4. Complex Integration: Cauchy Integral Theorem And
4. Complex integration: Cauchy integral theorem and Cauchyintegral formulasDefinite integral of a complex-valued function of a real variableConsider a complex valued function f (t) of a real variable t:f (t) u(t) iv(t),which is assumed to be a piecewise continuous function defined inthe closed interval a t b. The integral of f (t) from t a tot b, is defined asZ baf (t) dt Z bau(t) dt iZ bav(t) dt.1
Properties of a complex integral with real variable of integration1.ReZ bImZ baf (t) dt Z baRe f (t) dt Z bau(t) dt.2.af (t) dt Z baIm f (t) dt Z bav(t) dt.3.Z ba[γ1f1(t) γ2f2(t)] dt γ1Z baf1(t) dt γ2where γ1 and γ2 are any complex constants.Z baf2(t) dt,4.Z baf (t) dt Z ba f (t) dt.2
To prove (4), we considerZ bawhere φ ArgZ baf (t) dt e iφZ bf (t) dta!Z baf (t) dt f (t) dt . Since Re Z baZ baZ baZ bae iφf (t) dt,f (t) dt is real, we deduce thate iφf (t) dt e iφf (t) dt Z baZ baRe [e iφf (t)] dt f (t) dt.3
ExampleSuppose α is real, show that e2απi 1 2π α .SolutionLet f (t) eiαt, α and t are real. We obtainZ 2π0eiαt dt Z 2π0 eiαt dt 2π.The left-hand side of the above inequality is equal toZ 2π0iαtedt eiαtiα2π0 e2απi 1 . α Combining the results, we obtain e2απi 1 2π α , α is real.4
Definition of a contour integralConsider a curve C which is a set of points z (x, y) in the complexplane defined byx x(t),y y(t),a t b,where x(t) and y(t) are continuous functions of the real parametert. One may writez(t) x(t) iy(t), a t b. The curve is said to be smooth if z(t) has continuous derivativez ′(t) 6 0 for all points along the curve. A contour is defined as a curve consisting of a finite numberof smooth curves joined end to end. A contour is said to be asimple closed contour if the initial and final values of z(t) arethe same and the contour does not cross itself.5
Let f (z) be any complex function defined in a domain D in thecomplex plane and let C be any contour contained in D withinitial point z0 and terminal point z. We divide the contour C into n subarcs by discrete points z0, z1, z2,. . ., zn 1, zn z arranged consecutively along the direction of increasing t. Let ζk be an arbitrary point in the subarc zk zk 1 and form thesumn 1Xk 0f (ζk )(zk 1 zk ).6
Subdivision of the contour into n subarcs by discrete points z0, z1, · · · ,zn 1, zn z.7
We write zk zk 1 zk . Let λ max zk and take the limitklimn 1Xλ 0n k 0f (ζk ) zk .The above limit is defined to be the contour integral of f (z) alongthe contour C.If the above limit exists, then the function f (z) is said to be integrable along the contour C.If we writedx(t)dy(t)dz(t) i, a t b,dtdtdtthenZZ bdz(t)f (z) dz f (z(t))dt.dtCa8
Writing f (z) u(x, y) iv(x, y) and dz dx idy, we haveZCf (z) dz Zu dx v dy iCZ b"ZCu dy v dx#dx(t)dy(t) u(x(t), y(t))dt v(x(t), y(t))dtdta#Z b"dy(t)dx(t) iu(x(t), y(t))dt. v(x(t), y(t))dtdtaThe usual properties of real line integrals are carried over to theircomplex counterparts. Some of these properties are:(i)Z(ii)ZCf (z) dz is independent of the parameterization of C; Cf (z) dz ZCf (z) dz, where C is the opposite curve of C;(iii) The integrals of f (z) along a string of contours is equal to thesum of the integrals of f (z) along each of these contours.9
