Complex Analysis I. All My Students Version 1

Transcription

1 omplex Anlysis I All My Students Version 1 Fll, 009

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3 ontents 1 OMPLEX NUMBERS Sums nd Products Bsic Algebric Properties Further Properties Vectors nd Moduli omplex onjugtes Exponentil Form Products nd Powers in Exponentil Form Arguments of Products nd Quotients Roots of omplex Numbers Exmples Regions in the omplex Plne ANALYTI FUNTIONS 7.1 Functions of omplex Vrible Mppings Mppings by the Exponentil Function Limits Theorems on Limits Limits Involving the Point t Infinity ontinuity Derivtives Differentition Formuls uchy Riemnn Equtions Sufficient onditions for Differentibility Polr oordintes Anlytic Functions Exmples Hrmonic Functions Uniquely Determined Anlytic Functions Reflection Principle ELEMENTARY FUNTIONS The Exponentil Function The Logrithmic Function Brnches nd Derivtives of Logrithms Some Identities Involving Logrithms omplex Exponents Trigonometric Functions Hyperbolic Functions

4 3.8 Inverse Trigonometric nd Hyperbolic Functions INTEGRALS Derivtives of Functions w(t) Definite Integrls of Functions w(t) ontours ontour Integrls Some Exmples Exmples with Brnch uts Upper Bounds for Moduli of ontour Integrls Antiderivtives Proof of the Theorem uchy Gorst Theorem Proof of the Theorem Simply onnected Domins Multiply onnected Domins uchy Integrl Formul An Extension of the uchy Integrl Formul Some onsequences of the Extension Liouville s Theorem nd the Fundmentl Theorem of Algebr Mximum Modulus Principle of 84

5 hpter 1 OMPLEX NUMBERS 1.1 Sums nd Products. 1. Bsic Algebric Properties. 1.3 Further Properties. 1.4 Vectors nd Moduli. 1.5 omplex onjugtes. 1.6 Exponentil Form. 1.7 Products nd Powers in Exponentil Form. 1.8 Arguments of Products nd Quotients. 1.9 Roots of omplex Numbers Exmples Regions in the omplex Plne. 5

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7 hpter ANALYTI FUNTIONS.1 Functions of omplex Vrible. If D nd R re sets of rel numbers, then function f : D R is clled rel vlued function of rel vrible. It is customry to write such function s y = f(x). If D nd R re sets of complex numbers, then function f : D R is clled complex vlued function of complex vrible. It is customry to write such function s w = f(z). Representtion of w = f(z). 1. Using rectngulr coordintes on z nd w: f(x + iy) = u(x, y) + iv(x, y).. Using polr coordintes on z nd w: f(re iθ ) = u(r, θ) + iv(r, θ). Exmple.1.1. If f(z) = z, then So in the nottions bove, we hve When polr coordintes re used, Hence, f(x + yi) = (x + yi) = x y + xyi. u(x, y) = x y nd v(x, y) = xy. f(re iθ ) = (re iθ ) = r e iθ = r cos(θ) + ir sin(θ). u(r, θ) = r cos(θ) nd v(r, θ) = r sin(θ). We note tht if v = 0, then f(z) = u + iv becomes f(z) = u, which is rel vlued function of complex vrible. Exmple.1.. An importnt rel vlued function of complex vrible: for z = x + yi, Just s in Rel Anlysis, the function f(z) = z = x + y + i0. P (z) = z + z + + n z n, where n = 0 or positive integer nd n 0 nd 0, 1,..., n re complex numbers, is clled polynomil of degree n. Domin of P (z) is the whole complex plne. The quotient P (z)/q(z) of polynomils is clled rtionl function nd defined t the point z where Q(z) 0. Usully, when we sy function, it mens rule ssigning one vlue to one point in the domin. Explicitly, it is clled the single vlued function. We cn generlize this rule by ssigning more thn one vlues to one point in the domin. It is clled the multiple vlued function. We will study functions of complex vrible from the viewpoint of the multiple vlued function. Exmple.1.3. For z 0, we recll from Section 8. Roots of omplex Numbers, ( z 1/ = ± z exp i Arg(z) ), where π < Arg(z) π is the principl vlue of Arg(z). As we cn see, the function f(z) = z 1/ hs two vlues so it is multiple vlued function. However, if we choose the positive vlue of ± z nd write ( f(z) = z exp i Arg(z) ), then it becomes the single vlued function. 7

8 . Mppings. As we pproched complex number from the geometricl viewpoint, we investigte the geometricl work of complex function. Since complex function is defined on the complex plne nd mps point in the plne to nother point in the plne, we need to distinguish those two plnes (domin plne nd imge plne). Usully we sy xy plne (or z plne) for the domin nd uv plne (or w plne) for the imge, which should be cler from the expression f(x + iy) = u + iv introduced in the previous section. So complex function becomes mpping from the xy plne to the uv plne. (Recll tht for rel function, it is mpping from the x xis to the y xis.) Three Geometricl hrcteristics 1. Trnsltion: For exmple, the mpping w = z + 1 = (x + 1) + iy, where z = x + iy, cn be thought of s trnsltion of ech point z one unit to the right.. Rottion: Since i = e iπ/, the mpping w = iz = re i(θ+ π ), where z = re iθ, rottes the rdius vector for ech nonzero point z through right ngle bout the origin in the counterclockwise direction. 3. Reflection: The mpping w = z = x iy trnsforms ech point z = x + iy into its reflection in the rel xis. Exmple..1. Sketch the imge of the mpping w = z in the uv plne. Answer. Putting z = x + iy into the function, we hve u = x y, v = xy. (se 1) u = x y = c 1 (positive constnt): the grph of x y = c 1 > 0 in the xy plne is the hyperbol cutting the x xis. But in the uv plne u = c 1 represents verticl line. Tht is, hyperbols in the xy plne re mpped to the verticl lines in the uv plne. (se ) v = xy = c (positive constnt): the grph of xy = c > 0 in the xy plne is the hyperbol locted in the first nd the third qudrnts. But in the uv plne v = c represents horizontl line. Tht is, hyperbols in the xy plne re mpped to the horizontl lines in the uv plne. Exmple... Find the imge of S = {z = x + iy : x > 0, y > 0, xy < 1} under the mpping w = z. Answer. The mpping w = z implies v(x, y) = xy from the previous exmple. So if 0 < xy < 1, then 0 < v(x, y) = xy <. Agin from the previous exmple, the mpped imge of the hyperbol is horizontl line. So xy = 1 in the xy plne is mpped into horizontl line v = in the uv plne nd thus S is mpped to the infinite strip S = { w = u + iv 0 < v < }. Exercise..3. Sketch the domin in the xy plne whose imge by the mpping w = z is the squre domin S = {(u, v) : 1 u, 1 v }. Answer. The domin hving such n imge is S = {(x, y) : 1 x y, 1 xy } Exmple..4. The mpping w = z becomes w = r e iθ when z = re iθ. Hence, if w = ρe iϕ, then we hve ρe iϕ = r e iθ, i.e., ρ = r nd ϕ = θ + kπ, where k is n integer. So the first qudrnt in the xy plne is mpped to the upper hlf plne of the uv plne under the mpping w = z. Exercise..5. Sketch the imge of the sector S = {(r, θ) : r 1, 0 θ π 4 } by the mpping (1) w = z, () w = z 3, (3) w = z 4..3 Mppings by the Exponentil Function. 8 of 84

