Chpter 1 The Complex Plne 1.1. The Complex Numbers A complex number is n expression of the form z = x + iy = x + yi, where x, y re rel numbers nd i is symbol stisfying i 2 = ii = i i = 1. Here, x is clled the rel prt of z nd y the imginry prt of z nd we denote x = Rez, y = Imz. We identify two complex numbers z nd w if nd only if Rez = Rew nd Imz = Imw. We lso write x + 0i = x, 0 + yi = yi. In this wy, rel numbers re exctly those complex numbers whose imginry prt is zero. The modulus, or bsolute vlue, of z is defined by z = x 2 + y 2 if z = x + iy. The complex conjugte of z = x + iy is defined by Obviously, z = x iy. Rez = 1 (z + z) = Re z, 2 Imz = 1 (z z) = Im z 2i nd, for z = x + iy, x z, y z ; z = z. The ddition, subtrction, multipliction nd division of complex numbers re defined s follows: for z = x + iy nd w = s + it, () z + w = (x + s) + i(y + t); (b) zw = (xs yt) + i(xt + ys); z w = (x s) + i(y t); 1
2 1. The Complex Plne (c) if s 2 + t 2 0 then z w = wz ww = (xs + yt) + i(ys xt) s 2 + t 2. These opertions will follow the sme ordinry rules of rithmetic of rel numbers. Under these opertions, the set of ll complex numbers becomes field, with 0 = 0 + 0i nd 1 = 1 + 0i. Note tht for ll complex numbers z, w z z = z 2 ; z + w = z + w; zw = z w; zw = z w. Complex Numbers s Vectors in the Complex Plne. A complex number z = x+iy cn be identified s point P (x, y) in the xy-plne, nd thus cn be viewed s vector OP in the plne. All the rules for the geometry of the vectors cn be recst in terms of complex numbers. For exmple, let w = s + it be nother complex number. Then the point for z + w becomes the vector sum of P (x, y) nd Q(s, t), nd z w is exctly the distnce between P (x, y) nd Q(s, t). Henceforth, then, we refer to the xy-plne s the complex plne, nd the x-xis s the rel xis, y-xis the imginry xis. (See the figure below.) y z = x + iy z O x Figure 1.1. Complex numbers s vectors Polr Representtion. The identifiction of z = x + iy with the point P (x, y) lso gives the polr representtion of z: for z 0, z = r cos θ + ir sin θ = z (cos θ + i sin θ), where θ is ny ngle verifying this equlity. (See the figure below.) Given z 0, if θ is such ngle then θ + 2πk lso verifies this representtion for ll integers k. The set of ll such θ s is clled the rgument of z nd is denoted by rgz. A concrete choice of rgz cn be defined by requiring the ngle θ to be in the intervl [ π, π); we define this vlue to the principl rgument of z nd denote it s θ = Argz [ π, π). This is well-defined for ll z 0 s the unique ngle θ [ π, π) such tht z = z (cos θ + i sin θ).
1.1. The Complex Numbers 3 z = z (cos θ + i sin θ) y z θ O x Figure 1.2. Polr representtion The use of polr representtion of complex numbers gives simple nd esy wy for multipliction nd division nd powers. Exmple 1. 1 + i = ( 2 cos 3π 4 + i sin 3π 4 ) so Arg( 1 + i) = 3π 4. Multipliction nd division through polr representtion. Let z = z (cos θ + i sin θ), w = w (cos ψ + i sin ψ). Then, by the definition of multipliction nd conjugtion, using the trigonometric identities for the sine nd cosine of the sum nd difference of two ngles, we hve zw = z w (cos(θ + ψ) + i sin(θ + ψ)), z = z (cos θ i sin θ), nd, if w 0, z w = z (cos(θ ψ) + i sin(θ ψ)). w Theorem 1.1 (De Moivre s Theorem). Let z = z (cos θ + i sin θ) 0. Then, for n = 0, ±1, ±2,, z n = z n (cos nθ + i sin nθ). Proof. Use induction. Exmple 2. (1) Find the polr representtion of z = 1 + i nd w = 3 + i. (2) Find in polr representtion ( 1 + i)( 3 + i) nd 1 + i. 3 + i (3) Find in polr representtion ( 3+i) n for ll integers n. In prticulr, find ( 3+i) 6. Solution. (1) As bove, 1 + i = 2(cos 3π 4 + i sin 3π 4 ). The polr representtion of w is w = 3 + i = 2(cos π 6 + i sin π 6 ). (2) In polr representtion, ( 1 + i)( 3 + i) = 2 2 [cos( 3π4 + π6 ) + i sin(3π4 + π6 ] )
4 1. The Complex Plne nd = 2 [ 2 cos 11π ] 11π + i sin 12 12 1 + i 2 = [cos( 3π4 3 + i 2 π6 ) + i sin(3π4 π6 ] [ 2 ) = cos 7π ] 7π + i sin. 2 12 12 (3) In polr representtion, ( 3 + i) n = 2 n [ cos nπ 6 + i sin nπ 6 ]. In prticulr, find ( 3 + i) 6 = 2 6 (cos π + i sin π) = 2 6. Exercises. Problems 1, 2, 3, 5, 6, 11. 1.2. Some Geometry The Tringle Inequlity. From the vector representtion of complex numbers, we cn esily hve the following tringle inequlity: z + w z + w. This inequlity cn be proved directly s follows. Given z nd w, Hence From this inequlity, one lso obtins z + w 2 = (z + w)( z + w) = z z + z w + w z + w w which is lso clled tringle inequlity. = z 2 + w 2 + 2Re(z w) z 2 + w 2 + 2 z w = z 2 + w 2 + 2 z w = ( z + w ) 2. z + w z + w. ζ ξ ζ ξ ζ, ξ, Locus of Points. The locus of points is the set of points tht stisfy generl eqution F (z) = 0. It is sometime esier to use the xy-coordintes by setting z = x+iy nd to study the equtions defined by F (x + iy) = 0. Stright Lines nd Circles. The eqution of stright line cn be written s z p = z q, where p nd q re two distinct complex numbers. This line is the bisecting line of the line segment joining p nd q. This is the geometric wy for the line eqution. Also, the lgebric eqution for stright line is Re(z + b) = 0, where nd b re two complex numbers nd 0. Note tht, b re not unique nd we cn tke b to be rel. In the xy-coordintes, the line hs n eqution of the form Ax + By = C, where A, B, C re rel constnts nd A 2 + B 2 0.