ExampleEvaluate the integralI1dz,C z z0where C is a circle centered at z0 and of any radius. The path istraced out once in the anticlockwise direction.SolutionThe circle can be parameterized byz(t) z0 reit , 0 t 2π,where r is any positive real number. The contour integral becomes2π2π ireit1dz(t)1dz dt dt 2πi.C z z00 z(t) z0 dt0 reitThe value of the integral is independent of the radius r.IZZ10
ExampleEvaluate the integral(i)ZC z 2 dz and (ii)Z1dz,2Czwhere the contour C is(a) the line segment with initial point 1 and final point i;(b) the arc of the unit circle Im z 0 with initial point 1 and finalpoint i.Do the two results agree?11
Solution(i)ConsiderZC z 2 dz,(a) Parameterize the line segment byz 1 (1 i)t,0 t 1,so that z 2 ( 1 t)2 t2anddz (1 i) dt.The value of the integral becomesZC z 2 dz Z 1022(2t 2t 1)(1 i) dt (1 i).312
(b) Along the unit circle, z 1 and z eiθ , dz ieiθ dθ. The initialpoint and the final point of the path correspond to θ π andθ π2 , respectively. The contour integral can be evaluated asZC z 2 dz Zπ2πieiθ dθ eiθπ2 1 i.πThe results in (a) and (b) do not agree. Hence, the value of thiscontour integral does depend on the path of integration.13
(ii)ZConsider1dz.2Cz(a) line segment from 1 to iZ111111 idz dt 1 1 i.22 1 (1 i)t 0iCz0 [ 1 (1 i)t]Z(b) subarc from 1 to iZZπ2 π11 iθ iθ 2 1 i.dz iedθ e22iθCzπ eπ14
Estimation of the absolute value of a complex integralThe upper bound for the absolute value of a complex integral canbe related to the length of the contour C and the absolute value off (z) along C. In fact,ZCf (z) dz M L,where M is the upper bound of f (z) along C and L is the arc lengthof the contour C.15
We considerZCf (z) dz Z ba Z b Z bf (z(t))dz(t)dtdtdz(t) f (z(t)) dtdtadz(t)Mdtdtav!Z buu dx(t) 2t M dtady(t)dt!2dt M L.16
ExampleShow thatZ1dz 2, where C is the line segment joining 1 i and 1 i.2CzSolutionAlong the contour C, we have z x i, 1 x 1, so that 1 111 z 2. Correspondingly, 2 1. Here, M max 2 1z C z 2 z and the arc length L 2. We haveZ1dz M L 2.2Cz17
ExampleEstimate an upper bound of the modulus of the integralZLog zI dzC z 4iwhere C is the circle z 3.Log z Now,z 4iln z Arg z z 4i so thatLog zln 3 π ln 3 π; z C z 4i 3 4 maxHence,ZL (2π)(3) 6π.Log zdz 6π(π ln 3).C z 4i18
ExampleFind an upper bound forZΓez /(z 2 1) dz , where Γ is the circle z 2 traversed once in the counterclockwise direction.SolutionThe path of integration has length L 4π. Next we seek an upperbound M for the function ez /(z 2 1) when z 2. Writing z x iy, we have ez ex iy ex e2,for z and by the triangle inequality z 2 1 z 2 1 4 1 3qforHence, ez /(z 2 1) e2/3 for z 2, and sox2 y 2 2, z 2.eze2dz · 4π.23Γz 1Z19
Path independenceUnder what conditions thatZC1f (z) dz ZC2f (z) dz,where C1 and C2 are two contours in a domain D with the sameinitial and final points and f (z) is piecewise continuous inside D.The property of path independence is valid for f (z) 12 but it failsz2when f (z) z . The above query is equivalent to the question:When doesICf (z) dz 0hold, where C is any closed contour lying completely inside D? Theequivalence is revealed if we treat C as C1 C2.We observe that f (z) z12 is analytic everywhere except at z 0but f (z) z 2 is nowhere analytic.20
Cauchy integral theoremLet f (z) u(x, y) iv(x, y) be analytic on and inside a simple closedcontour C and let f ′(z) be also continuous on and inside C, thenICf (z) dz 0.ProofThe proof of the Cauchy integral theorem requires the Green theorem for a positively oriented closed contour C: If the two real functions P (x, y) and Q(x, y) have continuous first order partial derivatives on and inside C, thenICP dx Q dy ZZD(Qx Py ) dxdy,where D is the simply connected domain bounded by C.21