9 .4 Limits. The concepts of limits nd continuity re similr to tht of rel vribles. In this sense our discussion cn serve s brief review of mny previously understood notions. onsider function w = f(z) defined t ll points in some neighborhood of z = z 0, except possibly for z 0 itself. We sy f(z) hs the limit w 0 if s z pproches z 0, f(z) pproches w 0 (z 0, w 0 finite). Mthemticlly, we sy if for every (sufficiently smll) ε > 0 there is δ > 0 such tht lim f(z) = w 0, (.4.1) z z 0 f(z) w 0 < ε, whenever 0 < z z 0 < δ. (.4.) This definition is cler when z 0 is n interior point of region R in which f(z) is defined. If z 0 is boundry point of R, then we require Eq. (.4.) to hold only for those z R. Geometriclly, under the mpping w = f(z), ll points interior to the circle z z 0 = δ with z 0 deleted re mpped to points interior to the circle w w 0 = ε. The limit will exist only in the cse when z pproches z 0 (tht is, z z 0 ) in n rbitrry direction; then this implies tht w w 0. Exmple.4.1. Show tht ( ) z lim + iz + = 6i. (.4.3) z i z i Proof. We must show tht given ε > 0, there is δ > 0 such tht ( ) ( ) z + iz + 6i (z + i)(z i) z i = 6i z i < ε, (.4.4) whenever 0 < z i < δ. (.4.5) Since z = i, inequlity (.4.1) implies tht z i < ε. Thus if δ = ε/, Eq. (.4.5) ensures tht Eq. (.4.4) is stisfied. Therefore Eq. (.4.3) is demonstrted. Exercise.4.. ompute the limit in the open disk z < 3: lim z 3 iz 1. When limit of function f(z) exists t point z 0, the limit is unique. Tht is, if then w 0 = w 1. Proof. For every ε/ > 0 there is δ 0 > 0 such tht For every ε/ > 0 there is δ 1 > 0 such tht We observe for 0 < z z 0 < δ = min {δ 0, δ 1 }, lim z z 0 f(z) = w 0 nd lim z z 0 f(z) = w 1, f(z) w 0 < ε, whenever 0 < z z 0 < δ 0. f(z) w 1 < ε, whenever 0 < z z 0 < δ 1. 0 w 0 w 1 = (f(z) w 1 ) (f(z) w 0 ) f(z) w 1 + f(z) w 0 < ε/ + ε/ = ε. Since ε > 0 cn be chosen rbitrrily smll, hence we deduce w 0 w 1 = 0, i.e., w 0 = w 1. 9 of 84

10 We recll from Rel Anlysis or lculus, the limit exists only when two sided limit exist nd they re sme. However, in omplex Anlysis, if z 0 is n interior point of the domin of f nd the limit is to exist, the inequlity f(z) w 0 < ε must hold for ll points in the deleted neighborhood 0 < z z 0 < δ. Thus, the symbol z z 0 implies tht z is llowed to pproch z 0 in rbitrry mnner, not just from some prticulr direction. Exmple.4.3. onsider the limit of f(z) = z z s z pproch of 84

11 .5 Theorems on Limits. Theorem.5.1. Suppose tht f(z) = u(x, y) + iv(x, y), z 0 = x 0 + iy 0, w 0 = u 0 + iv 0. Then if nd only if lim u(x, y) = u 0, (x,y) (x 0,y 0 ) lim z z 0 f(z) = w 0 lim v(x, y) = v 0. (x,y) (x 0,y 0 ) Theorem.5.. Suppose tht lim f 0 (z) = w 0, nd lim F (z) = W 0. z z 0 z z0 Then lim (f(z) + F (z)) = w 0 + W 0 = lim f(z) + lim F (z), z z 0 z z0 z z0 ( ) ( ) lim (f(z)f (z)) = w 0 W 0 = lim f(z) lim F (z), z z 0 z z 0 z z 0 lim z z 0 f(z) F (z) = w 0 W 0 = lim z z 0 f(z) lim z z0 F (z), (if W 0 0). 11 of 84

12 .6 Limits Involving the Point t Infinity. We introduce n idel point which we cll the point t infinity, denoted by. The points in the plne together with the point t infinity form the extended complex plne. We gree tht every stright line shll pss through the point t infinity. By contrst, no hlf plne shll contin the idel point. How to visulize the point t infinity? onsider unit sphere S centered t the origin nd think of the complex plne s the plne pssing through the equtor of the unit sphere S. Let N be the north pole on the sphere S. (1) To ech point z in the complex plne, there corresponds exctly one point P on the surfce of the sphere, which cn be found s the intersection point between the sphere nd the stright line connecting z nd N. () On the other wy, to ech point P on the sphere, there corresponds exctly one point z in the complex plne, which cn be found s the intersection point between the complex plne nd the stright line connecting P nd N. (3) We let the north pole N correspond to the point t infinity,. From (1) to (3), we obtin one to one correspondence between the points of the sphere nd the points of the extended complex plne. The sphere is known s the Riemnn sphere nd the correspondence is clled stereogrphic 1 projection. In fct, except (0, 0, 1), to point (x 1, x, x 3 ) on the unit sphere S centered t the origin, we cn ssocite complex number z = x 1 + ix 1 x 3, nd this correspondence is one to one. Indeed, from the mp bove, we obtin ( ) ( ) z x1 + ix x1 + ix = z z = 1 x 3 1 x 3 where x 1 + x + x 3 = 1 is used. It implies Further computtion gives = x 1 + ix 1 x 3 x 1 ix 1 x 3 = x 1 + x (1 x 3 ) = 1 x 3 (1 x 3 ) = 1 + x 3 1 x 3, x 3 = z 1 z + 1. z + z = x 1 1 x 3 = ( 1 + z ) x 1, x 1 = z + z 1 + z, similrly, x = z z i ( 1 + z ). The correspondence cn be completed by letting the point t infinity correspond to (0, 0, 1), nd we cn thus regrd the sphere s representtion of the extended plne or of the extended number system. We note tht the hemisphere x 3 < 0 corresponds to the disk z < 1 nd the hemisphere x 3 > 0 to its outside z > 1. Moreover, for ech smll positive rel number ε, those points in the complex plne exterior to the circle z = 1/ε correspond to the points on the sphere close to N. We thus cll the set z > 1/ε n ε neighborhood, or just neighborhood, of. Unless specificlly mentioned, when we refer point z, it mens point in the finite plne. Since we hve the point t infinity,, we cn hve the following theorem. Theorem.6.1. If z 0 nd w 0 re points in the z nd w plnes, respectively, then lim f(z) = if nd only if lim z z 0 z z 0 1 lim f(z) = w 0 if nd only if lim f z z 0 lim f(z) = if nd only if lim z z 0 f(z) = 0, ( ) 1 = w 0, z 1 f(1/z) = 0. 1 stereogrphy: the representtion of three dimensionl things on two dimensionl surfce 1 of 84

