + uy )dt y y } udy Example: z i, z i, find i i. dz along the unit semicircular arc lying in Re(z) z Method (): f u + iv (x iy) x y ixy (x y ) + ixy (x y ) + 4x y x y (x + y ) + i xy (x + y )

41 . COMPLEX PATH INTEGRATION 39 i i f(z)dz x + y, Re(z) x x y i i xy (x + y ) dy x y (x + y ) dy f(z)dx 3 i udx xy (x + y ) dy i { vdy + i y y dy ( y )dy 3 Method (): x(t) cos(( t) π) ; y(t) sin(( t) π) ( ux (t) vy (t) [cos ( t) π ) ( sin ( t) π )] [ ( + cos ( t) π ) ( sin ( t) π ( π sin )] π cos vdx + x y (x + y ) dy } udy µdµ (µ y ) ) ( t) π ( ( t) π ) ux dt vy dt µ dµ (µ cos(( t) π )) µ dµ (µ cos(( t) π )) Exercise: Show vx + uy dt 3 Define ux vy dt dz dx + idy x + iy dt (x ) + (y ) dt Given γ : [, ] C a smooth path st γ() z, γ() z z z dz (x ) + (y ) dt L(γ) path length

42 4 4. THE COMPLEX INTEGRAL Now consider f(z) st f γ : [, ] C, and f γ(t) M γ, i.e., M γ is an upper bound for f on the image of γ, then z f(z)dz z z z f(z) dz f γ(t) (x ) + (y ) dt M γ (x ) + (y ) dt M γ L(γ) Example: Consider I + i without evaluating I precisely. i e i Log( z) dz along y x. Show that I (.)e π/ y x x t y t (x ) + (y ) + t π/4 π/4 L(γ) π/4 + t dt sec 3 θdθ t tan(θ) + t sec(θ) dt sec (θ)dθ π/4 sec θdθ ln(sec θ + tan θ)] π 4 ln( + ) tan θ sec θdθ tan θ sec θ] π/4 π/4 sec θ + tan θ sec θdθ sec 3 θdθ

43 . CAUCHY S THEOREM 4 π/4 sec 3 θ d θ (ln( + ) + ) L(γ) Now f(z) e i Log( z) f(z) e i log z Arg( z) e Arg( z) e Arg(z) e π/ M γ L(γ) e π/ { ln( + ) + }.5e π/. Cauchy s Theorem Definition.. A simple closed contour is a loop γ (γ() γ() z ) such that γ([, ]) forms the boundary of a simply connected bounded domain Ω C

44 4 4. THE COMPLEX INTEGRAL We say γ has positive orientation with respect to Ω if interior of Ω always left of γ(t) as t moves from to. For positively oriented γ, we define the contour integral f(z) dz, and recall γ Green s theorem: If P (x, y), Q(x, y), P x, P y, Q x, Q are all cts fns on Ω Ω, where y Ω γ is a positively oriented simple closed contour, then ( Q P dx + Q dy x P ) dx dy. y γ Now f(z) u + iv analytic u x v y ; u y v x γ f(z) dz and we have proved γ, Ω u dx v dy + i Ω ( v x + u ) y γ v dx + u dy dx dy + i Ω ( u x v ) y dx dy Cauchy s theorem: Let C be a closed simple contour, and let f(z) be a complex function which is analytic on some domain containing C. Then C Example: f(z) dz. () f(z) Log(z), analytic on Ω C \ {Im(z) ; Re(z) }

45 C { z C z i. CAUCHY S THEOREM 43 } C Log(z) dz () f(z) tanh(z) sinh(z) cosh(z) analytic on Ω C \ {z i ( π ± πk), k,,, } tanh(z) dz z i (3) f(z) z analytic on C \ {} cannot apply Cauchy theorem to z z(t) e iπt z dz z. But z dz z πie iπt e πit dt πi dt πi Exercise: Show z dz for all integer n. zn