Suppose we write f (z) u(x, y) iv(x, y), z x iy; we haveICf (z) dz ICu dx v dy iICv dx u dy.One can infer from the continuity of f ′(z) that u(x, y) and v(x, y)have continuous derivatives on and inside C. Using the Green theorem, the two real line integrals can be transformed into doubleintegrals.ICf (z) dz ZZD( vx uy ) dxdy iZZD(ux vy ) dxdy.Both integrands in the double integrals are equal to zero due to theCauchy-Riemann relations, hence the theorem.In 1903, Goursat was able to obtain the same result without assuming the continuity of f ′(z).22
Goursat TheoremIf a function f (z) is analytic throughout a simply connected domainD, then for any simple closed contour C lying completely inside D,we haveICf (z) dz 0.Corollary 1The integral of a function f (z) which is analytic throughout a simplyconnected domain D depends on the end points and not on theparticular contour taken. Suppose α and β are inside D, C1 and C2are any contours inside D joining α to β, thenZC1f (z) dz ZC2f (z) dz.23
ExampleIf C is the curve y x3 3x2 4x 1 joining points (1, 1) and (2, 3),find the value ofZC(12z 2 4iz) dz.Method 1 . The integral is independent of the path joining (1, 1)and (2, 3). Hence any path can be chosen. In particular, let uschoose the straight line paths from (1, 1) to (2, 1) and then from(2, 1) to (2, 3).Case 1 Along the path from (1, 1) to (2, 1), y 1, dy 0 so thatz x iy x i, dz dx. Then the integral equalsZ 212{12(x i)2 4i(x i)} dx {4(x i)3 2i(x i)2} 20 30i.124
Case 2 Along the path from (2, 1) to (2, 3), x 2, dx 0 so thatz x iy 2 iy, dz idy. Then the integral equalsZ 313{12(2 iy)2 4i(2 iy)}i dy {4(2 iy)3 2i(2 iy)2} 176 8i.1Then adding, the required value (20 30i) ( 176 8i) 156 38i.Method 2 .Z 2 3i1 iThe given integral equals2 3i(12z 2 4iz) dz (4z 3 2iz 2)1 i 156 38i.It is clear that Method 2 is easier.25
Corollary 2Let f (z) be analytic throughout a simply connected domain D. Consider a fixed point z0 D; by virtue of Corollary 1,F (z) Z zz0f (ζ) dζ, for any z D,is a well-defined function in D. Consideringz zF (z z) F (z)1 f (z) [f (ζ) f (z)] dζ. z z zZBy the Cauchy Theorem, the last integral is independent of the pathjoining z and z z so long as the path is completely inside D. Wechoose the path as the straight line segment joining z and z zand choose z small enough so that it is completely inside D.26
By continuity of f (z), we have for all points u on this straight linepath f (u) f (z) ǫwhenever u z δ.Note that z δ is observed implicitly.27
We haveZ z zz[f (u) f (z)] du ǫ z so that1F (z z) F (z) f (z) z z Z z zz[f (u) f (z)] du ǫfor z δ. This amounts to sayF (z z) F (z) f (z), z 0 zthat is, F ′(z) f (z) for all z in D. Hence, F (z) is analytic in Dsince F ′(z) exists at all points in D (which is an open set).lim28
This corollary may be considered as a complex counterpart ofthe fundamental theorem of real calculus. If we integrate f (z) along any contour joining α and β inside D,then the value of the integral is given byZ βαf (z) dz Z βz0f (ζ) dζ Z α F (β) F (α),z0f (ζ) dζα and β D.29