1.2. Some Geometry 5 A circle is the set (locus) of points equidistnt from given point (center); the distnce is clled the rdius of the circle. The eqution for circle of rdius r nd center z 0 is z z 0 = r. A useful chrcteriztion of circles nd lines. A circle is lso locus of points stisfying the eqution (1.1) z p = ρ z q, where p, q re distinct complex numbers nd ρ 1 is positive rel number. To see this, suppose 0 < ρ < 1. Let z = w + q nd c = p q; then the eqution (1.1) becomes w c = ρ w. Upon squring nd trnsposing terms, this cn be written s w 2 (1 ρ 2 ) 2Re(w c) + c 2 = 0. Dividing by 1 ρ 2, completing the squre of the left side, nd tking the squre root will yield tht w c 1 ρ 2 = c ρ 1 ρ 2. Therefore (1.1) is equivlent to z q p q ρ 1 ρ 2 = p q 1 ρ 2. This is the eqution of the circle centered t the point z 0 = p ρ2 q 1 ρ 2 of rdius R = ρ p q 1 ρ 2. Note tht if we llow ρ = 1 in (1.1) we hve the eqution of lines s well. Therefore (1.1) with ρ > 0 represents the eqution for circles or lines. Exmple 3. The locus of points z with z i = 1 2 z 1. Solution. Here p = i, q = 1 nd ρ = 1 2 ; so we know the locus is circle of center z 0 = p ρ2 q = 1 1 ρ 2 3 + 4 ρ p q 3i nd rdius R = = 2 2 1 ρ 2 3. To confirm this, we multiply the eqution by 2 nd then squre both sides to obtin 4[ z 2 2Re(zī) + i 2 ] = z 2 2Rez + 1, which simplifies to nd to tht 3 z 2 8y + 2x = 3 3x 2 + 2x + 3y 2 8y = 3. This is the eqution of the circle (x + 1 3 )2 + (y 4 3 )2 = 8 9 of center z 0 = 1 3 + 4 3 i nd rdius 2 2 3.
6 1. The Complex Plne Roots of Complex Numbers. Let n = 1, 2, nd z 0. Any number w stisfying w n = z is clled n nth root of z. We shll see there exist exctly n distinct nth roots of z if n 2; the nottion z 1/n is usully to denote the set of ll nth roots of z, not prticulr one. To find ll nth roots of z 0, write z in polr representtion: z = z (cos θ + i sin θ). Let w = w (cos ψ + i sin ψ) be n nth root of z; tht is, w n = z. By De Moivre s Theorem, w n (cos nψ + i sin nψ) = z (cos θ + i sin θ). Hence w n = z nd nψ = θ + 2πk for some integer k = 0, ±1, ±2,. Therefore w = n z ; ψ = ψ k = θ n + 2πk n. Here, for positive number t > 0 we use s = n t to denote the unique positive number stisfying s n = t. Note tht for ny integer k the vlue of ψ k differs from one of the n vlues {ψ 0, ψ 1,, ψ n 1 } by n integer multiple of 2π. Therefore, there exist exctly n distinct vlues of w for the nth roots of z given by where ψ k is defined bove. w k = n z (cos ψ k + i sin ψ k ), k = 0, 1,, n 1, Exmple 4. (1) Find ll 12th roots of 1. (2) Find ll 5th roots of 1 + i. (3) Solve the eqution z 4 4z 2 + 4 2i = 0. Solution. (1) Since 1 = 1(cos 0 + i sin 0), it follows tht ll 12th roots of 1 re given by cos( 2πk 2πk + i sin 12 12 ) = cos(πk 6 which re 12 different numbers. πk + i sin ), k = 0, 1,, 11, 6 (2) Since 1 + i = 2(cos π 4 + i sin π 4 ), it follows tht ll 5th roots of 1 + i re given by [ 2 1/10 cos( π 20 + 2πk 5 ) + i sin( π 20 + 2πk ] 5 ), k = 0, 1, 2, 3, 4, which re five points on the circle of rdius 2 1/10 tht strt t the ngle π 20 nd hve equl distnces on the circle. (3) By completing the squre, z 4 4z 2 + 4 2i = (z 2 2) 2 2i nd so the eqution is equivlent to (z 2 2) 2 = 2i = (1 + i) 2, which is equivlent to tht is, z 2 2 = 1 + i or z 2 2 = 1 i, z 2 = 3 + i or z 2 = 1 i. Hence the solutions re the squre roots of 3 + i nd 1 i. They re in the form of z = ±z 1, z = ±z 2, where z1 2 = 3 + i nd z2 2 = 1 i cn be found in polr representtion. Exercises. Pge 20. Problems 1, 2, 3, 7, 11, 12, 15, 21, 23, 24, 25.