13 Proof. 1. Suppose lim f(z) = nd we prove lim = 0. By the hypothesis, for ech ε > 0, there z z 0 f(z) exists δ > 0 such tht f(z) > 1 whenever 0 < z z 0 < δ. ε It implies z z f(z) 0 < ε whenever 0 < z z 0 < δ, 1 1 tht is, lim = 0. By the similr rgument, we cn prove lim = 0 implies lim f(z) =. z z 0 f(z) z z 0 f(z) z z ( ) 0 1. Suppose lim f(z) = w 0 nd we prove lim f = w 0. By the hypothesis, for ech ε > 0, there z z 0 z exists δ > 0 such tht f(z) w 0 < ε whenever z > 1 δ. When we replce z by 1/z, the inequlities bove become ( ) 1 f w 0 < ε whenever 0 < z 0 < δ, z ( ) 1 tht is, lim f = w 0. By the similr rgument, we cn prove the other impliction. z 0 z 3. Suppose lim f(z) = nd we prove lim z z 0 1 f(1/z) exists δ > 0 such tht f(z) > 1 whenever z > 1 ε δ. When we replce z by 1/z, the inequlities bove become 1 f(1/z) 0 < ε whenever 0 < z 0 < δ, = 0. By the hypothesis, for ech ε > 0, there 1 tht is, lim z 0 f(1/z) = 0. By the similr rgument, we cn prove the other impliction. Exmple.6.. ompute the limits: z + i 1. lim z z + 1,. lim z 1 z 1 z 1, lim z 3 1 z z + 1, lim z 1 1 z 1, lim z 1 4z (z 1), lim z + 1 z z 1, 1 (z 1), lim iz z 1 z + 1 lim z lim z z + iz 13 of 84

14 .7 ontinuity. Definition.7.1. A function f is sid to be continuous t z 0 if lim f(z) = z 0. z z 0 If f is continuous t ech point in region R, then the function is sid to be continuous in the region. If complex function f is not continuous t point z 0, then we sy tht f is discontinuous t z 0. The eqution bove contins the existence of the limit lim z z 0 f(z) nd the vlue f(z 0 ). Exmple.7.. (1). Is the function f(z) = 1 continuous t z = ±i? 1 + z (). Is f(z) = z iz + continuous t z = 1 i? (3). Is the principl squre root function f(z) = z 1/ = z e i Arg(z)/ continuous t z = 1? The following results re esily deduced: 1. For continuous function f nd complex number c, cf is continuous.. Sum/Product/Quotient of two continuous functions is lso continuous. 3. If f is continuous, then so re Re (f(z)), Im (f(z)), complex conjugte f(z), nd f(z). Specificlly, for continuous f(z) = f(x, y) = u(x, y) + iv(x, y), the functions Re (f(z)) = u(x, y) nd Im (f(z)) = v(x, y) nd f(z) = u(x, y) iv(x, y) nd f(z) = u (x, y) + v (x, y) re continuous. 4. A complex polynomil is continuous in the entire complex plne. More importnt results re given in the theorems. Theorem.7.3. A composition g f of two continuous functions f nd g is itself continuous. Since the proof is lmost similr to the one for the rel vlued functions of rel vribles, we skip it. Theorem.7.4. If function f is continuous nd nonzero t point z 0, then there exists neighborhood of z 0 where f(z) 0 t ny point in tht neighborhood. Proof. Let f is continuous nd nonzero t z 0. Then, by the continuity, for ε = f(z 0) > 0, there exists δ > 0 such tht f(z) f(z 0 ) < ε = f(z 0) whenever z z 0 < δ. So if there is point z in the neighborhood z z 0 < δ t which f(z) = 0, then we hve the contrdiction f(z 0 ) < f(z 0). Therefore, we should hve f(z) 0 t ny point in the neighborhood z z 0 < δ. We recll in rel nlysis: function tht is continuous in compct (i.e., closed nd bounded) region R ttins both mximum nd minimum vlues on R. This result is lso vilble in complex nlysis through the following theorem, which is very useful in pplictions. Theorem.7.5. If function f is continuous throughout compct (i.e., closed nd bounded) region R, then there exists nonnegtive rel number M such tht where equlity holds for t lest one such z. f(z) M for ll points z in R, 14 of 84

15 Proof. We observe f(z) = u (x, y) + v (x, y) is continuous throughout R. Since the functions of rel vribles, u(x, y) nd v(x, y), ttin both mximum nd minimum vlues on R from rel nlysis, so f(z) should be bounded on R. Exmple.7.6. heck the continuity of the given function: f(z) = { z + 1 Re(z) z + 4z + 5, f(z) = for z 0 z 0 for z = 0, f(z) = x + i x + y for z 0 i for z = of 84

16 .8 Derivtives. Suppose z = x + iy nd z 0 = x 0 + iy 0. Then the chnge in z 0 is the difference z = z z 0 or z = x x 0 + i (y y 0 ) = x + i y. If complex function w = f(z) is defined t z nd z 0, then the corresponding chnge in the function is the difference w = f(z 0 + z) f(z 0 ). The derivtive of the function f is defined in terms of limit of the difference quotient w/ z s z 0. Definition.8.1. Suppose the complex function f is defined in neighborhood of point z 0. The derivtive of f t z 0, denoted by f (z 0 ), is provided this limit exists. f (z 0 ) = lim z 0 f(z 0 + z) f(z 0 ), z If the limit bove exists, then the function f is sid to be differentible t z 0. Two other symbols denoting the derivtive of w = f(z) re w nd dw. If the ltter nottion is used, then the vlue of derivtive t dz specified point z 0 is written dw dz. z=z0 Exmple.8.. Use the definition to find the derivtive, if it exists: f(z) = z, f(z) = z 5z. Answer. (1). f(z) = z hs the derivtive, f (z) = z. (heck by yourself.) (). f(z) = z 5z hs the derivtive, f (z) = z 5. (heck by yourself.) Exmple.8.3. Use the definition to find the derivtive of f(z) = z, if it exists. Answer. f(z) = z does not hve derivtive t ny point, i.e., it is not differentible nywhere. The reson is s follows. w z + z z z + z z = = = z z z z z. As z pproches 0 long the rel xis, i.e., z = ( x, 0) (0, 0), we hve nd so in this cse, z = x + i0 = x i0 = x + i0 = z, w z = z z = z z = 1, i.e., the limit long the rel xis is 1. Similrly, s z pproches 0 long the imginry xis, i.e., z = (0, y) (0, 0), we hve nd so in this cse, z = 0 + i y = 0 i y = (0 + i y) = z, w z = z z = z z = 1, i.e., the limit long the imginry xis is 1. Since the limit should be unique, it follows dw/dz does not exist nywhere. Exmple.8.4. Use the definition to find the derivtive of f(z) = z, if it exists. 16 of 84