46 44 4. THE COMPLEX INTEGRAL Deformation of Contours: Let C, C be closed simple contours such that C j Ω j, j,, C Ω Divide Ω \ Ω into two simply connected domains with boundary Γ, Γ as follows Γ Now consider f(z) analytic on Ω \ Ω f(z) dz f(z) dz by Cauchy s theorem. Note Γ Γ Γ C ( C ) f(z) dz Γ Γ f(z) dz C ( C ) f(z) dz C C f(z) dz C f(z) dz C f(z) dz. More generally, consider C, C as follows:

47 . CAUCHY S THEOREM 45 Then C 3 such that C f(z) dz C f(z) dz C f(z) dz C 3 C 3 C f(z) dz f(z) dz f(z) dz Theorem (deformation of contours) If C, C are closed simple contours such that C can be continuously deformed into C without passing through a singularity of some analytic function f(z), then C f(z) dz f(z) dz. Example: f(z) z ; C { z } ; C { z i } ; C 3 { z } C

48 46 4. THE COMPLEX INTEGRAL C f(z) dz πi ; C f dz C 3 f dz 3. The Fundamental Theorem of Calculus Principal of Path Independence: If γ, γ : [, ] C are two paths such that γ j () z, γ j () z, j, and the region Ω bounded by γ γ is simply connected, then any function f(z), analytic on a domain containing Ω, satisfies f(z) dz γ γ f(z) dz, z ie z f(z) dz does not depend on γ st γ() z, γ() z (Proof: f(z) dz γ γ γ f(z) dz γ f(z) dz.) Now suppose f(z) df on some simply connected domain Ω C, then for any dz γ : [, ] Ω stγ() z, γ() z Ω we have the Fundamental Theorem of Calculus: Example: z z f(z) dz z z d F (z) dz dz df dz dz dt dt d(f γ) dt dt F (γ()) F (γ()) F (z ) F (z ).

49 3. THE FUNDAMENTAL THEOREM OF CALCULUS 47 () f(z) z d Log(z) on C \ {Im(z), Re(z) } dz +i i z dz Log( + i) Log( i) log + i π 4 ( log i 3π 4 iπ ) () f(z) Solution: (a) (a) +i i (b) z+3 (z )(z + 3). Compute f(z) dz f(z) dz (z )(z + 3) A (z ) + B (z + 3) (A + B)z + 3A B, ie +i i f(z) dz 4 +i i A + B 3A B ] z dz 4 4 Log(z ) ] +i i A 4 B 4 +i i z + 3 dz 4 Log(z + 3) ] +i {log + i + i Arg( + i) log + i i Arg ( + i) 4 log 5 + i i Arg(5 + i) + log 3 + i + i Arg (3 + i)} { ( ) ( )} 5 ( + i)(3 + i) log + i Arg 4 9 ( + i)(5 + i) { ( ) ( )} 5 4 log + i Arg i5 58 ( ) 5 4 log + i 9 4 arctan( 5 4 ) i

50 48 4. THE COMPLEX INTEGRAL (b) Now z+3 z+3 f(z) dz 4 z w z + 3 z+3 z dz 4 z+3 z + 3 dz dz by Cauchy s theorem, while z+3 z + 3 dz w w dw πi z+3 dz (z )(z + 3) πi 4 πi 4. The Cauchy Integral Formula Theorem: Let f(z) be analytic on a domain Ω C. Let C be a closed simple contour in Ω, and z o Ω lying in the interior of C, then f(z ) f(z)dz πi z z. Proof: of contours For r sufficiently small, let z z r lie in the interior of C, then deformation C C f(z)dz f(z)dz, z z z z r z z since f(z) z z analytic in the region between z z r and C. dz dz Similarly πi. z z z z z z r