Corollary 3Let C, C1, C2, . . ., Cn be positively oriented closed contours, whereC1, C2, . . ., Cn are all inside C. For C1, C2, . . ., Cn, each of thesecontours lies outside of the other contours. Let int Ci denote thecollection of all points bounded inside Ci. Let f (z) be analytic onthe set S : C int C \ int C1 \ int C2 \ · · · \ int Cn (see the shadedarea in Figure), thenICf (z) dz n IXk 1 Ckf (z) dz.30
The proof for the case when n 2 is presented below.31
Proof The constructed boundary curve is composed of C C1 C2and the cut lines, each cut line travels twice in opposite directions. To explain the negative signs in front of C1 and C2, we note thatthe interior contours traverse in the clockwise sense as parts ofthe positively oriented boundary curve. With the introduction of these cuts, the shaded region boundedwithin this constructed boundary curve becomes a simply connected set.We haveICf (z) dz so thatICZ C1f (z) dz f (z) dz ZC1Z C2f (z) dz Zf (z) dz 0,C2f (z) dz.32
ExampleLet D be the domain that contains the whole complex plane exceptthe origin and the negative real axis. Let Γ be an arbitrary contourlying completely inside D, and Γ starts from 1 and ends at a pointα. Show thatZdz Log α.Γ zSolutionLet Γ1 be the line segment from 1 to α along the real axis, andΓ2 be a circular arc centered at the origin and of radius α whichextends from α to α. The union Γ1 Γ2 Γ forms a closedcontour. Since the integrand 1z is analytic everywhere inside D, bythe Cauchy integral theorem, we haveZZZdzdzdz .Γ zΓ1 zΓ2 z33
The contour Γ starts from z 1 and ends at z α. The arc Γ2 ispart of the circle z α .34
Since α cannot lie on the negative real axis, so Arg α cannot assumethe value π. If we write α α eiArg α ( π Arg α π), thenZ α dtdz ln α ztΓ11ZZ Arg αdzireiθ dθ i Arg α.iθreΓ2 z0Combining the results,ZZdz ln α i Arg α Log α.Γ zNote that the given domain D is the domain of definition of Log z,the principal branch of the complex logarithm function.35
Poisson integral2 zConsider the integration of the function earound the rectangularcontour Γ with vertices a, a ib and oriented positively. By lettinga while keeping b fixed, show thatZ Z 222 x 2ibx x beeedx cos 2bx dx eπ. y( a, b)Γ3(a, b)Γ2Γ4( a, 0)Γ1(a, 0)xThe configuration of the closed rectangular contour Γ.36
Solution2 zSince eis an entire function, we haveI2e z dz 0,Γby virtue of the Cauchy integral theorem. The closed contour Γconsists of four line segments: Γ Γ1 Γ2 Γ3 Γ4, whereΓ1 {x : a x a},Γ2 {a iy : 0 y b},Γ3 {x ib : a x a},Γ4 { a iy : 0 y b},and Γ is oriented in the anticlockwise direction.37
The contour integral can be split into four contour integrals, namely,IΓ z 2edz ZΓ1 z 2edz ZΓ2 z 2edz ZΓ3 z 2edz ZΓ42e z dz.The four contour integrals can be expressed as real integrals asfollows:ZZZΓ1Γ2 z 2e z 2edz dz 2Γ3e z dz Z a aZ beea2Γ4Z 0bi dy,2e (x ib) dx,b2e z dz dx, (a iy)20Z a eZ x2 Z a a x2e2cos 2bx dx ie ( a iy) i dy.Z a a x2e sin 2bx dx ,38
First, we consider the bound on the modulus of the second integral.ZΓ2 z 2edz Z b022 e (a y 2iay)i dy a2 e a e Therefore, the value ofbebZ b0Z b20Γ2b2e22eaZ2ey dydy(since 0 y b) 0 as a and b is fixed. z 2edz 0 as a .By similar argument, the fourth integral can be shown to be zeroas a .39
limIa Γ z 2edz lima Z a x2e a i lima ebso thatZ x2ecos 2bx dx iZ x2eb2dx eZ2 a aZ a a x2 x2cos 2bx dxe sin 2bx dx 0,e b2sin 2bx dx eZ x2e b2 dx eπ.Either by equating the imaginary parts of the above equation or2observing that e x sin 2bx is odd, we deduceZ Hence, we obtainZ x2e 2ibxe x2edx sin 2bx dx 0.Z x2e b2 cos 2bx dx eπ.40