17 Answer. f(z) = z hs derivtive, f (z) = 0, only t z = 0, i.e., f (0) = 0 nd it is not differentible except t z = 0. The reson is s follows. w z = z + z z z = (z + z) ( z + z ) z z z = z + z + z z z. As the solution of the exmple bove, the horizontl nd verticl pproches of z towrd the origin gives us, respectively, z = z, nd z = z. It implies w z w z = z + z + z z z = z + z + z z z = z + z + z, when z = ( x, 0), = z z z, when z = (0, y). Agin, by the uniqueness of the limit, s z pproches 0, we should hve Therefore, dw/dz cnnot exist when z 0. However, when z = 0, we hve z + z = z z, i.e., z = 0. w z w = z, nd lim z 0 z = lim z 0 z = 0. Tht is, dw/dz does exist when z = 0 nd its vlue t z = 0 is 0. Now, we my rise the following questions: I. (Reltion between ontinuity nd Differentibility) I.1 If f is continuous t point z 0, then is it differentible t tht point? Answer: No. See the Exmple.8.3 bove. f(z) = z is continuous everywhere but it is not differentible nywhere. I. If f is differentible t point z 0, then is it continuous t tht point? Answer: Yes. We hve the following fct: if f is differentible t z 0, then f is continuous t z 0. Proof. Suppose f (z 0 ) exists. Then, [ ] f(z) f(z0 ) lim [f(z) f(z 0 )] = lim (z z 0 ) z z 0 z z 0 z z [ 0 ] [ ] f(z) f(z 0 ) = lim lim (z z 0 ) = f (z 0 ) 0 = 0, z z 0 z z 0 z z 0 which implies i.e., f is continuous t z 0. lim f(z) = f(z 0 ), z z 0 II. (Reltion between Function nd Its Rel/Imginry Prts on Differentibility) II.1 If f(z) = u(x, y) + iv(x, y) is differentible t point z 0 = (x 0, y 0 ), then re its rel nd imginry prts, u(x, y) nd v(x, y), differentible t the point (x 0, y 0 )? Answer: Yes. 17 of 84

18 II. If both u(x, y) = Re (f(z)) nd v(x, y) = Im (f(z)) re differentible t point z 0 = (x 0, y 0 ), then is the function f(z) differentible t the point z 0 = (x 0, y 0 )? Answer: No. The reltions (especilly II.) in the prt II bove re very importnt in omplex Anlysis nd so it will be discussed in Section 1 uchy Riemnn Equtions in detil. We my briefly discuss the reltions through the Exmple.8.4 bove. The Exmple.8.4 sys tht function w = f(z) cn be differentible t point but nowhere else in ny neighborhood of tht point. From f(z) = z = x + y = u(x, y) + iv(x, y), we hve u(x, y) = x + y, v(x, y) = 0, which re differentible everywhere. So the Exmple.8.4 gives us function which is not differentible t point but its rel prt u(x, y) nd imginry prt v(x, y) re differentible t tht point. Moreover, if function f is differentible t point z, then both u(x, y) nd v(x, y) should be differentible there. Geometric interprettion of derivtive of complex function is not esy to study nd thus it will be postponed until hpter 9 onforml Mpping. 18 of 84

19 .9 Differentition Formuls. By the definition, we cn deduce the following formuls: for differentible functions f nd g nd complex number c, 1. (onstnt Rules) d dz c = 0; d dz (cf) = cf. (Power Rule) d dz zn = nz n 1, where n 0 is n integer; 3. (Sum Rule) d dz (f ± g) = f ± g 4. (Product Rule) d dz (fg) = f g + fg 5. (Quotient Rule) d ( ) f = f g fg, where g 0. dz g g 6. (hin Rule) d dz (g (f)) = g (f) f d dz z = 1 Proof of hin Rule. Let point z 0 be fixed t which f (z 0 ) exists. We write w 0 = f(z 0 ) nd ssume tht g (w 0 ) exists. Then there is some ε neighborhood w w 0 < ε of w 0 such tht for ll points w in tht neighborhood, we cn define function Φ hving the vlues Φ(w 0 ) = 0 nd Φ(w) = g(w) g(w 0) w w 0 g (w 0 ), when w w 0. (.9.1) Since g (w 0 ) exists, by tking the limit on both sides, we cn deduce i.e., Φ(w) is continuous t w 0. The eqution (.9.1) cn be rewritten by lim w w 0 Φ(w) = 0 = Φ(w 0 ), g(w) g(w 0 ) = [g (w 0 ) + Φ(w)] (w w 0 ) ( w w 0 < ε) (.9.) which is lso vlid when w = w 0. Since f is differentible t z 0 nd so continuous there, we cn choose δ > 0 such tht f(z) f(z 0 ) < ε whenever 0 z z 0 < δ. It llows us to replce w in the eqution (.9.) by f(z) where z lies in the δ deleted neighborhood of z 0. Hence, we get g(f(z)) g(f(z 0 )) = [g (f(z 0 )) + Φ(f(z))] (f(z) f(z 0 )) (0 < z z 0 < δ) Dividing the whole eqution by z z 0, which does not vnish in the δ deleted neighborhood, we hve g(f(z)) g(f(z 0 )) z z 0 = [g (f(z 0 )) + Φ(f(z))] f(z) f(z 0) z z 0 (0 < z z 0 < δ) Tking the limit s z pproches z 0, we deduce [ ] [ ] [ ] d d (g (f (z))) g (f(z 0 )) + lim Φ(f(z)) dz z z 0 dz f(z) z=z 0 = Since f is continuous t z 0 nd Φ is continuous t the point w 0 = f(z 0 ) nd Φ(w 0 ) = 0, so we hve lim Φ(f(z)) = Φ(f(z 0 )) = Φ(w 0 ) = 0. z z 0 Therefore, finlly we conclude [ ] [ ] d d (g (f (z))) = g (f(z 0 )) dz z=z 0 dz f(z), i.e., [g(f(z))] = g (f(z))f (z) z=z 0 t z = z 0. Since z 0 is rbitrry, the eqution holds for ny z. 19 of 84 z=z 0

20 Exmple.9.1. Find the derivtives. 1. f(z) = (z + i) 5. f(z) = sin z 3. f(z) = cos z 4. f(z) = e z 5. f(z) = 3z 4 5z 3 + z 6. f(z) = z 4z f(z) = ( iz + 3z ) 5 Exmple.9. (L Hopitls Rule). Estblish specil cse of l Hopitls rule. Suppose tht f(z) nd g(z) hve forml power series bout z = (e.g., f(z) = n n(z ) n ), nd f() = f () = f () = = f (k) () = 0, g() = g () = g () = = g (k) () = 0. If f (k+1) () nd g (k+1) () re not simultneously zero, show tht Wht hppens if g (k+1) () = 0? f(z) lim z g(z) = f (k+1) () g (k+1) (). Exmple.9.3. If g(z) = M, M > 0 for ll z in neighborhood of z = z 0, show tht if lim z z0 f(z) = 0, then lim z z 0 (f(z)g(z)) = 0. Exmple.9.4. Find the set of points where following functions re differentible. 1. f(z) = sin z. f(z) = tn z 3. f(z) = z 1 z f(z) = e 1/z 5. f(z) = z We end the section with nowhere differentible functions. In lculus/rel Anlysis, is there function which is not differentible nywhere, i.e., nowhere differentible? If so, wht is it? If not, why not? We rise the sme question in omplex Anlysis. Is there function which is not differentible nywhere? Exmple.9.5. Prove tht the following functions re nowhere differentible. 1. f(z) = x + 4iy. f(z) = Re (z) 3. f(z) = Im (z) 0 of 84