1.3. Subsets of the Plne 7 1.3. Subsets of the Plne The definitions re the sme s for the subsets of the Eucliden spce R 2, such s interior points, boundry points, open sets, closed sets, closure, connected sets, etc. An open disc in the complex plne is the set of complex numbers defined by A hlf open plne is the set defined by {z : z z 0 < r}. {z : Re(z + b) > 0}. Given two distinct points p nd q, the directed line segment pq with strting point p nd ending point q is the set defined by pq = {(1 t)p + tq : 0 t 1}. Essentilly, we need to know the following definitions nd fcts: (1) We sy tht z 0 is n interior point of set D if there is number r > 0 such tht the open disc z : z z 0 < r is contined in D. We sy set D is open if every point of D is n interior point of D. (2) We sy tht z 0 is boundry point of set D if every open disc centered t z 0 contins both points in D nd points not in D. The set of ll boundry points of D is clled the boundry of D, usully denoted by D. (3) For set D in the complex plin, the union D D is clled the closure of D, usully denoted by D. Set D is clled closed if D = D; tht is, if D D. (4) Theorem: A set D is open if nd only if it contins no boundry points. A set C is closed if nd only if its complement D = {z : z / C} is open. (5) A polygonl curve is the union of finite number of directed line segments P 1 P 2, P 2 P 3,, P n 1 P n, where P 1 is clled the strting point nd P n is clled the ending point. (6) An open set D is clled connected if for ech pir of points p, q in D, there exists polygonl curve lying entirely in D with strting point p nd ending point q. (This definition works only for open sets; for generl sets, the connectedness is defined differently!) (7) A domin in the complex plne is n open nd connected set. (8) A set S is clled convex if for ech pir of points p, q in S the line segment pq lies lso in S. (9) set D is sid to contin the point t infinity in its interior if there exists number M > 0 such tht { z > M} D. Exmple 5. (1) Ech open disc D = {z : z z 0 < r} is open nd connected. The boundry of disc D is the circle {z : z z 0 = r}. (2) The set R = {z : Rez > 0} is open. The set {z : Imz 1} is closed, so is the set {z : Rez 6}. (3) The boundry of the set {z = x + iy : x 2 < y} is the prbol {z = x + iy : y = x 2 }. (4) The open set {z : Rez > 0} is connected, so is the open set {z : 0 < z z 0 < r}. (5) The set {z : Rez 0} is open but not connected.
8 1. The Complex Plne (6) The open set {z : Rez > 0} is convex, but the open set {z : 0 < z z 0 < r} is not convex. (7) The set {z : Rez > 0} does not contin the point t infinity in its interior, for, given ny M > 0, there re points z with z > M but Rez 0. But the set D = {z : z + 1 + z 1 > 1} does contin the point t infinity in its interior becuse, for ll z with z > 1.5, it follows from the tringle inequlity tht z +1 + z 1 ( z 1) + ( z 1) = 2 z 2 > 1 nd hence these z re contined in the set D. Exercises. Pge 28. Problems 1 8, 10, 11. 1.4. Functions nd Limits A function of the complex vrible z, written s w = f(z), is rule tht ssigns complex number w to ech complex number z in given subset D of the complex plne. The set D is clled the domin of definition of the function. The collection of ll possible vlues w of the function is clled the rnge of the function. A function w = f(z) is clled one-to-one on set D if from f(z 1 ) = f(z 2 ) with z 1, z 2 D it must follow z 1 = z 2. A function w = f(z) is clled onto set R if R is subset of the rnge of this function. Exmple 6. Show the rnge of the function w = T (z) = (1 + z)/(1 z) on the disc z < 1 is the set of those w whose rel prt is positive. Proof. Let D = {z : z < 1} nd S = {w : Rew > 0}. We show T : D S is onto (surjective). (1) Given ny z D, let w = T (z) = 1+z 1 z. Compute Rew = Re 1 + z 1 z since z < 1. Hence the rnge of T is inside S. = Re(1 + z)(1 z) 1 z 2 = 1 z 2 1 z 2 > 0 (2) Given ny w S, we wnt to show tht there is z D such tht w = T (z). Solve for z from T (z) = w nd we hve z = w 1 w+1 (this shows tht T is one-to-one). We need to show tht this z belongs to D; tht is, z < 1. This is the sme s w 1 < w + 1 or w 1 2 < w + 1 2. We expnd w 1 2 = (w 1)( w 1) nd w + 1 2 = (w + 1)( w + 1) to obtin w 1 2 = w 2 + 1 2Rew; w + 1 2 = w 2 + 1 + 2Rew. Since Rew > 0, it follows esily tht w 1 2 < w + 1 2 ; this proves z < 1, hence w = T (z) is in the rnge of T. Limits, Continuity nd Convergence. The concepts of the limit of sequence of complex numbers nd the limit nd continuity of complex vrible function nd the convergence of n infinite series of complex numbers re ll identicl to those for rel vrible. All the rules bout the limit nd the convergence for rel vrible theory re lso true for the complex vrible theory. We mention some of these below s review of such mterils lerned in the clculus. A sequence {z n } is list of complex numbers, usully strting with n = 1, 2, 3,. We sy {z n } converges to complex number A nd write lim z n = A or simply z n A, n