Cauchy integral formulaLet the function f (z) be analytic on and inside a positively orientedsimple closed contour C and z is any point inside C, thenI1f (ζ)f (z) dζ.2πi C ζ zProofWe draw a circle Cr , with radius r around the point z, small enough(ζ)to be completely inside C. Since fζ zis analytic in the region lyingbetween Cr and C, we haveII11f (ζ)f (ζ)dζ dζ2πi C ζ z2πi Cr ζ zII1f (ζ) f (z)f (z)1 dζ dζ.2πi Crζ z2πi Cr ζ zThe last integral is seen to be equal to f (z). To complete the proof,it suffices to show that the first integral equals zero.41
yCCr zxThe contour C is deformed to the circle Cr , which encircles thepoint z.42
Since f is continuous at z, for each ǫ 0, there exists δ 0 suchthat f (ζ) f (z) ǫwhenever ζ z δ.Now, suppose we choose r δ (it is necessary to guarantee thatCr lies completely inside the contour C), the modulus of the firstintegral is bounded byI1f (ζ) f (z)dζ2πi Crζ zI1 f (ζ) f (z) dζ 2π Cr ζ z I1 f (ζ) f (z) dζ 2πr ICrǫǫ dζ 2πr ǫ.2πr Cr2πrSince the modulus of the above integral is less than any positivenumber ǫ, however small, so the value of that integral is zero.43
By the Cauchy integral formula, the value of f (z) at any point insidethe closed contour C is determined by the values of the functionalong the boundary contour C.ExampleApply the Cauchy integral formula to the integralIto show thatekzdz, z 1 zZ 2π0Z 2π0k is a real constant,ek cos θ sin(k sin θ) dθ 0ek cos θ cos(k sin θ) dθ 2π.44
SolutionBy Cauchy’s integral formula:ekzdz (2πi)ekz z 1 zIOn the other hand,2πi I i 2πi.z 02π ek(cos θ i sin θ)ekziθdz iedθeiθ z 1 z0Z 2π0Zek cos θ [cos(k sin θ) i sin(k sin θ)] dθ.Equating the real and imaginary parts, we obtain0 2π Z 2π0Z 2π0ek cos θ sin(k sin θ) dθek cos θ cos(k sin θ) dθ.45
ExampleEvaluatesin πz 2 cos πz 2dz,C (z 1)(z 2)Iwhere C is the circle: z i 3.Solutionsin πz 2 cos πz 2sin πz 2 cos πz 2sin πz 2 cos πz 2dz dz dzz 2z 1C (z 1)(z 2)CCIIIBy Cauchy’s integral formula, we havesin πz 2 cos πz 2dz 2πi{sin π(2)2 cos π(2)2} 2πiz 2CIsin πz 2 cos πz 2dz 2πi{sin π(1)2 cos π(1)2} 2πiz 1CIsince z 1 and z 2 are inside C and sin πz 2 cos πz 2 is analyticon and inside C. The integral has the value 2πi ( 2πi) 4πi.46
RemarkAlternately, by Corollary 3 of the Cauchy Integral Theorem, we havesin πz 2 cos πz 2dz C (z 1)(z 2)(sin πz 2 cos πz 2)/(z 2)dzz 1C1I(sin πz 2 cos πz 2)/(z 1) dz,z 2C2where C1 and C2 are closed contours completely inside C, C1 encircles the point z 1 while C2 encircles the point z 2.II47
sin πz 2 cos πz 2By the Cauchy Integral formula, choosing f (z) ,z 2we obtainIf (z)sin π cos πdz 2πif (1) 2πi 2πi. 1C1 z 1In a similar manner(sin πz 2 cos πz 2)/(z 1)sin πz 2 cos πz 2dz 2πi 2πi.z 2z 1C2z 2Hence, the integral is equal to 2πi 2πi 4πi.I48
The Cauchy integral formula can be extended to the case where thesimple closed contour C can be replaced by the oriented boundaryof a multiply connected domain.Suppose C, C1, C2, . . . , Cn and f (z) are given the same conditionsas in Corollary 3, then for any point z C int C \ int C1 \int C2 \ · · · \ int Cn, we havenX1f (ζ)1f (ζ)f (z) dζ dζ.2πi C ζ z2πi Ck ζ zk 1II49