21 .10 uchy Riemnn Equtions. We will study one of the most importnt theorems in omplex Anlysis: Theorem with uchy Riemnn Equtions. Simply speking, the theorem gives the reltions on differentibility between complex function f nd its rel/imginry prts Re(f), Im(f). Theorem The function f(z) = u(x, y) + iv(x, y) is differentible t point z = x + iy of region in the complex plne if nd only if the prtil derivtives u x, u y, v x nd v y re continuous nd stisfy the uchy Riemnn equtions u x = v y, u y = v x (.10.1) t the point z = x + iy. Proof on Necessry ondition. We strt with the definition of the derivtive of f. f (z) = lim z 0 f(z + z) f(z). z We write the rel nd imginry prts of f(z), f(z) = u(x, y) + iv(x, y), nd compute the right hnd side of the eqution bove for(1) z = x rel nd () z = i y pure imginry (i.e., we tke the limit long the rel nd then long the imginry xis). Then, for cse (1), f (z) = For cse (), f (z) = lim x 0 lim y 0 ( u(x + x, y) u(x, y) + i x ( u(x, y + y) u(x, y) + i y Setting equtions (.10.) nd (.10.3) equl yields v(x + x, y) v(x, y) x ) = u x (x, y) + iv x (x, y). (.10.) ) i (v(x, y + y) v(x, y)) = iu y (x, y) + v y (x, y). (.10.3) i y u x = v y, u y = v x, which re clled the uchy Riemnn equtions. The proof on sufficient condition is skipped. Exmple.10.. Show tht the function is differentible everywhere nd tht f (z) = z. f(z) = z = x y + ixy Using the necessry condition in the theorem for the differentibility of function f t point, we find the points t which f is not differentible. Exmple Find the set of points t which the given function is differentible. 1. f(z) = z. f(z) = z 3. f(z) = x + y + i(y x) 4. P n (z) = z + + n z n = n k z k ( polynomil of degree n) k=0 Using the sufficient condition, we cn find the derivtive of f. Exmple Find the derivtive of the given function t given point. 1. f(z) = e z t ny point z 1 of 84

22 . f(z) = z t z 0 = 0 Exmple Find rel constnts, b, c nd d so tht the given function is differentible. 1. f(z) = 3x y i (x + by 3). f(z) = x + xy + by + i ( cx + dxy + y ) We end with the following theorem, which is direct consequence of the uchy Riemnn equtions. Theorem.10.6 (onstnt Functions). Suppose the function f(z) = u(x, y) + iv(x, y) is differentible in region R in the complex plne. 1. If f(z) is constnt in R, then so is f(z).. If f (z) = 0 in R, then f(z) = c in R, where c is constnt..11 Sufficient onditions for Differentibility. of 84

23 .1 Polr oordintes. We cn restte the Theorem in the previous section using polr coordintes x = r cos θ nd y = r sin θ. Theorem.1.1. The function f(z) = u(r, θ) + iv(r, θ) is differentible t point z = re iθ of region in the complex plne if nd only if the prtil derivtives u r, u θ, v r nd v θ re continuous nd stisfy the polr coordintes version of uchy Riemnn equtions t the point z = re iθ. The derivtive of f(z) = u(r, θ) + iv(r, θ) is ru r = v θ, u θ = rv r (.1.1) f (z) = e iθ (u r + iv r ). Key Ide of Proof. We prove only the key ide, the polr coordintes version of uchy Riemnn equtions. From x = r cos θ, y = r sin θ, r = x + y nd tn θ = y, we get x ( ) r = ( x + y ), r r x x x = x, ( y ) (tn θ) =, sec θ θ x x x x = y x, ( ) r = ( x + y ), r r y y y = y, y (tn θ) = ( y, sec y x) θ θ y = 1 x, r x = x r = r cos θ = cos θ r θ x = y x sec θ = r sin θ r cos θ sec θ = sin θ r r y = y r = r sin θ = sin θ r θ y = 1 x sec θ = 1 r cos θ sec θ = cos θ r. ombining the results bove with the chin rule for differentiting rel vlued functions of two vribles implies the differentil reltionships: By the reltionships, we get u x = cos θ u r sin θ r v x = cos θ v r sin θ r x = r r x + θ θ x = cos θ r sin θ r θ y = r r y + θ θ y = sin θ r + cos θ r θ. u θ, v θ, u y = sin θ u r + cos θ r v y = sin θ v r + cos θ r The uchy Riemnn equtions sys u x = v y nd u y = v x. Applying them bove, we deduce cos θ u r sin θ r sin θ u r + cos θ r u θ = u x = v y = sin θ v r + cos θ r v θ, u θ = u y = v x = cos θ v r + sin θ r v θ. u θ, (.1.) v θ. (.1.3) Multiplying the first of these equtions by cos θ, the second by sin θ, nd dding yields ru r = v θ. Similrly, multiplying the first by sin θ, the second by cos θ, nd dding yields u θ = rv r. Tht is, ru r = v θ, u θ = rv r. Using the reltions of (.1.) nd (.1.3) in f (z) = u x + iv x with the results bove yields f (z) = cos θ u r sin θ ( u θ + i cos θ v r sin θ ) v θ r r = (cos θ i sin θ) (u r + iv r ) = e iθ (u r + iv r ). 3 of 84

24 Exmple.1.. onsider the function where z 0. f(z) hs the component nd we hve We observe f(z) = 1 z = 1 re = 1 iθ r e iθ = 1 (cos θ i sin θ), r u(r, θ) = cos θ θ, v(r, θ) = sin r r u r = cos θ r, u θ = sin θ r, v r = sin θ r, v θ = cos θ r. ru r = cos θ r = v θ u θ = sin θ r t ny nonzero point. Therefore f (z) exists when z 0 nd it is in fct ( f (z) = e iθ (u r + iv r ) = e iθ cos θ + i sin θ ) r r Exmple.1.3. onsider the function = rv r = 1 r e iθ (cos θ i sin θ) = 1 r e iθ e iθ = 1 (re iθ ) = 1 z. f(z) = z 1/3 = [ 3]re iθ/3, (r > 0, α < θ < α + π), where α is fixed rel number. Prove tht f is differentil everywhere. Proof. We observe f(z) = [ 3]re iθ/3 = [ 3]r ( cos θ 3 + i sin θ ) = [ 3]r cos θ i [ 3]r sin θ 3 = u(r, θ) + iv(r, θ), i.e., u(r, θ) = [ 3]r cos θ 3, v(r, θ) = [ 3]r sin θ 3. We compute u r = r /3 3 cos θ 3, [ 3]r u θ = sin θ 3 3, v r = r /3 3 sin θ 3, v θ = [ 3]r cos θ 3 3, nd observe ru r = [ 3]r cos θ 3 3 = v θ t ny point. Therefore f (z) exists everywhere nd it is in fct [ 3]r u θ = sin θ 3 3 = rv r ( r f (z) = e iθ (u r + iv r ) = e iθ /3 cos θ i r /3 sin θ ) 3 3 ( = r /3 e iθ cos θ i sin θ ) = r /3 e iθ e iθ/3 = 1 1 ( [ ) = 1 3]re iθ/3 3f (z). 4 of 84

25 The exmple bove shows us d dz z1/3 = 1 3(z 1/3 ), which is n exmple of Power Rule with non integer power. Be creful: The power rule in Section 0. Differentition Formuls sys d dz zn = nz n 1, for integer n. Exercise.1.4. Discuss the differentibility of the given function in the indicted domin of definition. If the derivtive exists, then find it. 1. f(z) = z = e iθ ( = the whole complex plne). f(z) = r e iθ/ (r > 0 nd α < θ < α + π) 3. f(z) = e θ cos(ln r) + ie θ sin(ln r) (r > 0 nd 0 < θ < π) 5 of 84