1.4. Functions nd Limits 9 if, given ny positive number ɛ, there exists n integer N such tht z n A < ɛ n N. Fct. If z n = x n + iy n nd A = s + it, then z n A if nd only if x n s nd y n t. Fct. If z n A then z n A. Theorem. Let z n A nd w n B. Then, for ny constnts λ, µ, λz n + µw n λa + µb, z n w n AB, z n w n A B if B 0. Suppose next tht f is function defined on subset S of the complex plne. Let z 0 be point either in S or in the boundry of S. We sy tht f hs limit L t the point z 0 nd we write lim z z 0 f(z) = L or f(z) L s z z 0 if, given ny ɛ > 0, there is δ > 0 such tht f(z) L < ɛ whenever z S nd 0 < z z 0 < δ. We sy tht function f hs limit L t, nd we write if lim f(z) = L z lim f(1 z 0 z ) = L, which mens tht, given ny ɛ > 0, there is lrge number M such tht It is esy to see for ll m = 1, 2,. f(z) L < ɛ whenever z S nd z M. lim z 1 z m = 0 Exmple 7. 1) The function f(z) = z 2 hs limit 4 t the point z 0 = 2i. 2) The function g(z) = 1 z 1 3) The function f(z) = z4 1 z i f(z) = z4 1 z i hs limit L = 1+i 2 t z 0 = i. is not defined t z = i, but hs limit 4i t z 0 = i, since = (z + 1)(z 1)(z + i)(z i) z i = (z + 1)(z 1)(z + i) = (z 2 1)(z + i), nd so f(z) = (z 2 1)(z + i) (i 2 1)(2i) = 4i s z i. 4) The function f(z) = z z hs no limit t z 0 = 0. For if z is rel, f(z) = 1, while if z = iy (purely imginry), then f(z) = f(iy) = 1. Such function cnnot hve limit t z 0 = 0. 5) lim z z 4 + 1 2z 4 + 5z 2 + 3 = lim z 1 + 1 z 4 2 + 5 z 2 + 3 z 4 = 1 2.
10 1. The Complex Plne 6) Let z = x + iy nd f(z) = x+y3 x 2 +y 3. Then lim z f(z) does not exist; for 1 lim f(z) = lim z=x x x = 0; lim f(z) = lim 1 = 1. z=iy y Suppose f is function defined on set S. Let z 0 S. Then we sy tht f is continuous t z 0 if lim z z 0 f(z) = f(z 0 ). If f is continuous t every point of S then we sy f is continuous on S. The function f is continuous t if f( ) is defined nd lim z f(z) = f( ). Polynomils re continuous functions on the whole plne. Rtionl functions re quotients of two polynomils. All rtionl functions re continuous wherever the denomintor is not zero. Infinite Series. Like rel vribles, n infinite series of complex numbers is written s z j. j=1 We cn define the prtil sums, convergence, sum, divergence, bsolute convergence of such series in the sme wy s the rel vrible theories. A specil kind of infinite series is the power series of the form c n (z z 0 ) n. n=0 Fcts. 1. If z n converges, then z n converges. 2. Let z n = x n +iy n. Then z n converges if nd only if both x n nd y n converge, nd the sum is given by z n = x n + i y n. n=1 n=1 In this wy, the convergence problems for z n become the corresponding problems for two rel series x n nd y n, or become the problem for one rel series z n. For exmple, we cn use the rtio test nd root test for these rel series. However, the rtio nd root tests cn lso be pplied directly to the complex series z n (see Exercises 42 nd 43 in this section). n=1 A useful identity is the geometric series formul: k α n = 1 αk+1 α 1. 1 α Therefore Exmple 8. 1) The series n=0 n=0 α n = 1 1 α n=1 n( 1 + 2i ) n 3 α < 1.
1.5. The Exponentil, Logrithm, nd Trigonometric Functions 11 converges, since nd hence z n converges. 2) The series ( ) 1 + 2i n ( 5 n = n 3 3 i n n ) n cn be written s n=1 i n n = ( 1 2 + 1 4 1 6 + ) + i(1 1 3 + 1 5 1 7 + ) n=1 nd hence converges by the lternting-series test. 3) If z n converges then z n 0. This is very useful for showing the divergence of series. Exercises. Pge 41. Problems 1 10, 13, 14, 31, 33, 36. 1.5. The Exponentil, Logrithm, nd Trigonometric Functions The Exponentil Function. The exponentil function e z is one of the most importnt functions in complex nlysis nd is defined s follows: Given z = x + iy, e z = e x (cos y + i sin y). We lso use exp(z) to denote e z, especilly when z itself is complicted expression. From the definition, it is direct to verify the following importnt properties for the exponentil function: e z+w = e z e w (using the sum formuls for sine nd cosine); e z = e Rez > 0 (in prticulr, e iy = 1 for ll rel numbers y); e z+2πki = e z k = 0, ±1, ±2, (tht is, e z is periodic of period 2πi). We lso hve e it = cos t + i sin t for ll rel t; hence we hve the interesting formul relting five most importnt numbers: e iπ + 1 = 0. Also for z 0, z = z e iθ, where θ rgz, or z = z e i Arg z. The function w = f(z) = e z is never zero nd it mps the z-plne onto the w-plne with the origin 0 removed but is not one-to-one (see the logrithm function below). The function w = e z mps the verticl line x = Rez = x 0 onto the circle w = e x 0 nd mps the horizontl line y = Imz = y 0 onto ry from origin with fixed rgument y 0. The function w = e z crries ech strip y 0 Imz < y 0 + 2π, < Rez <, one-toone nd onto the w-plne with the origin removed. For, if e z 1 = e z 2 nd z 1, z 2 re in the strip, then z 1 z 2 = 2πik for some integer k; but then 2π k = Imz 1 Imz 2 < 2π for z 1, z 2 in the strip, so k must be zero nd hence z 1 = z 2.
