We identify two complex numbers z and w if and only if Rez = Rew and Imz = Imw. We also write
|
|
- Chester Smith
- 6 years ago
- Views:
Transcription
1 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 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).
3 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 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 4 1. The Complex Plne nd = 2 [ 2 cos 11π ] 11π + i sin i 2 = [cos( 3π4 3 + i 2 π6 ) + i sin(3π4 π6 ] [ 2 ) = cos 7π ] 7π + i sin (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, 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 z w = z 2 + w 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.
5 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 ρ p q 3i nd rdius R = = 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 )2 + (y 4 3 )2 = 8 9 of center z 0 = i nd rdius
6 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 i = 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 ) = 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 πk 5 ) + i sin( π πk ] 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 i = (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.
7 1.3. Subsets of the Plne 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 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 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, 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 We expnd w 1 2 = (w 1)( w 1) nd w = (w + 1)( w + 1) to obtin w 1 2 = w Rew; w = w Rew. Since Rew > 0, it follows esily tht w 1 2 < w ; 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
9 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 z 4 + 5z = lim z z z z 4 = 1 2.
10 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.
11 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 = ( ) + i( ) 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, 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 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,.
13 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 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 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
15 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, 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 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.
17 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 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 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.
19 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 = π 38 3 i. (2) Show if R > 2. z =R 1 z dz 2πR R 2 4 Proof. On the circle z = R, by the tringle inequlity, z z 2 4 = R 2 4, so mx 1 z =R z 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 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.
21 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 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 πi 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 As expected, we got the sme nswer. Exercises. Pge 73. Problems. 1, 2, 3, 7, 8, 9, 12. Homework Problems for Chpter , 2, 3, 5, 6, , 2, 3, 7, 11, 12, 15, 21, 23, 24, , 10, , 13, 14, 31, 33, , 3, 4, 10, 23, 27, , 2, 3, 7, 8, 9, 12
The usual algebraic operations +,, (or ), on real numbers can then be extended to operations on complex numbers in a natural way: ( 2) i = 1
Mth50 Introduction to Differentil Equtions Brief Review of Complex Numbers Complex Numbers No rel number stisfies the eqution x =, since the squre of ny rel number hs to be non-negtive. By introducing
More informationINDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Analysis Autumn 2012
Lecture 6: Line Integrls INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Anlysis Autumn 2012 August 8, 2012 Lecture 6: Line Integrls Lecture 6: Line Integrls Lecture 6: Line Integrls Integrls of complex
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationII. Integration and Cauchy s Theorem
MTH6111 Complex Anlysis 2009-10 Lecture Notes c Shun Bullett QMUL 2009 II. Integrtion nd Cuchy s Theorem 1. Pths nd integrtion Wrning Different uthors hve different definitions for terms like pth nd curve.
More informationNotes on length and conformal metrics
Notes on length nd conforml metrics We recll how to mesure the Eucliden distnce of n rc in the plne. Let α : [, b] R 2 be smooth (C ) rc. Tht is α(t) (x(t), y(t)) where x(t) nd y(t) re smooth rel vlued
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7. - Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More informationMath Advanced Calculus II
Mth 452 - Advnced Clculus II Line Integrls nd Green s Theorem The min gol of this chpter is to prove Stoke s theorem, which is the multivrible version of the fundmentl theorem of clculus. We will be focused
More informationIntroduction to Complex Variables Class Notes Instructor: Louis Block
Introuction to omplex Vribles lss Notes Instructor: Louis Block Definition 1. (n remrk) We consier the complex plne consisting of ll z = (x, y) = x + iy, where x n y re rel. We write x = Rez (the rel prt
More informationMORE FUNCTION GRAPHING; OPTIMIZATION. (Last edited October 28, 2013 at 11:09pm.)