26 .13 Anlytic Functions. Definition Let f be complex function. 1. If f hs derivtive t ech point in some neighborhood of z, then f is nlytic (or regulr, holomorphic) t z.. If f is nlytic t ech point in domin D or region R, then f is nlytic in D or R. (Recll: domin D mens connected open set nd region R mens domin with some, none, or ll of its boundry points.) 3. If f is nlytic in the entire complex plne, then f is n entire function. 4. If f is not nlytic t z 0 but it is nlytic t some point in every neighborhood of z 0, then z 0 is singulr point or singulrity of f. Exmple f(z) = z is nowhere nlytic. It is differentible only t z = 0. So there is no neighborhood of z = 0 in which f is differentible. Moreover, f(z) = z hs no singulr point.. f(z) = 1/z is nlytic t ech point z 0 nd z = 0 is singulr point of f. 3. A polynomil P n (z) = z + + n z n is n entire function. (Are trigonometric functions nd exponentil function entire? The nswer will be discussed lter.) Using the theorems on previous sections, we cn deduce the following results. Theorem Let f nd g be nlytic nd c ny complex number. Then, cf, f ± g, fg, nd f/g with g 0, nd f g (composition) re nlytic. The following property of nlytic functions is especilly useful. Theorem If f (z) = 0 everywhere in domin D, then f(z) must be constnt throughout D. Proof. Step 1: Let f(z) = u(x, y) + iv(x, y) stisfy f (z) = 0 in D. Tht is, f (z) = u x + iv x = 0, u x = 0, v x = 0. Since f is differentible in D, it stisfies the uchy Riemnn equtions in D nd so we hve u y = 0 = v y. Tht is, t ech point in D, u x = 0 = u y = v x = v y. Step : We show tht u(x, y) is constnt long ny line segment L from point P to point P nd lying entirely in D. Let s denote the distnce long L from the point P nd let U denote the unit vector long L in the direction of incresing s. Then from lculus, the directionl derivtive of u long L is given by du ds = ( u) U, where u = u x i + u y j is the grdient of u. Since u x = 0 = u y, the grdient vnishes nd so the directionl derivtive becomes zero long L. It mens u is constnt long L. By generlizing this rgument, we deduce tht u(x, y) = is constnt in D. By the similr rgument, constnt v(x, y) = b cn be obtined. Therefore, we find tht f(z) = u(x, y) + iv(x, y) = + ib t ech point in D. Before we end this section, we summrize the reltion between differentibility nd nlyticity. 1. If f is nlytic t z 0, then f is differentible t z 0.. On the other hnd, even if f is differentible t z 0, f my not be nlytic t z 0. (ex: f(z) = z t z = 0.) However, if f is differentible in some neighborhood of z 0, then by definition, f is nlytic t z 0. If f is nlytic t point on the boundry, then it mens f is nlytic in n open set contining the boundry point. 6 of 84

27 .14 Exmples. We study vrious exmples. Exmple Since f(z) = z nd g(z) = (z 3)(z + 1) re entire (i.e., nlytic everywhere), so the rtionl function h = f/g is nlytic everywhere except t z = ± 3 nd z = ±i. Moreover, those points re singulrities of the rtionl function h. The derivtive of h is obtined by the Quotient Rule. Since the nlyticity is deduced from the differentibility, we use the uchy Riemnn equtions to check the differentibility nd thus eventully the nlyticity. Exmple.14.. Trigonometric functions (sin z nd cos z) re entire. We prove this for f(z) = sin z. Proof. Let us tke the followings formlly (will be discussed lter): f(z) = sin (x + iy) = sin x cos (iy) + cos x sin (iy) = sin x cosh y + i cos x sinh y, where sinh y nd cosh y re hyperbolic trigonometric functions with properties sinh z = i sin (iz), cosh z = cos (iz), Hence, from f(z) = u(x, y) + iv(x, y), we hve d dz sinh z = cosh z, d dz cosh z = sinh z. u(x, y) = sin x cosh y, v(x, y) = cos x sinh y, which re differentible for ny x nd y. Now we check the uchy Riemnn equtions: u x = cos x cosh y = v y, u y = sin x sinh y = v x, t ny point. Since the component functions of f stisfy the equtions t ny point, thus f is differentible everywhere, i.e., nlytic everywhere, i.e., entire. Exmple Suppose f(z) = u(x, y) + iv(x, y) nd its conjugte f(z) = u(x, y) iv(x, y) re both nlytic in domin D. Then f must be constnt in D. Proof. We write f(z) s f(z) = U(x, y) + iv (x, y), where U(x, y) = u(x, y) nd V (x, y) = v(x, y). (.14.1) Becuse of the nlyticity of f, the uchy Riemnn equtions hold in D. Also the nlyticity of f(z) tells us u x = v y, u y = v x (.14.) U x = V y, U y = V x. In view of reltions (.14.1), these lst two equtions cn be written u x = v y, u y = v x. (.14.3) By dding corresponding sides of the first of equtions (.14.) nd (.14.3), we find tht u x = 0 in D. Similrly, subtrction involving corresponding sides of the second of equtions (.14.) nd (.14.3) revels tht v x = 0. Hence, f (z) = u x + iv x = 0 + i0 = 0, nd by the Theorem in the previous section, f is constnt throughout D. 7 of 84

28 Exmple Let f be nlytic in domin D nd let its modulus f be constnt in D. Then f must be constnt in D. Proof. By the hypothesis, f(z) = c, for ll z in D, where c is rel constnt. If c = 0, then f = 0 t ll points in D, which implies constnt f = 0 in D. Now let c 0. Tking squred the eqution gives f = c, f f = c. Since c 0, so neither f nor f should be zero, which llows us to hve f = c f for ll points in D. Since f nd c re nlytic in D, the quotient f is lso nlytic there. The Exmple.14.3 bove sys tht if both g nd ḡ re nlytic in domin, then g is constnt there. By this result, we conclude tht f is lso constnt in D. ombining Exmples.14.3 nd.14.4, we cn sy for n nlytic function f in domin D, 1. If f is nlytic, then f is constnt.. If f is constnt, then f is constnt. Exercise Find ech region on which f(z) = xy + i(x y ) is differentible nd nlytic, respectively. 8 of 84