51 4. THE CAUCHY INTEGRAL FORMULA 49 Now consider z z r f(z) dz πif(z ) z z z z r f(z) f(z ) dz. z z Note that f(z) cts at z ε >, δ > such that z z < δ f(z) f(z ) < ε. choose r < δ so that f(z) f(z ) f(z) f(z z z sup ) z z r < ε. r r z z r Hence M γ L(γ) inequality, cf. p. 35 z z r f(z) f(z ) dz M dz z z z z r < ε πr πε r Now C f(z)dz πif(z ) z z z z r z z r f(z) f(z )dz z z ( ) f(z) f(z ) dz < πε. ( ) z z But (*) does not depend on r >, for any ε >, choose r < δ, and conclude ( ) f(z) C z z dz πif(z ). Example: () z+3 g(z) z + 3z z + z 3 z + 3z g(z)dz (z )(z + 3) dz 4 z+3 z πif( 3) where f(z) z + 3z πi ( ) πi. z + 3z dz z 4 z+3 z + 3z dz z + 3

52 5 4. THE COMPLEX INTEGRAL () z 4 g(z) z + 3z z + z 3 g(z)dz g(z)dz Γ Γ g(z)dz + g(z)dz Γ Γ g(z)dz + g(z)dz C C z + 3z dz z + 3z dz 4 C z 4 C z + 3 πi {f() f( 3)} πi. f(w) w z dw Rewrite Cauchy Formula: f(z) πi C differentiating under sign, e.g., when df Moreover df dz πi C d n n! f(z) dzn πi f(w) d dz C ( w z f(w) dw (w z) n+ cts on C, dz ) dw πi C f(w) (w z) dw

53 Example: () z 4. THE CAUCHY INTEGRAL FORMULA 5 z + (z ) 4 (z + ) g(z)dz z (z ) πi 4 3! g () g(z) z + z + ; g (z) z + z + z + (z + ) z + 5z + (z + ) g (z) z + 5 (z + ) (z ) z + 8 (z + ) 3 (z + ) 3 g 3(z 8) (z) + (z + ) 3 (z + ) z 6 4 (z + ) 4 () z i sinh(z)dz (z i) 7. z { Note f(z) sinh(z) dn dz f(z) n And sinh(z) d6 (z i) 7 dw 6 ( ) sinh(z) (z w) wi g(z)dz (z ) πi πi 3 5 sinh(z) if n even cosh(z) if n odd (3) C z i sinh(z) πi dz (z i) 7 6! d 6 (sinh(w)) dw6 wi πi πi sinh(i) 6! 6! (ei e i ) e z dz, where C is the rectangle with corners z ± i, ± i. (z ) (z ) (z ) (z + ) let C C C, where C has corners ± i, ±i C has corners ± i, ±i

54 5 4. THE COMPLEX INTEGRAL On C, let f(z) (z + ) (z ) dz f(z) d dz πi C (z ) dw f(w) w ( ) e w πi (w + ) ew (w + ) w 3 ( e πi 4 e ) 8 C e z e z C On C, let f(z) e z dz (z ) C e z (z ) f(z) (z + ) C dz d πi dw f(w) ( πi e z dz (z ) iπe + w ) e (w ) ew πi (w ) w 3 ( e 4 + e ) 8 Remark: Cauchy s formula implies that if f(z) is analytic, then dn dz n f(z) exists for all n. 5. Applications of the Cauchy Formula Let f(z) be analytic on a simply connected domain Ω C. Consider z Ω, and a circle z z r contained in Ω. Then f(z ) f(z) dz πi z z r z z πi π f(z + re πit ) re πit f(z + re πit )dt π f(z + re iθ )dθ πire πit dt f(z ) average value of f(z) on the circle z z r. (Gauss Mean Value Theorem) Can use this to prove:

55 5. APPLICATIONS OF THE CAUCHY FORMULA 53 Maximum Modulus Principle: If f is analytic on a closed, bounded region R C, and f is not constant then Also have M f sup f(z) f(ζ) for some ζ R, z R f(ζ) > f(z) for all z R\ R. Liouville s Theorem: If f(z) is entire, and f(z) M for some finite M R and for all z C, then f(z) must be constant. Proof: and Consider C : z z r, some fixed z C, then f (z ) f(z) πi C (z z ) dz f (z ) f(z) π C (z z ) dz π M r πr where M r sup f(re πit ) (re πit ) sup f(re πit ) t [,] r t [,] Now, we assumed that f(z) M for all z C sup f(re πit ) M M r M t [,] r for any z C, we have f (z ) M r r M r independently of r > f (z ). Corollary (Fundamental Theorem of Algebra): The polynomial equation P (z) a + a z + + a n z n, a n, always has n solutions z C. Proof: Suppose P (z) has no solutions for some non constant polynomialp. Then P (z) for all z C. Now P entire and nowhere zero f(z) P (z) Moreover, P (z) constant P (z) constant. P (z) z n a n + + a z + a n z n lim P (z) lim z z z n a n lim z P (z) f(z) M also entire.

56

57 CHAPTER 5 Complex Power Series. Convergence of Complex Series A complex series is a summation S(z) a n z n where a n C for all n. We n denote by S N (z) the N-th Partial sum, i.e., S N (z) sequence {S N (z)} N of partial sums. N a n z n, and hence define the Definition.. () Given z C, we say S(z ) converges at z if σ lim S N(z ) N exists, i.e., for all ε >, there exists N such that σ S N (z ) < ε when N > N. () The Domain of Convergence of S(z) is the set Ω C such that z Ω S(z ) converges. (3) Conversely, S(z ) diverges if lim N S N(z ) does not exist. Example: Find Domain of convergence of S(z) Note z N ( z)( + z + + z N ) S N (z) zn z Guess Now z re iθ z N r N e inθ z n. n lim S z N+ N(z) lim N N z ( But σ(z) S N (z) z N+ z z n lim N zn if r z <. if z <. z ( σ(z)) ) z N+ z < ε z N+ < ε z (N + ) log z < log(ε z ) N > 55 log(ε z ) log z

58 56 5. COMPLEX POWER SERIES noting log z < if z <. lim S N(z) for all z st z <. N z Now, by analogy with real series there is the Divergence Test: If lim a n z n, then a n z n diverges. n Example: Note z lim z n lim z n lim r n n n n S(z) z n diverges if z. n n Definition.. S(z ) a n z n converges absolutely if the real series n a n z n converges. S(z ) converges conditionally otherwise. n Recall: () Absolute convergence ordinary convergence. () If S(z ) absolutely convergent, then order of summation not important. (3) S (z ) and S (z ) abs convergent S S (z ) abs.cgt. where S (z) S S (z) c n a n z n, S (z) c n z n n a j b n j j where b n z n Example: z ( z) z n for z < n ( ) z n ( k m+nk z n z m ) (k + )z k k

59 . CONVERGENCE OF COMPLEX SERIES 57 Change of Variable: If S(w) a n w n converges for all w Ω and f : C C n is an analytic function such that w f(z), then S(f(z)) z f (Ω) {z C f(z) Ω}. Example: S(w) w n converges for w < : n a n f(z) n converges for all n w e iz S(e iz ) e inz converges for e iz <, where e iz e Re(iz) e Im(z). n e Im(z) < Im(z) < Im(z) > Conversely, Im(z) lim e iz lim e Im(z) n n e inz diverges for Im(z). n Domain of Convergence of e inz ( ): eiz Ratio Test: Consider ρ(z) lim Recall n a n+ z n+ a n z n lim n a n+ a n z (i) If ρ(z ) < then S(z ) converges absolutely at z C. (ii) If ρ(z ) > then S(z ) diverges at z.