MORE FUNCTION GRAPHING; OPTIMIZATION FRI, OCT 25, 203 (Lst edited October 28, 203 t :09pm.) Exercise. Let n be n rbitrry positive integer. Give n exmple of function with exctly n verticl symptotes. Give
More informationMATH 185: COMPLEX ANALYSIS FALL 2009/10 PROBLEM SET 5 SOLUTIONS. cos t cos at dt + i
MATH 85: COMPLEX ANALYSIS FALL 9/ PROBLEM SET 5 SOLUTIONS. Let R nd z C. () Evlute the following integrls Solution. Since e it cos t nd For the first integrl, we hve e it cos t cos t cos t + i t + i. sin
More informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
More informationMath 113 Fall Final Exam Review. 2. Applications of Integration Chapter 6 including sections and section 6.8
Mth 3 Fll 0 The scope of the finl exm will include: Finl Exm Review. Integrls Chpter 5 including sections 5. 5.7, 5.0. Applictions of Integrtion Chpter 6 including sections 6. 6.5 nd section 6.8 3. Infinite
More informationOptimization Lecture 1 Review of Differential Calculus for Functions of Single Variable.
Optimiztion Lecture 1 Review of Differentil Clculus for Functions of Single Vrible http://users.encs.concordi.c/~luisrod, Jnury 14 Outline Optimiztion Problems Rel Numbers nd Rel Vectors Open, Closed nd
More informationSpace Curves. Recall the parametric equations of a curve in xy-plane and compare them with parametric equations of a curve in space.
Clculus 3 Li Vs Spce Curves Recll the prmetric equtions of curve in xy-plne nd compre them with prmetric equtions of curve in spce. Prmetric curve in plne x = x(t) y = y(t) Prmetric curve in spce x = x(t)
More informationSection 17.2 Line Integrals
Section 7. Line Integrls Integrting Vector Fields nd Functions long urve In this section we consider the problem of integrting functions, both sclr nd vector (vector fields) long curve in the plne. We
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationPhil Wertheimer UMD Math Qualifying Exam Solutions Analysis - January, 2015
Problem 1 Let m denote the Lebesgue mesure restricted to the compct intervl [, b]. () Prove tht function f defined on the compct intervl [, b] is Lipschitz if nd only if there is constct c nd function
More information13.3 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS
33 CLASSICAL STRAIGHTEDGE AND COMPASS CONSTRUCTIONS As simple ppliction of the results we hve obtined on lgebric extensions, nd in prticulr on the multiplictivity of extension degrees, we cn nswer (in
More informationChapter 6 Techniques of Integration
MA Techniques of Integrtion Asst.Prof.Dr.Suprnee Liswdi Chpter 6 Techniques of Integrtion Recll: Some importnt integrls tht we hve lernt so fr. Tle of Integrls n+ n d = + C n + e d = e + C ( n ) d = ln
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More informationComplex variables lecture 5: Complex integration
omplex vribles lecture 5: omplex integrtion Hyo-Sung Ahn School of Mechtronics Gwngju Institute of Science nd Technology (GIST) 1 Oryong-dong, Buk-gu, Gwngju, Kore Advnced Engineering Mthemtics omplex
More informationUS01CMTH02 UNIT Curvature
Stu mteril of BSc(Semester - I) US1CMTH (Rdius of Curvture nd Rectifiction) Prepred by Nilesh Y Ptel Hed,Mthemtics Deprtment,VPnd RPTPScience College US1CMTH UNIT- 1 Curvture Let f : I R be sufficiently
More informationChapter 8: Methods of Integration
Chpter 8: Methods of Integrtion Bsic Integrls 8. Note: We hve the following list of Bsic Integrls p p+ + c, for p sec tn + c p + ln + c sec tn sec + c e e + c tn ln sec + c ln + c sec ln sec + tn + c ln
More informationu(t)dt + i a f(t)dt f(t) dt b f(t) dt (2) With this preliminary step in place, we are ready to define integration on a general curve in C.
Lecture 4 Complex Integrtion MATH-GA 2451.001 Complex Vriles 1 Construction 1.1 Integrting complex function over curve in C A nturl wy to construct the integrl of complex function over curve in the complex
More informationIntegrals along Curves.
Integrls long Curves. 1. Pth integrls. Let : [, b] R n be continuous function nd let be the imge ([, b]) of. We refer to both nd s curve. If we need to distinguish between the two we cll the function the
More information. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. cos(2θ) = sin(2θ) =.