29 .15 Hrmonic Functions. Definition A rel vlued function H(x, y) of rel vribles is clled to be hrmonic in domin D if 1. first nd second derivtives, H x, H y, H xx, H yy, H xy nd H yx, re continuous. H xx nd H yy stisfies which is clled the Lplce eqution. H H xx + H yy = 0 for ll (x, y) in D Exmple.15.. The function T (x, y) = e y sin x is hrmonic in ny domin of the xy plne, prticulrly, in the semi infinite verticl strip 0 < x < π nd y > 0. Lplce eqution is stisfied nd the followings hold T (x, 0) = sin x, lim T (x, y) = 0, y T (0, y) = 0 = T (π, y). Theorem If f(z) = u(x, y) + iv(x, y) is nlytic in domin D, then its component functions u nd v re hrmonic in D. Proof. We borrow result from hpter 4 Integrls: If function w = f(z) = u(x, y) + iv(x, y) of complex vrible is nlytic t z 0 = x 0 + iy 0, then both u nd v hve continuous prtil derivtives of ll orders t (x 0, y 0 ). Assume f is nlytic in D. Then its component functions u nd v should stisfy the uchy Riemnn equtions, u x = v y, u y = v x t ny point in D. By the Theorem bove from hpter 4 Integrls, we hve t ny point in D, u xx = (u x ) x = (v y ) x = v yx, u xy = (u x ) y = (v y ) y = v yy, u yx = (u y ) x = ( v x ) x = v xx, u yy = (u y ) y = ( v x ) y = v xy. Agin by the Theorem bove, u x, u y, v x nd v y re continuous. From Advnced lculus, Hence, we deduce u xy = u yx, v xy = v yx. u xx + u yy = v yx v xy = v xy v xy = 0, v xx + v yy = u yx + u xy = u xy + u xy = 0. Tht is, u nd v re hrmonic in D. Be creful! The converse of the Theorem is not true. Exmple The functions p(x, y) = xy nd q(x, y) = x y re hrmonic through the plne, becuse p xx + p yy = 0 nd q xx + q yy = 0. However, the function f(z) = iz = xy + i ( x y ) = p(x, y) + iq(x, y) is not nlytic nywhere, becuse its component functions do not stisfy the uchy Riemnn equtions in n open set: p x = y = y = q y, p y = x = x = q x, only t z = (x, y) = 0 Tht is, f(z) is differentible only t z = 0 nd so it is nlytic nowhere. 9 of 84

30 Exmple onsider f(z) = ie iz = e y sin x ie y cos x = u(x, y) + iv(x, y). We observe f(z) is entire, i.e., nlytic everywhere, by checking the uchy Riemnn equtions with u nd v. The Theorem bove implies tht u(x, y) = e y sin x nd v(x, y) = e y cos x re hrmonic everywhere. We lredy checked in Exmple.15. tht u(x, y) is hrmonic everywhere nd one cn esily check tht v is lso hrmonic everywhere. Exmple onsider f(z) = i where z 0. z f(z) = i z = iz z 4 = xy + i (x y ) (x + y ) = u(x, y) + iv(x, y), u(x, y) = xy (x + y ), v(x, y) = x y (x + y ). Since u nd v stisfy the uchy Riemnn equtions t ny point except t (x, y) = (0, 0), so f(z) is nlytic everywhere except t z = 0 nd by the Theorem, u nd v re hrmonic everywhere except t (x, y) = (0, 0). Definition Suppose two rel vlued functions p(x, y) nd q(x, y) of rel vribles re hrmonic in domin D nd their first prtil derivtives stisfy the uchy Riemnn equtions through D, i.e., Then q is sid to be hrmonic conjugte of p. Let us remrk three points. p x = q y, p y = q x. 1. (Existence) For hrmonic function p in domin of certin type, does its hrmonic conjugte exist? Yes, it does. onfer Section 104 Hrmonic onjugtes in hpter 9 onforml Mpping.. (Uniqueness) For given function, its hrmonic conjugte is unique except for n dditive constnt. 3. (ommuttivity) If q is hrmonic conjugte of p, then in generl, p my not be hrmonic conjugte of q. See the Exmple.15.9 following the Theorem below. Theorem A function f(z) = u(x, y) + iv(x, y) is nlytic in domin D if nd only if v is hrmonic conjugte of u. Proof. Suppose v is hrmonic conjugte of u in D. Then first derivtives of u nd v stisfy the uchy- Riemnn equtions in D. By the Theorem in Section, f = u + iv is nlytic in D. Suppose f = u + iv is nlytic in D. Then the Theorem.15.3 implies u nd v re hrmonic in D. The Theorem in Section 1 gives tht first derivtives of u nd v stisfy the uchy-riemnn equtions. Hence, v is the hrmonic conjugte of u. Exmple Suppose Then, we observe: u(x, y) = x y, v(x, y) = xy. 1. The function u(x, y) + iv(x, y) = x y + xyi = z is entire, i.e., nlytic everywhere. So by the Theorem.15.8, v is the hrmonic conjugte of u.. However, v(x, y) + iu(x, y) = xy + i (x y ) = iz is not nlytic nywhere by the Exmple By the Theorem.15.8, u is not hrmonic conjugte of v. For hrmonic function u, there exists unique (up to constnt) hrmonic conjugte of u. This implies tht every hrmonic function is the rel prt of n nlytic function. As the lst subject in this section, for given hrmonic function, how cn we find its hrmonic conjugte nd n nlytic function of which rel prt is the given hrmonic function? 30 of 84

31 Exmple The function u(x, y) = y 3 3x y is hrmonic everywhere, becuse u = u xx + u yy = = 0 t ny point (x, y). Find the hrmonic conjugte v of u nd the nlytic function f such tht Re (f) = u. Answer. First derivtives of u nd its hrmonic conjugte v should stisfy the uchy Riemnn equtions: u x = v y, u y = v x. (.15.1) From the first eqution, v y = x ( y 3 3x y ) = 6xy, v = ( 6xy) dy = 3xy + ϕ(x), where ϕ(x) is n rbitrry function of x. Using the second eqution in (.15.1), we hve 3y 3x = ( y 3 3x y ) = u y = v x = ( 3xy + ϕ(x) ) = 3y ϕ (x) y x ϕ (x) = 3x, ϕ(x) = x 3 +, where is rel constnt of integrtion. Therefore, we deduce v(x, y) = 3xy + x 3 +. The corresponding nlytic function is f(z) = u(x, y) + iv(x, y) = y 3 3x y + i ( 3xy + x 3 + ). When we express it in terms of z, the function becomes f(z) = y 3 3x y + i ( 3xy + x 3 + ) = i ( z 3 + ). Exercises: Solve ll the problems on pge 81 8 in the book. The followings re chllengeble. onsider the function u(x, y) = x 3 3xy 5y. 1. Verify tht u is hrmonic in the entire complex plne.. Find the hrmonic conjugte function v of u. 3. Find the nlytic function f hving u s its rel prt. onsider the function u(x, y) = e x y cos (xy). 1. Verify tht u is hrmonic in n pproprite domin.. Find the hrmonic conjugte function v of u. 3. Find the nlytic function f hving u s its rel prt nd stisfying f(0) = 1. onsider the function v(x, y) = x x + y. 1. Verify tht v is hrmonic in n pproprite domin.. Find the nlytic function f hving v s its imginry prt. 3. Express f in terms of z. onsider the polynomil of form: u(x, y) = x 3 + bx y + cxy + dy 3, where, b, c nd d re constnts. 31 of 84

32 1. Find the most generl hrmonic polynomil of the given form.. Find the hrmonic conjugte function v of u. 3. Find the nlytic function f hving u s its rel prt. Suppose f(z) = u(r, θ) + iv(r, θ) is nlytic. Use the uchy Riemnn equtions in polr coordintes form to show tht u stisfies the Lplce eqution in polr coordintes: r u rr + ru r + u θθ = 0. Prove the proposition: If u is hrmonic nd v is its hrmonic conjugte, then ϕ = uv is lso hrmonic. Hints. Suppose f = u + iv is nlytic. Then f is lso nlytic. Exmine f..16 Uniquely Determined Anlytic Functions..17 Reflection Principle. 3 of 84