12 1. The Complex Plne y y 0 + 2π τ y y 0 w = e z O y y0 e x 0 σ O x 0 x y 0 + 2π Figure 1.3. The mpping w = e z The Logrithmic Function. The logrithm function is the inverse of the exponentil function (but it is not well-defined function). For nonzero complex number z, ny number w stisfying e w = z is clled logrithm of z; the set of ll logrithms of z is denoted by log z (this is not single vlued function). We now compute log z. Write z = z (cos θ + i sin θ) with ny θ being n rgument of z nd let w = s + it be such tht e w = z. Then Hence Therefore ll vlues of log z re given s e s (cos t + i sin t) = z (cos θ + i sin θ). s = ln z, t = θ + 2πk rg z. log z = ln z + i rg z. Since rg z is not single-vlued, neither is log z; but we cn define the principl logrithm function using Argz to be Logz = ln z + iargz. This is well-defined function for ll z 0. There is wy to mke log z well-defined (nd nice, sy, continuous) function if we delete ry strting the origin from the the z-plne nd use continuous rnge of rguments for rg z in the definition of log z. For exmple, if D is the open domin in the complex plne with the origin nd the negtive x-xis deleted, then Logz becomes continuous (even lter nlytic) function in D. Given two complex numbers z nd with z 0, the power z is defined by z = e log z, which is not single-vlued (unless is n integer). When = 1 n for positive integers n this definition grees with the nth roots of z defined before. Exmple 9. Find ( 1) i. Solution. Since log( 1) = ln 1 + i rg( 1) = (2n + 1)πi, n = 0, ±1, ±2,, we obtin ( 1) i = e i log( 1) = e (2n+1), n = 0, ±1, ±2,.
1.5. The Exponentil, Logrithm, nd Trigonometric Functions 13 Exmple 10. Solve z 1+i = 4. Solution. Write the eqution s e (1+i) log z = 4, so (1 + i) log z = log 4 = ln 4 + 2πni, n = 0, ±1,. Hence log z = log 4 1 + i = 1 i [ln 4 + 2πni] 2 = (1 i)[ln 2 + πni] = (ln 2 + πn) + i(πn ln 2). Thus z = e log z = e ln 2+πn [cos(πn ln 2) + i sin(πn ln 2)] = 2e πn [( 1) n cos ln 2 + i( 1) n+1 sin ln 2] = ( 1) n 2e πn [cos ln 2 i sin ln 2], n = 0, ±1, ±2,. Exmple 11. Estblish the formul lim n ( 1 + z ) n = e z n z. Proof. Look t n Log(1 + z/n) for lrge n. Write The rel prt stisfies n Log(1 + z n ) = n ln 1 + z n + inarg(1 + z n ). n ln 1 + z n = 1 2 n ln(1 + 2x n + x2 + y 2 ) x s n. Next, to hndle the imginry prt, let z = r(cos θ + i sin θ) nd ψ n = Arg(1 + z/n); then from the geometry, from which it follows tht ψ n 0 nd tn ψ n = r n sin θ 1 + r n cos θ, n tn ψ n r sin θ = y s n. Hence nψ n y s n. Consequently, ( n Log 1 + z ) x + iy = z n s n. This proves tht ( 1 + z ) n [ ( = exp n Log 1 + z )] e z n n n 2 s n, s intended.
14 1. The Complex Plne Trigonometric Functions. The trigonometric functions of z re defined in terms of the exponentil function e z s follows: cos z = 1 ( e iz + e iz) ; sin z = 1 ( e iz e iz). 2 2i tn z = sin z cos z ; cos z cot z = sin z ; sec z = 1 cos z ; csc z = 1 sin z, wherever the denomintor is not zero. It is esy to see tht cos(z + 2πk) = cos z, sin(z + 2πk) = sin z for z nd ll k = 0, ±1, ±2,. Furthermore, 2πk is the only numbers α for which cos(z + α) = cos z holds for ll z; the sme is true for sin z. For ifcos(z + α) = cos z for ll complex numbers z then e iz e iα + e iz e iα = e iz + e iz z. So e iz (e iα 1) = e iz e iα (e iα 1) for ll z. Set z = 0 to obtin (e iα 1) 2 = 0 nd hence e iα = 1, which gives α = 2kπ for some integer k. For this reson, 2π is clled the bsic period of sin z nd cos z. The mpping property of w = sin z. Note tht sin(x + iy) = sin x cosh y + i cos x sinh y, where, sinh u nd cosh u re hyperbolic-sine nd hyperbolic-cosine functions for rel numbers u, sinh u = 1 2 (eu e u ), cosh u = 1 2 (eu + e u ). These hyperbolic functions cn lso be defined for ll complex numbers u by the sme formuls. y z = x 0 + iy τ y 0 S w = sin z sinh(y 0 ) w = σ + iτ O x 0 π 2 x O sin(x 0 ) 1 cosh(y 0 ) σ Figure 1.4. The mpping w = sin z We now restrict z = x + iy to the hlf-strip S = {z : 0 < x < π/2, y > 0}. Note tht sin(iy) = i sinh y, sin( π + iy) = cosh y. 2 The function w = sin z is one-to-one from S onto the open first qudrnt of the w-plne, with the boundry of S being mpped onto the boundry of the first qudrnt in n interesting