60 58 5. COMPLEX POWER SERIES (iii) If ρ(z ) or ρ(z ) does not exist, then don t know. Example: S(z) n z n n! ρ(z) lim n! n (n + )! z lim n n + for all z C. Now consider n z n n! S ( ) ( ) ρ z z lim n n + z for all z z n n! converges for all z. Definition.. S(z). Uniform Convergence a n z n converges uniformly on R C if for all ε > there exists N such that S(z ) S N (z ) < ε for all N N independently of z R. Recall Weierstrass M-test : Consider S(z) a n z n. If there exists a sequence {M n } n of positive real numbers M n such that a n z n M n for all z R C, and uniformly on R. M n <, then S(z) converges Example: S(z) z n converges absolutely on z <, e.g., S ( + i) converges n absolutely since ( ) n + i < <. ( ) n Now z z n z n Mn. In fact, z n converges uniformly on R { z }. z n converges uniformly on { z r} for r <, but not on { z < }.

61 Theorem: Suppose S(w) analytic, w f(z), such that R f(ω). Then () If γ is a smooth path in f (R) Ω S(f(z))dz d () dz S(f(z)) d a n dz (f(z)n ) (3) S(f(z)) Example. UNIFORM CONVERGENCE 59 a n w n converges uniformly on R C, and f : Ω C is γ γ a n f(z) n dz n df na n f(z) dz on f (R). a n f(z) n is analytic on f (R) Ω. In particular, S(z) analytic on R (cf. () plus Morera s theorem ). () S(w) w n converges uniformly on w r < F (z) S(e iz ) e inz analytic on Im(z) log r > Domain of analyticity of F (z) {z C Im(z) > } () w w n on w < F (z) on {Im(z) > }. e iz Hence if γ {Im(z) > } st γ() z, γ() z γ dζ z e iζ z z i S(x) e inζ dζ i n einz i x n n d ne n dx S(x) z z e inζ dζ ] z [ i n einζ + (z z ) z i ( e inz e ) inz + (z z ) n x n x if z <

62 6 5. COMPLEX POWER SERIES x ) ) z i e inz n S(x) ( z i i dζ e iz i eiζ e dt ln t ( x S ( e dζ i [ ln(e )] (z i) eiζ dw w( w) ) ( ln ) ln(e ) e { e iz } dw e iz i w + dw w e e i {Log(e iz ) + Log( e iz ) + log( e } ) e inz n i ([ + iz Log( e iz ) ] ( i) z + i ) i Log( e iz ) on {Im(z) > } { < Re(z) < π} Theorem : If S(z) 3. Taylor Series a n z n converges at z, then S(z) converges uniformly on the set of all z such that z r, where r < z. Hence S(z) is analytic on { z < z }. Proof: S(z ) converges lim a n z n C > such that a n z n C for all n. n Now consider z C st z r < z, write a n z n a n z n ( z z ) n C ( ) n r z for all n. ( ) n r ( ) ( ) n r Let M n C, then M n C C z z r. z Since r <. Hence Weierstrass z a n z n converges uniformly on { z r}. Change of Variable: If S(z a) a n (z a) n converges at z a C then S(z a) converges uniformly on { z a r}, where r < z a. Hence S(z a) analytic on { z a < z a }. Definition 3.. : The largest r such that a n(z a) n converges for all z { z a < r}, given a C, is called the radius of convergence of S(z a). The set

63 3. TAYLOR SERIES 6 { z a < r max }, where r max is the radius of convergence, is called the disc of convergence of S(z a). Theorem : Suppose f(z) is analytic on a disc of radius r about a C then there exists a power series S(z a) a n (z a) n which converges to f(z) for all z D r (a). Moreover, a n f (n) (a) n! for each n. Proof: First consider a C, and z C such that z < r. Now choose r such that z < r < r, and define C { z r } in C. Cauchy s formula f(z ) πi πi C C f(z) dz z z f(z)dz z( z z ). Note z r z < z ( z ) z f(z ) πi πi n ( z ) n z f(z) C z n ( z n ( z ) n dz z ) f(z)dz C z n+ since f analytic and ( z ) n converges uniformly on z < r z r < r. z