Review of some needed Trig Identities for Integrtion Your nswers should be n ngle in RADIANS rccos( 1 2 ) = rccos( - 1 2 ) = rcsin( 1 2 ) = rcsin( - 1 2 ) = Cn you do similr problems? Review of Bsic Concepts
More information4.4 Areas, Integrals and Antiderivatives
. res, integrls nd ntiderivtives 333. Ares, Integrls nd Antiderivtives This section explores properties of functions defined s res nd exmines some connections mong res, integrls nd ntiderivtives. In order
More informationDisclaimer: This Final Exam Study Guide is meant to help you start studying. It is not necessarily a complete list of everything you need to know.
Disclimer: This is ment to help you strt studying. It is not necessrily complete list of everything you need to know. The MTH 33 finl exm minly consists of stndrd response questions where students must
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 2-3, Sections 4.1-4.8, nd Sections 5.1-5.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationMath 113 Exam 1-Review
Mth 113 Exm 1-Review September 26, 2016 Exm 1 covers 6.1-7.3 in the textbook. It is dvisble to lso review the mteril from 5.3 nd 5.5 s this will be helpful in solving some of the problems. 6.1 Are Between
More information. Double-angle formulas. Your answer should involve trig functions of θ, and not of 2θ. sin 2 (θ) =
Review of some needed Trig. Identities for Integrtion. Your nswers should be n ngle in RADIANS. rccos( 1 ) = π rccos( - 1 ) = 2π 2 3 2 3 rcsin( 1 ) = π rcsin( - 1 ) = -π 2 6 2 6 Cn you do similr problems?
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More informationMA 124 January 18, Derivatives are. Integrals are.
MA 124 Jnury 18, 2018 Prof PB s one-minute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,
More informationMATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1
MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further
More informationg i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f
1. Appliction of functionl nlysis to PEs 1.1. Introduction. In this section we give little introduction to prtil differentil equtions. In prticulr we consider the problem u(x) = f(x) x, u(x) = x (1) where
More informationTHE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrd-lindelof-theorem/ This document is proof of the existence-uniqueness theorem
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More informationODE: Existence and Uniqueness of a Solution
Mth 22 Fll 213 Jerry Kzdn ODE: Existence nd Uniqueness of Solution The Fundmentl Theorem of Clculus tells us how to solve the ordinry differentil eqution (ODE) du = f(t) dt with initil condition u() =
More information1 Line Integrals in Plane.
MA213 thye Brief Notes on hpter 16. 1 Line Integrls in Plne. 1.1 Introduction. 1.1.1 urves. A piece of smooth curve is ssumed to be given by vector vlued position function P (t) (or r(t) ) s the prmeter
More informationThis is read as: Q is the set of all things of the form a/b where a and b are elements of Z and b is not equal to 0.
1. Some nottion 1.1. Nottion from set theory: We wont worry bout wht set is. If S is set nd x is member of S, we write x S (red s x belongs to S ). If x is not member of S we write x / S. The empty set
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationMATH 144: Business Calculus Final Review
MATH 144: Business Clculus Finl Review 1 Skills 1. Clculte severl limits. 2. Find verticl nd horizontl symptotes for given rtionl function. 3. Clculte derivtive by definition. 4. Clculte severl derivtives
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationHigher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors
Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector
More informationIndefinite Integral. Chapter Integration - reverse of differentiation
Chpter Indefinite Integrl Most of the mthemticl opertions hve inverse opertions. The inverse opertion of differentition is clled integrtion. For exmple, describing process t the given moment knowing the
More informationChapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...
Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationa < a+ x < a+2 x < < a+n x = b, n A i n f(x i ) x. i=1 i=1
Mth 33 Volume Stewrt 5.2 Geometry of integrls. In this section, we will lern how to compute volumes using integrls defined by slice nlysis. First, we recll from Clculus I how to compute res. Given the
More informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationAbsolute values of real numbers. Rational Numbers vs Real Numbers. 1. Definition. Absolute value α of a real
Rtionl Numbers vs Rel Numbers 1. Wht is? Answer. is rel number such tht ( ) =. R [ ( ) = ].. Prove tht (i) 1; (ii). Proof. (i) For ny rel numbers x, y, we hve x = y. This is necessry condition, but not
More informationSUMMER KNOWHOW STUDY AND LEARNING CENTRE
SUMMER KNOWHOW STUDY AND LEARNING CENTRE Indices & Logrithms 2 Contents Indices.2 Frctionl Indices.4 Logrithms 6 Exponentil equtions. Simplifying Surds 13 Opertions on Surds..16 Scientific Nottion..18
More informationUNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE
UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence
More informationMath 6455 Oct 10, Differential Geometry I Fall 2006, Georgia Tech
Mth 6455 Oct 10, 2006 1 Differentil Geometry I Fll 2006, Georgi Tech Lecture Notes 12 Riemnnin Metrics 0.1 Definition If M is smooth mnifold then by Riemnnin metric g on M we men smooth ssignment of n
More informationWe partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.