1.6. Line Integrls nd Green s Theorem 15 wy. The verticl hlf line {z = x 0 + iy, y > 0} in S is mpped onto the portion of the hyperbol σ 2 sin 2 τ 2 x 0 cos 2 = 1 x 0 in the first qudrnt of the w plne, while the horizontl line segment {z = x + iy 0, 0 < x < π/2} in S is mpped onto the portion of the ellipse σ 2 cosh 2 + τ 2 y 0 sinh 2 = 1 y 0 in the first qudrnt of the w plne. Note tht the hyperbol nd ellipse given bove is perpendiculr t their intersection point z 0 = x 0 + iy 0. Exercises. Pge 53. Problems 1, 3, 4, 10, 23, 27, 28. 1.6. Line Integrls nd Green s Theorem The fundmentl theorems of complex vribles re built on Cuchy s Theorem nd Formul, which depend on line integrls nd Green s Theorem. So we review nd discuss these mterils now. A curve is defined to be continuous function from n finite closed intervl [, b] to the complex plne. This function (t) is sometime lso clled prmeteriztion of the curve. The nturl orienttion of the curve is defined by trcing the point (t) strting with t = nd ending with t = b. A curve is clled simple if (t 1 ) (t 2 ) for ll t 1 < t 2 < b. A curve is clled closed if () = (b); tht is, if its strting nd ending points coincide. Theorem 1.2 (Jordn s Theorem). Let be simple nd closed curve. Then the complement of its rnge consists of two domins, one of which is bounded nd the other is unbounded. The bounded one is clled the inside of nd the unbounded one the outside of. Exmple 12. (1) Given ny two distinct complex numbers z 0 nd z 1, the (directed) line segment from z 0 to z 1 is curve with prmeteriztion (2) The curve with function (t) = (1 t)z 0 + tz 1, 0 t 1. (t) = z 0 + Re it, 0 t 2π represents the simple closed positively oriented circle with center z 0 nd rdius R. However, if we chnge the prmeter intervl to 0 t 4π, this curve is lso closed but not simple; but the rnge (imgine) of the curve is the sme circle. Let f(t) = x(t)+iy(t) be complex-vlued function defined on [, b], where x(t), y(t) re the rel nd imginry prts of f(t). If both x nd y re differentible t point t 0 [, b] then we sy f is differentible t t 0, nd define f (t 0 ) = x (t 0 ) + iy (t 0 ). We sy f is differentible on [, b] if f is differentible t every point of [, b]; we sy f is smooth on [, b] or C 1 on [, b] if f is differentible on [, b] nd f is continuous on [, b].
16 1. The Complex Plne The differentition of complex-vlued functions of one rel vrible is consistent with the differentition of vector-vlued functions studied in the vector clculus. Mny differentition rules lso hold for this differentition. But we wnt to emphsize the following product rule: (f(t)g(t)) = f(t)g (t) + f (t)g(t) holds for differentible complex-vlued functions f nd g. Therefore we hve by induction [(f(t)) m ] = m(f(t)) m 1 f (t), m = 1, 2,. A curve is sid to be smooth curve if its prmetriztion (t) is smooth on its intervl [, b]. A curve cn lso be defined by joining finite number of curves together; this is piecewisely defined curve. The prmetriztion my hve severl independent prmeters on different intervls, but the only requirement is tht the end point of one piece coincides with the strting point of the following piece. Therefore, if is piece-wisely defined curve if there exist intervls [ k, b k ] (k = 1, 2,, n) nd continuous function k on [ k, b k ] such tht (b k ) = ( k+1 ) k = 1, 2,, n 1. The strting point of the curve is 1 ( 1 ) nd the ending point is n (b n ). A curve is clled piecewise smooth if ech piece of the curve k [k,b k ] is smooth for ll k = 1, 2,, n. If (t), t b, is curve, then the curve defined by (t) = ( + b t), t b, is clled the reversed curve of. This is the curve with the sme rnge s but strting with (b) nd ending with (); the orienttion is exctly reversed. A simple closed curve is clled positively oriented if you trce the points (t) while incresing the prmeter from to b the inside of is lwys on your left; this is equivlent to sying tht the is trced counterclockwise. Exmple 13. (1) The squre with vertices t points z 0, iz 0, z 0 nd iz 0 cn be mde simple closed curve by the following prmeteriztion: tiz 0 + (1 t)z 0, 0 t 1 (t 1)( z 0 ) + (2 t)iz 0, 1 t 2 (t) = (t 2)( iz 0 ) + (3 t)( z 0 ), 2 t 3 (t 3)z 0 + (4 t)( iz 0 ), 3 t 4. Of course, one cn lso define the sme squre s piecewisely defined curve s follows: tiz 0 + (1 t)z 0, 0 t 1 t( z 0 ) + (1 t)iz 0, 0 t 1 (t) = t( iz 0 ) + (1 t)( z 0 ), 0 t 1 tz 0 + (1 t)( iz 0 ), 0 t 1. (2) The curve : { z = Re iθ, z = t, 0 θ π R t R is piece-wisely defined nd represents the simple closed piecewise smooth curve which consists semicircle nd dimeter.