33 hpter 3 ELEMENTARY FUNTIONS 3.1 The Exponentil Function. Definition The function e z defined by is clled the (complex) exponentil function. exp z = e z = e x e iy, or e z = e x cos y + ie x sin y When y = 0, i.e., z = x + i0, the definition gives e z = e x. So for z = 1/n + i0 = 1/n with n Z +, we hve e 1/n = e x, which is one rel number with x = 1/n. Tht is, the positive n th root e 1/n of the Euler number e is defined s one rel number, i.e., e 1/n = e x, where x = 1/n with n Z +. Be creful: ccording to Section 9 Roots of omplex Numbers, the positive n th root z 1/n of z is the set of n different complex numbers. However, the definition in Section 9 does not hold for e 1/n. Tht is, e 1/n is exceptionlly not the set of n different complex numbers. Now let us see some properties. Theorem 3.1. (Algebric Properties of e z ). If z 1 nd z re complex numbers, then 1. e 0 = 1. e z 0 for ny z 3. e z 1 e z = e z 1+z e z 1 4. e = z ez 1 z 5. (e z 1 ) n = e nz 1, n Z 6. e z = e z Proof. 1. e 0 = e 0+i0 = e 0 cos 0 + ie 0 sin 0 = 1 + i0 = 1.. e z = e x cos y + ie x sin y implies e z = e x 0. Hence, e z 0 for ny z. (Recll: z = 0 if nd only if z = 0.) 3. With z 1 = x 1 + iy 1 nd z = x + iy, we hve e z 1 e z = e x 1+iy 1 e x +iy = e x 1 e iy 1 e x e iy = e x 1+x e i(y 1+y ) = e x 1+x +i(y 1 +y ) = e z 1+z. Another wy is s follows. e z 1 e z = (e x 1 cos x 1 + ie x 1 sin x 1 ) (e x cos x + ie x sin x ) = e x 1 e x cos x 1 cos x e x 1 e x sin x 1 sin x + i (e x 1 e x cos x 1 sin x + e x 1 e x sin x 1 cos x ) = e x 1+x (cos x 1 cos x sin x 1 sin x ) + ie x 1+x (cos x 1 sin x + sin x 1 cos x ) = e x 1+x cos (x 1 + x ) + ie x 1+x sin (x 1 + x ) = e z 1+z. 4. Using the result 3 bove, we hve e z 1 z e z = e z 1 z +z = e z 1. Dividing by e z, we get ez 1 5. De Moivre s Formul implies, for n Z, 6. (e z 1 ) n = (e x cos y + ie x sin y) n = [e x (cos y + i sin y)] n = (e x ) n (cos y + i sin y) n = e nx (cos (ny) + i sin (ny)) = e nx cos (ny) + ie nx sin (ny) = e nz 1. e z = ez 1 z. e z = e x cos y + ie x sin y = e x cos y ie x sin y = e x cos ( y) + ie x sin ( y) = e x e iy = e x iy = e z. 33

34 Theorem (Anlyticity of e z ). The exponentil function e z is entire nd its derivtive is given by d dz ez = e z. It cn be proved by using the uchy Riemnn equtions. Exercise Find the derivtive of ech of the following functions: (1) iz 4 ( z e z) nd () e z (1+i)z+3. Anlogous to rel periodic functions, we sy tht complex function f is periodic with period T if for ll complex z, f(z + T ) = f(t ). Theorem (Properties Only for omplex Exponentil Function). The following properties does not hold for rel exponentil function. 1. e z my be negtive rel number.. e z is periodic with pure imginry period πi. 3. For given w 0 0, there my mny vlues of z such tht e z = w 0. It implies tht the inverse of the exponentil function cn hve mny vlues. Proof. 1. When y = nπ, n Z, we hve e z = e x cos y + ie x sin y = e x ( 1) n + i0 = ( 1) n e x = Tht is, e (m+1)πi = 1, nd e mπi = 1, m Z. From 1 bove, we hve e mπi = 1, m Z. It implies e z+πi = e z e πi = e z 1 = e z. Hence, e z is periodic with the pure imginry period πi. 3. For given w 0 = w 0 e i Arg(w 0) 0, the eqution implies e x e iy = e z = w 0 = w 0 e i Arg(w 0) { e x for n = m e x for n = m + 1, w 0 = e x, Arg (w 0 ) = y + nπi, i.e., x = ln w 0, y = Arg (w 0 ) + nπ, m Z. where n Z. Tht is, z = x + iy stisfying x = ln w 0 nd y = Arg (w 0 ) + nπ is the solution of the eqution e z = w 0. Since y hve infinitely mny vlues with n Z, there re mny solutions of the eqution. From the result 3 in the Theorem bove, we my expect the inverse (clled logrithmic function) of the exponentil function is multiple vlued function, which will be studied in next section. Exmple Find z such tht e z = 1 + i. Answer. Since 1 + i hs the modulus 1 + i = nd the principl rgument Arg (1 + i) = π/4, so we hve e x e iy = e z = 1 + i = e πi/4. It implies e x =, y = π 4 + nπ, i.e., x = ln = ln ( ) 1, y = 4 + n π, where n Z. Thus, we conclude the solution of e z = 1 + i is obtined by z = x + iy = ln ( ) 1 + i 4 + n π, which is not unique, becuse of n Z. Exercise Determine the complex function e z is nlytic. 34 of 84

35 3. The Logrithmic Function. The min gol of this section is to study the inverse of the exponentil function, i.e., for given z 0, the solution w of the eqution e w = z. Let us write z = re iθ, π < Θ π nd w = u + iv. Then the eqution e w = z becomes where n Z. It implies e u e iv = e u+iv = z = re iθ, i.e., e u = r, v = Θ + nπ, u = ln r, v = Θ + nπ, i.e., w = u + iv = ln r + i (Θ + nπ), where ln r = log e r. Since z = r nd Arg (z) = Θ, thus, we cn rewrite the solution w of the eqution e w = z s follows w = ln z + i (Arg (z) + nπ), i.e., e ln z +i(arg(z)+nπ) = z, n Z. Becuse of n Z, there re infinitely mny such w s. If we write log z := ln z + i (Arg (z) + nπ), we hve e log z = e ln z +i(arg(z)+nπ) = z, i.e., e log z = z. For this reson, we hve the following definition. Definition 3..1 (omplex Logrithm). The multiple vlued function log z, z 0, defined by log z = ln z + i Arg (z) = ln z + i (Arg (z) + nπ), n Z, is clled the complex logrithm or (complex) logrithmic function. Exmple 3... Find log z for z = 1 i 3. Answer. We compute z = nd Arg (z) = π/3. Putting them into the definition, we get log z = ln + i ( π3 ) + nπ = ln + i ( 3 ) + n π, n Z. Exercise Find ll complex solutions to ech of the following equtions: (1) e w = i, () e w = 1 + i nd (3) e w =. Be creful: log (e z ) z. Proof. We compute e z = e x+iy = e x nd Arg (e z = e x+iy ) = y + nπ. Putting them into the definition, log (e z ) = ln e z + i Arg (e z ) = ln e x + i (y + nπ) = x + iy + nπi = z + nπi, n Z. The logrithmic function is multiple vlued. We restrict the rgument in the logrithmic function nd deduce single vlued function. Definition 3..4 (Principl Vlue of omplex Logrithm). The complex function ln z, z 0, defined by ln z = ln z + i Arg (z), π < Arg (z) π, is clled the principl vlue of the complex logrithm log z. We observe (1) ln z is single vlued function. () log z = ln z + i (Arg (z) + nπ) = ln z + nπi, n Z. 35 of 84