Mth 255 - Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn
More informationAdvanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015
Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n
More informationAnti-derivatives/Indefinite Integrals of Basic Functions
Anti-derivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationLine Integrals. Chapter Definition
hpter 2 Line Integrls 2.1 Definition When we re integrting function of one vrible, we integrte long n intervl on one of the xes. We now generlize this ide by integrting long ny curve in the xy-plne. It
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More informationLecture 0. MATH REVIEW for ECE : LINEAR CIRCUIT ANALYSIS II
Lecture 0 MATH REVIEW for ECE 000 : LINEAR CIRCUIT ANALYSIS II Aung Kyi Sn Grdute Lecturer School of Electricl nd Computer Engineering Purdue University Summer 014 Lecture 0 : Mth Review Lecture 0 is intended
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 2009-10 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationf(a+h) f(a) x a h 0. This is the rate at which
M408S Concept Inventory smple nswers These questions re open-ended, nd re intended to cover the min topics tht we lerned in M408S. These re not crnk-out-n-nswer problems! (There re plenty of those in the
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationMath Calculus with Analytic Geometry II
orem of definite Mth 5.0 with Anlytic Geometry II Jnury 4, 0 orem of definite If < b then b f (x) dx = ( under f bove x-xis) ( bove f under x-xis) Exmple 8 0 3 9 x dx = π 3 4 = 9π 4 orem of definite Problem
More information( dg. ) 2 dt. + dt. dt j + dh. + dt. r(t) dt. Comparing this equation with the one listed above for the length of see that
Arc Length of Curves in Three Dimensionl Spce If the vector function r(t) f(t) i + g(t) j + h(t) k trces out the curve C s t vries, we cn mesure distnces long C using formul nerly identicl to one tht we
More informationAQA Further Pure 1. Complex Numbers. Section 1: Introduction to Complex Numbers. The number system
Complex Numbers Section 1: Introduction to Complex Numbers Notes nd Exmples These notes contin subsections on The number system Adding nd subtrcting complex numbers Multiplying complex numbers Complex
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums - 1 Riemnn
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationAP Calculus Multiple Choice: BC Edition Solutions
AP Clculus Multiple Choice: BC Edition Solutions J. Slon Mrch 8, 04 ) 0 dx ( x) is A) B) C) D) E) Divergent This function inside the integrl hs verticl symptotes t x =, nd the integrl bounds contin this
More informationHow can we approximate the area of a region in the plane? What is an interpretation of the area under the graph of a velocity function?
Mth 125 Summry Here re some thoughts I ws hving while considering wht to put on the first midterm. The core of your studying should be the ssigned homework problems: mke sure you relly understnd those
More information4. Calculus of Variations
4. Clculus of Vritions Introduction - Typicl Problems The clculus of vritions generlises the theory of mxim nd minim. Exmple (): Shortest distnce between two points. On given surfce (e.g. plne), nd the
More information1. On some properties of definite integrals. We prove
This short collection of notes is intended to complement the textbook Anlisi Mtemtic 2 by Crl Mdern, published by Città Studi Editore, [M]. We refer to [M] for nottion nd the logicl stremline of the rguments.
More informationMath Solutions to homework 1
Mth 75 - Solutions to homework Cédric De Groote October 5, 07 Problem, prt : This problem explores the reltionship between norms nd inner products Let X be rel vector spce ) Suppose tht is norm on X tht
More informationAPPM 4360/5360 Homework Assignment #7 Solutions Spring 2016
APPM 436/536 Homework Assignment #7 Solutions Spring 6 Problem # ( points: Evlute the following rel integrl by residue integrtion: x 3 sinkx x 4 4, k rel, 4 > Solution: Since the integrnd is even function,
More informationMath 100 Review Sheet
Mth 100 Review Sheet Joseph H. Silvermn December 2010 This outline of Mth 100 is summry of the mteril covered in the course. It is designed to be study id, but it is only n outline nd should be used s
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More informationLinear Inequalities: Each of the following carries five marks each: 1. Solve the system of equations graphically.