1.6. Line Integrls nd Green s Theorem 17 (3) The curve shown in the figure below hs prmetriztion given by z = Re iθ, 0 θ π z = x, R x ɛ : z = ɛe i(π θ), 0 θ π z = x, ɛ x R. Note tht the third formul is different from the text becuse we trce the curve lwys long the incresing direction of the prmeter. R ɛ O ɛ R Figure 1.5. A piecewise simple closed curve with seprte prmeteriztions Definite Integrls. Suppose g(t) = σ(t) + iτ(t) is continuous complex-vlued function on the intervl [, b]. We define the definite integrl of g over [, b] by b g(t) dt = b σ(t) dt + i b τ(t) dt. This definition is consistent with the definition of the definite integrl of vector-vlued function of the rel vrible t in the vector clculus, nd so mny rules of integrtion lso hold. From this definition, it is esy to hve ( b ) b ( b ) b Re g(t) dt = Re(g(t)) dt, Im g(t) dt = Im(g(t)) dt. We lso hve the following estimte: b b (1.2) g(t) dt g(t) dt. Proof. The inequlity is obviously true if b g(t) dt = 0, so we my ssume z 0 = b g(t)dt 0 nd let θ = Argz 0. Define h(t) = e iθ g(t), t b. Then b g(t) dt = z 0 = e iθ z 0 ( b ) b = e iθ g(t) dt = h(t) dt ( b ) b = Re h(t) dt = (Reh(t)) dt b h(t) dt = b g(t) dt.
18 1. The Complex Plne Line Integrls. Suppose tht is piece-wise smooth curve with smooth prmeteriztions on intervls [ k, b k ], k = 1,, n, nd tht u is continuous function on the rnge of. We define the line integrl long of u by n bk (1.3) u(z) dz = u((t)) (t) dt. k=1 k Recll tht if smooth plne curve is prmetrized by (t) = (x(t), y(t)) on [, b] then for rel vlued continuous function p = p(x, y) we hve the definition of line integrl: b b p dx = p(x(t), y(t))x (t) dt, p dy = p(x(t), y(t))y (t) dt. Then it is esy to see tht the definition (1.3) bove grees with this definition under the convention tht dz = dx + idy nd (t) = x (t) + iy (t). Line integrls hve the fmilir properties of definite integrls studied in clculus. For exmple (1.4) [Au(z) + Bv(z)] dz = A u(z) dz + B v(z) dz if A, B re constnts of complex numbers nd u, v re continuous functions on the rnge of curve. For the reversed curve we hve u(z) dz = u(z) dz. Furthermore, we hve the following importnt estimte: ) (1.5) u(z) dz (mx u(z) Length (). z Proof. We cn ssume consists of simply one smooth curve with prmeteriztion (t) = x(t) + iy(t), t [, b]. Then, by (1.2) nd definition (1.3), b u(z) dz = u((t)) (t) dt b ( ) b u((t)) (t) dt mx u(z) (t) dt z ( ) = mx u(z) Length (), z becuse b b (t) dt = (x (t)) 2 + (y (t)) 2 dt = Length of. Exmple 14. (1) Compute (z2 3 z + Imz) dz, where (t) = 2e it, 0 t π/2. Hence Let u(z) = z 2 3 z + Imz. Then u((t)) = 4e 2it 3 2 + 2 sin t; (t) = 2ie it (informlly, but true). u((t)) (t) = 8ie 3it 12ie it + 4ie it sin t = 8ie 3it 12ie it + 2ie it (e it e it ) = 8ie 3it 12ie it + 2e 2it 2.
1.6. Line Integrls nd Green s Theorem 19 So, by definition, u(z) dz = π 2 0 u((t)) (t) dt = = π 2 0 (8ie 3it 12ie it + 2e 2it 2) dt ( 8 3 e3it 12e it + 1 ) π i e2it 2 2t = 28 0 3 π 38 3 i. (2) Show if R > 2. z =R 1 z 2 + 4 dz 2πR R 2 4 Proof. On the circle z = R, by the tringle inequlity, z 2 + 4 z 2 4 = R 2 4, so mx 1 z =R z 2 + 4 1 R 2 4 nd the length of the circle is 2πR. Hence this estimte follows from (1.5) bove. (3) Let u be continuous in z z 0 < r nd ɛ be the simple closed positively oriented circle z z 0 = ɛ with 0 < ɛ < r. Show tht 1 u(z) lim dz = u(z 0 ). ɛ 0 2πi z z 0 ɛ ɛ Proof. The curve ɛ is prmeterized by z = z 0 + ɛe it, 0 t 2π; hence ɛ(t) = ɛie it. So u(z) 2π u(z 0 + ɛe it ) dz = z z 0 ɛe it ɛie it dt 0 = i 2π 0 2π = i u(z 0 + ɛe it ) dt 0 [u(z 0 + ɛe it ) u(z 0 )] dt + i 2πiu(z 0 ) s ɛ 0. 2π 0 u(z 0 ) dt (4) Let be ny piece-wise smooth curve with strting point A nd ending point B. Then, for ll m = 0, 1,, z m dz = 1 m + 1 (Bm+1 A m+1 ). Therefore, the line integrl is independent of the curve but only dependent on the endpoints of the curve.