64 6 5. COMPLEX POWER SERIES f(z ) πi c n z n for all z D r (), where c n f(z)dz C z n+ πi n! f (n) (), f(z) f (n) () z n on D r (). n! Change of variable: z w a, f(w) f(w a) f(z) f(w) f (n) (a) f(w a) (w a) n. n! f (n) (a) Remark: If r is the largest radius for which (z a) n converges to f(z) on n! D r (a), then f(z) must have a singularity at some pt z such that z a r. Example: f(z) z f (n) n! (z) ( z) f (n) () n+ n! f (n) () Hence z n z n converges to f(z) on D (), with singularity at z. n! 4. Laurent series Recall the Riemann sphere S C { } (cf., e.g., Churchill and Brown, p.38). Define a complex coordinate near infinity to be the set of all complex numbers w C such that w, where z C is the usual complex coordinate near C. Hence z w corresponds to z on S. Now consider D r () {z C z < r } and D r ( ) {w C w < r } such that r > r. Then D r () D r ( ) {z C r < z < r }.

65 4. LAURENT SERIES 63 Suppose S (z) a n z n converges on D r () and S (w) D r ( ). Then on D r () D r ( ) we can write ( ) S (w) S z b n w n converges on b n z n. ( ) Write L(z) S (z) + S z c k z k a k if k > where c k b k if k < a + b if k. Note L(z) is convergent on the annulus D r () D r ( ) D r () D () r Laurent s theorem: Suppose f(z) is analytic on an annular region A {r < z < r }, then for all z A there exists a power series c k z k L(z) such that L(z ) converges to f(z ). Moreover c k πi Proof: C f(z)dz z k+ for all k, where C { z s} for some s (r, r ).

66 64 5. COMPLEX POWER SERIES Cauchy formula Now f(z ) πi z ρ πi f(z)dz z ρ z z z ρ Hence f(z ) deformation of contours γ f(z) z z dz z ρ f(z) z z dz f(z) dz z z πi f(z) z ρ z n z n f(z)dz z ( z z ) (m (n + )) ( c k z, k where c k z ρ z ρ z ρ ( f(z)dz z ρ z n+ f(z) z ρ z n f(z) z z dz ) ( z ) n dz z ), while ( ( ) ) n z dz z ( ) z (n+) z n f(z)dz z ρ m z m f(z)dz z ρ z m+ f(z)dz πi z ρ if k z k+ f(z) πi z ρ dz if k z k+ z s, s (r, r ).

67 4. LAURENT SERIES 65 Moreover c k z k converges uniformly inside A. Change of Variable: f(z) analytic on A {r < z a < r } f(z ) such that c k πi C f(z)dz (z a) k+. c k (z a) k

68

69 CHAPTER 6 Residue Theory and Applications. Complex Residues Definition.. : Let f(z) be analytic on the punctured disc D r (z )\{z } {z < z z < r}. The Residue of f at z is given by Res(f, z ) f(z)dz πi z z r for any < r < r. Consider L f (z, z ) c k (z z ) k, c k f(z)dz πi z z r (z z ). k+ Then But f(z)dz c k (z z ) k dz πi z z r πi c k (z z ) πi z z k dz. r { (z z ) k πi if k dz z z r k f(z)dz c πi z z r Theorem: Res(f, z ) c. 67

70 68 6. RESIDUE THEORY AND APPLICATIONS Example: f(z) z + i z + i i z i. Res(f, i)? Note z + i i z + i 4 L f (z, i) 4 (z i) + i i( i (z i)) ( ) k i (z i) ( i ) k (z i) k i (z i) c i Res(f, i) Now consider domain Ω C, such that f(z) analytic on Ω\{z,, z k }, z i Ω. Ω f(z)dz C C f(z)dz + Cauchy Theorem: k i Residue Theorem: z z i ε f(z)dz C C f(z)dz If Ω is a closed simple contour, and f(z) is analytic on Ω\{z,, z k } for z i Ω, then f(z)dz πi Ω k Res(f, z i ) i