Liner Inequlities: Ech of the following crries five mrks ech:. Solve the system of equtions grphiclly. x + 2y 8, 2x + y 8, x 0, y 0 Solution: Considerx + 2y 8.. () Drw the grph for x + 2y = 8 by line.it
More informationBefore we can begin Ch. 3 on Radicals, we need to be familiar with perfect squares, cubes, etc. Try and do as many as you can without a calculator!!!
Nme: Algebr II Honors Pre-Chpter Homework Before we cn begin Ch on Rdicls, we need to be fmilir with perfect squres, cubes, etc Try nd do s mny s you cn without clcultor!!! n The nth root of n n Be ble
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationMathematics. Area under Curve.
Mthemtics Are under Curve www.testprepkrt.com Tle of Content 1. Introduction.. Procedure of Curve Sketching. 3. Sketching of Some common Curves. 4. Are of Bounded Regions. 5. Sign convention for finding
More informationMain topics for the Second Midterm
Min topics for the Second Midterm The Midterm will cover Sections 5.4-5.9, Sections 6.1-6.3, nd Sections 7.1-7.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the
More informationAPPLICATIONS OF THE DEFINITE INTEGRAL
APPLICATIONS OF THE DEFINITE INTEGRAL. Volume: Slicing, disks nd wshers.. Volumes by Slicing. Suppose solid object hs boundries extending from x =, to x = b, nd tht its cross-section in plne pssing through
More informationUnit 5. Integration techniques
18.01 EXERCISES Unit 5. Integrtion techniques 5A. Inverse trigonometric functions; Hyperbolic functions 5A-1 Evlute ) tn 1 3 b) sin 1 ( 3/) c) If θ = tn 1 5, then evlute sin θ, cos θ, cot θ, csc θ, nd
More informationEdexcel GCE Core Mathematics (C2) Required Knowledge Information Sheet. Daniel Hammocks
Edexcel GCE Core Mthemtics (C) Required Knowledge Informtion Sheet C Formule Given in Mthemticl Formule nd Sttisticl Tles Booklet Cosine Rule o = + c c cosine (A) Binomil Series o ( + ) n = n + n 1 n 1
More informationComplex integration. L3: Cauchy s Theory.
MM Vercelli. L3: Cuchy s Theory. Contents: Complex integrtion. The Cuchy s integrls theorems. Singulrities. The residue theorem. Evlution of definite integrls. Appendix: Fundmentl theorem of lgebr. Discussions
More informationSection 14.3 Arc Length and Curvature
Section 4.3 Arc Length nd Curvture Clculus on Curves in Spce In this section, we ly the foundtions for describing the movement of n object in spce.. Vector Function Bsics In Clc, formul for rc length in
More informationConvex Sets and Functions
B Convex Sets nd Functions Definition B1 Let L, +, ) be rel liner spce nd let C be subset of L The set C is convex if, for ll x,y C nd ll [, 1], we hve 1 )x+y C In other words, every point on the line
More informationMA123, Chapter 10: Formulas for integrals: integrals, antiderivatives, and the Fundamental Theorem of Calculus (pp.
MA123, Chpter 1: Formuls for integrls: integrls, ntiderivtives, nd the Fundmentl Theorem of Clculus (pp. 27-233, Gootmn) Chpter Gols: Assignments: Understnd the sttement of the Fundmentl Theorem of Clculus.
More informationSOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set
SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL 28 Nottion: N {, 2, 3,...}. (Tht is, N.. Let (X, M be mesurble spce with σ-finite positive mesure µ. Prove tht there is finite positive mesure ν on (X, M such
More informationLine and Surface Integrals: An Intuitive Understanding
Line nd Surfce Integrls: An Intuitive Understnding Joseph Breen Introduction Multivrible clculus is ll bout bstrcting the ides of differentition nd integrtion from the fmilir single vrible cse to tht of
More informationES.182A Topic 32 Notes Jeremy Orloff
ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More information