20 1. The Complex Plne Proof. Note tht for complex-vlued differentible functions z(t) nd w(t) the product rule of differentition is lso vlid: Therefore the power chin rule is vlid: [z(t)w(t)] = z (t)w(t) + z(t)w (t). [z m (t)] = mz m 1 (t) z (t), m = 1, 2,. Therefore, if (t) is smooth, then [ ] m (t) 1 (t) = m + 1 m+1 (t), m = 0, 1, 2,. Let be piece-wise smooth curve with smooth prmeteriztions on intervls [ k, b k ], k = 1, 2,, n, nd ( 1 ) = A nd (b n ) = B. Then n bk z m dz = m (t) (t) dt = = n k=1 bk k k=1 n [ 1 m + 1 m+1 (t) k=1 k [ ] 1 m + 1 m+1 (t) dt ] bk k = 1 m + 1 (Bm+1 A m+1 ). Green s Theorem. The most importnt result on line integrls is Green s Theorem, which is reformultion of the theorems in vector clculus. To stte the theorem, we need to consider domins with certin properties. Γ 1 Γ 2 Γ n Figure 1.6. The domin Ω with positively oriented boundries.
1.6. Line Integrls nd Green s Theorem 21 Let Ω be domin whose boundry Γ = Ω consists of finite number of disjoint, piece-wise smooth, simple nd closed curves 1, 2,, n. We sy tht the boundry Γ is positively oriented if Ω remins lwys on our left if we wlk long Γ. Thus, if positively oriented, the outer piece of Γ is oriented counterclockwise, nd ech of the inner pieces of Γ is oriented clockwise. In this cse, if f is continuous complex-vlued function on Γ, we define the line integrl of f over Γ by n f(z) dz = f(z) dz, Γ j where, of course, we lredy know how to compute ech j f(z) dz. j=1 Green s Theorem reltes the line integrl of function f over Γ to the (re) integrl of certin combintion of prtil derivtives of f over domin Ω. In order to stte it properly, we introduce the following nottion. Let f(z) = p(z) + iq(z) be function of complex vrible z = x + iy in some domin D contining Ω = Γ Ω, where p, q re rel nd imginry prts of f, considered s functions of (x, y). We then define f x = p x + iq x, f y = p y + iq y, where p x, p y, q x, nd q y re prtil derivtives, tht re ssumed to exist. Theorem 1.3 (Green s Theorem). Let f(z) be function of complex vrible which hs continuous prtil derivtives in some domin D contining Ω = Γ Ω, where Ω is domin s described bove with positively oriented boundry Γ. Then (1.6) f(z) dz = i (f x + if y ) dxdy. Γ Recll tht in multi-vrible clculus, Green s Theorem ws stted s (1.7) (u dx + v dy) = (v x u y ) dxdy Γ Ω for ny two continuously differentible rel-vlued functions u, v of two rel vribles x, y. It is good exercise to verify tht the complex vrible formul (1.6) in Green s Theorem bove is equivlent to Formul (1.7). (See Exercises #10, #11.) Exmple 15. Let be piece-wise smooth positively oriented simple closed curve nd p be point not on. Compute 1 1 2πi z p dz. Proof. Let Ω be the inside of. Let f(z) = 1 z p f(z) = 1 x+iy p nd one cn compute hence f x + if y = 0 t ll points z p. 1 f x = (x + iy p) 2, f i y = (x + iy p) 2 ; If p / Ω nd thus p / Ω (since p / ), then by Green s Theorem, f(z) dz = i (f x + if y ) dxdy = 0. Ω Ω for ll z p. If z = x + iy p then
22 1. The Complex Plne Therefore 1 1 dz = 0 if p / Ω. 2πi z p Now ssume p Ω. Let D ɛ be the closed disc centered p of rdius ɛ > 0. Assume ɛ is sufficiently smll so tht D ɛ Ω. Let Ω ɛ = Ω \ D ɛ. The positively oriented boundry Γ ɛ of Ω ɛ consists of two prts: nd circle δ ɛ = { z p = ɛ} oriented clockwise. Since p / Ω ɛ, by Green s Theorem, f(z) dz = i ɛ (f x + if y ) dxdy = 0. Ω ɛ Hence f(z) dz = f(z) dz = δ ɛ f(z) dz. δ ɛ We cn prmeterize δ ɛ (which is the circle oriented counter clockwise) by z = p + ɛe it with 0 t 2π, nd hence Therefore δ ɛ f(z) dz = 2π 0 1 ɛe it iɛeit dt = 2πi. 1 1 2πi z p dz = 1 if p Ω. Exmple 16. Let be the simple positively oriented unit circle z = 1. Compute z dz. Solution. Let Ω be the unit disc z < 1 nd f(z) = z = x iy. Then f x + if y = (x iy) x + i(x iy) y = 1 i 2 = 2. Hence, by Green s Theorem, f(z) dz = i (f x + if y ) dxdy = i Ω Ω 2 dxdy = 2i Are of Ω = 2πi. We cn lso compute the line integrl directly by definition. Let (t) = e it, 0 t 2π. Then 2π 2π 2π z dz = e it (e it ) dt = e it ie it dt = i dt = 2πi. 0 0 0 As expected, we got the sme nswer. Exercises. Pge 73. Problems. 1, 2, 3, 7, 8, 9, 12. Homework Problems for Chpter 1. 1.1 1, 2, 3, 5, 6, 11 1.2 1, 2, 3, 7, 11, 12, 15, 21, 23, 24, 25 1.3 1 8, 10, 11 1.4 1-10, 13, 14, 31, 33, 36 1.5 1, 3, 4, 10, 23, 27, 28 1.6 1, 2, 3, 7, 8, 9, 12