71 Example: Compute z 3 dz z. COMPLEX RESIDUES 69 using residues z 3 ( ) ( ) dz z Res z, + Res z, Now z z z + ( ) ( ) Res z, Res z, 4πi ( ) ( ) Res z, Res z +, 4πi dz z. z 3 z ε dz z z+ ε dz z + Example: f(z) sin(z). Compute Res(f, ). z sinh(z) e z + z + z! + z3 3! + e iz + iz + (iz) + (iz)3 +! 3! sin(z) eiz e iz {iz + (iz)3 i i 3! + (iz)5 5! } + z { z3 3! + z5 5! z z sinh(z) ez e z } {z + z3 3! + z5 + 5! } {z + z3 3! + z5 5! + z sin(z) z sin h(z) z + z4 3! 5! z{ + z + z4 + } 3! 5! z { + } z 3! c Res(f, ). c k z k 3! + z4 5! } } { + z 3! + z4 5! +

72 7 6. RESIDUE THEORY AND APPLICATIONS. Isolated Singularities Definition.. : Given L f (z, z ) c k (z z ) k, define the Principal Part [L f (z, z )] c k (z z ) k. () f(z) has a pole of order N at z C if [L f (z, z )] c k (z z ) k, i.e. c k k < N. N Example: f(z) has poles of order 3 at z ± (z ) 3 f(z) sin(z) z sinh(z) has pole of order at z. () f(z) has an isolated essential singularity at z C if [L f (z, z ))] has infinitely many terms. Example: e z + z +!z + 3!z 3 + c k k! k > (3) f(z) has a Removable singularity at z C if f(z ) may be redefined to make f analytic at z. Example: f(z) sin(z) z + z4 z 3! 5! - not defined at z, but analytic if we define { sin(z) z z f(z) z. Remarks: () f(z) has a pole or order N at z lim z z (z z ) N f(z) < and lim f(z) z z

73 . ISOLATED SINGULARITIES 7 since L f (z, z ) c N (z z ) N + + c (z z ) + c + c (z z ) + ( c N + c (z z ) N N+ (z z ) + c (z z ) N + ) cf. Graph of f(z) : () f(z) has a removable singularity at z lim z z (z z )f(z). (3) f(z) has essential singularity at z lim z z f(z) does not exist ( ). (cf. Cassorati-Weierstrass thm) Example: () f(z) (z ) lim(z 3 z )3 f(z) lim z (z + ) 3 8 L f (z, ) c k (z ) k c 3 8. () f(z) sin(z) z 3 lim z zf(z) lim z sin(z) Definition.. : Order of f(z) at z, N if f(z) (z z ) N a n (z z ) n ord f (z ) N if f(z) (z z ) N if z essential. if f(z ) < a ( zero ) a n (z z ) n a ( pole )

74 7 6. RESIDUE THEORY AND APPLICATIONS Suppose ord f (z ) k, ord g (z ) l, then [ ] [ ] f g (z z ) k a n (z z ) n (z z ) l b m (z z ) m (z z ) k+l c p (z z ) p where c a b. c a b + a b.. ord f g (z ) k + l Moreover g Define ϕ(z z ) Now (z z ) l [b + b (z z ) + ] (b ) b m (z z ) m b g (z z ) l b ( ϕ) ϕ ϕ n if ϕ < (i.e., z z ) < ε) g (z z ) l b [ + ϕ + ϕ + ] ord (z ) l, g ord f/g (z ) k l. Example: Find ord h () for h(z) z + 3z sin 3 (z). f(z) 3z + z ord f (). sin(z) z z3 3! + z5 5! z( z 3! + z4 5! ) g(z) ( ) 3 sin 3 (z) z 3 z 3! + z4 5! ord g () 3 ord h () 3

75 In general f(z) c N (z z ) N + + c (z z ) + c +. ISOLATED SINGULARITIES 73 (z z ) N f(z) c N + c N+ (z z ) + + c (z z ) N + c (z z ) N + d N (N )! dz [(z z ) N f(z)] c N + c (z z ) + { Res(f, z ) lim z z (N )! Example: Compute Now z z d N [ (z z ) N f(z) ] dz N f(z)dz for f(z) z sin(z). ( ) dz z sin(z) πires z sin(z), } ord f () f(z) c z + c z + c + c z + d [ Res(f, ) lim z f(z) ] z! dz [ ] d z lim z dz sin(z) { lim z sin(z) z } sin (z) cos(z) { } sin(z) z cos(z) lim z sin (z) { } cos(z) cos(z) + z sin(z) (L Hôpital) lim z sin(z) cos(z) { } z lim z cos(z) Example: z sin ( z ) dz z ( ( sin z Res z ), ) πi

76 74 6. RESIDUE THEORY AND APPLICATIONS ( ) sin z ( ) z sin z z z 3 3! + z 5 5! z z 4 3! + z 6 5! L sin( (z, ) z z) Res ( z sin ( z ), ) c. 3. Applying Residues to Real Integration. Trigonometric Integrals: Consider integrals of the form: π P (sin θ, cos θ) dθ ; P, Q polynomials. Q(sin θ, cos θ) Example: π Recall sinθ z z i z z i if z cos θ + i sin θ e iθ since z z when z. Similarly cos θ z + z. Now z e iθ dz ie iθ dθ izdθ dθ iz dz P (sin θ, cos θ)dθ Q (a R) π z dθ a + sin θ P Q ( i (z z ), ) dz (z + z ) iz z z dz iz ( a + ( )) z z i dz iaz + z Note z + iaz (z + ia) + a (z + ia) (i a ) (z + i(a + a ))(z + i(a a ))

77 π dθ a + sin θ 3. APPLYING RESIDUES TO REAL INTEGRATION 75 z dz (z + i(a + a ))(z + i(a a )). Now a + a (a a ) a + a a a a + a a a But a + a a assume w.l.o.g. a > a + a > } π dθ a + sin θ Res(f(z), (a a )i) πi (f(z) : {[z + i(a + a )][z + i(a a )]} ) Let g(z) z + i(a + g(z), then f(z) a ) Cauchy formula z + i(a a ). Improper Integrals: πi Res(f(z), (a a )i) z g(z)dz z + i(a a ) πig( i(a a )) 4πi i a π a Example: x x 4 + dx. Note z4 + when z e i( π 4 + πk 4 ) (k,,, 3) Write x x 4 + dx lim a a x dx b x lim b x dx x 4 + Consider contour C : [, b] [, ib] γ such that γ {be iθ ϑ π } and b >.

78 76 6. RESIDUE THEORY AND APPLICATIONS Note C ( ) z dz z z 4 + πires z 4 +, eiπ/4 [ ] πi lim (z e iπ/4) z ) z e iπ/4 z 4 + (simple pole: N ) [ ] z lim z e iπ/4 (πi) lim z e iπ/4 4z 3 (L Hôpital s Rule) [ ] iπ z iπe iπ/4 Now z [, b] z tb (t [, ]) z [, ib] z tbi (t [, ]) z be iθ (dz izdθ) z dz C z 4 + b b t dt b 4 t 4 + ib Recall M max z θ π/ z 4 + z dz i γ z 4 + Moreover z b z z 4 + z z 4 πb 3 lim ML lim b b (b 4 ) iπ π ( i) 4 ( + i) b t dt b 4 t i γ ML M πb b b 4 z dz z 4 +. π b x dx ( + i) ( + i) lim b x 4 + (x bt)

79 3. APPLYING RESIDUES TO REAL INTEGRATION 77 Finally x dx x 4 + π + i + i π. x x 4 + even x x 4 + dx π x x 4 + dx x x 4 + dx