Derivatives of contour integralsSuppose we differentiate both sides of the Cauchy integral formulaformally with respect to z (holding ζ fixed), assuming that differentiation under the integral sign is legitimate, we obtainIIIf (ζ)1d f (ζ)1f (ζ)1 d′dζ dζ dζ.f (z) 2πi dz C ζ z2πi C dz ζ z2πi C (ζ z)2How to justify the legitimacy of direct differentiation of the Cauchyintegral formula? First, consider the expressionIf (z h) f (z)1f (ζ) dζ2h "2πi C (ζ z)(#)I11f (ζ)f (ζ)f (ζ) hdζ2h 2πi C ζ z h ζ z(ζ z)Ihf (ζ) dζ.22πi C (ζ z h) (ζ z)50
It suffices to show that the value of the last integral goes to zeroas h 0. To estimate the value of the last integral, we draw thecircle C2d: ζ z 2d inside the domain bounded by C and chooseh such that 0 h d.For every point ζ on the curve C, it is outside the circle C2d so that ζ z dand ζ z h d.Let M be the upper bound of f (z) on C and L is the total arclength of C. Using the modulus inequality and together with theabove inequalities, we obtainIhf (ζ) h M Ldζ .232πi C (ζ z h) (ζ z)2π d51
In the limit h 0, we observe thatIf (ζ) h M Lhlimdζ lim 0;h 0 2πi C (ζ z h)(ζ z)2h 0 2π d3therefore,If (z h) f (z)1f (ζ)′f (z) lim dζ.h 0h2πi C (ζ z)2By induction, we can show the general resultIf (ζ)k!f (z) dζ,k 1, 2, 3, · · · ,2πi C (ζ z)k 1for any z inside C. This result is called the generalized CauchyIntegral Formula.(k)52
TheoremIf a function f (z) is analytic at a point, then its derivatives of allorders are also analytic at the same point.ProofSuppose f is analytic at a point z0, then there exists a neighborhoodof z0 : z z0 ǫ throughout which f is analytic. Take C0 to bea positively oriented circle centered at z0 and with radius ǫ/2 suchthat f is analytic inside and on C0. We then haveIf (ζ)1dζf ′′(z) 3πi C0 (ζ z)at each point z interior to C0. The existence of f ′′(z) throughoutthe neighborhood: z z0 ǫ/2 means that f ′ is analytic at z0.Repeating the argument to the analytic function f ′, we can concludethat f ′′ is analytic at z0.53
Remarks(i) The above theorem is limited to complex functions only. Infact, no similar statement can be made on real differentiablefunctions. It is easy to find examples of real valued functionf (x) such that f ′ (x) exists but not so for f ′′(x) at certain points.(ii) Suppose we express an analytic function inside a domain D asf (z) u(x, y) iv(x, y), z x iy. Since its derivatives ofall orders are analytic functions, it then follows that the partial derivatives of u(x, y) and v(x, y) of all orders exist and arecontinuous.54
To see this, since f ′′(z) exists, we consider2v2v2u2u v u′′′ i 2 i if (z) from f (z) x2 x y x y x x xorf ′′(z) 2v x y i 2u x y 2u 2v "# v u′ ifromf(z) i.22 y y y yThe continuity of f ′′ implies that all second order partials of u andv are continuous at points where f is analytic. Continuing with theprocess, we obtain the result.The mere assumption of the analyticity of f (z) on and inside C issufficient to guarantee the existence of the derivatives of f (z) of allorders. Moreover, the derivatives are all continuous on and insideC.55
ExampleSuppose f (z) is defined by the integralfind f ′ (1 i).3ζ 2 7ζ 1f (z) dζ,ζ z ζ 3ISolutionSetting k 1 in the generalized Cauchy integral formula,3ζ 2 7ζ 1f (z) dζ(ζ z)2 ζ 3I3(ζ z)2 (6z 7)(ζ z) 3z 2 7z 1 dζ(ζ z)2 ζ 3II13 dζ (6z 7)dζ ζ 3 ζ 3 ζ zI1 (3z 2 7z 1)dζ.2 ζ 3 (ζ z)′I56
The first integral equals zero since the integrand is entire (a constantfunction). For the second integral, we observe thatI1dζ ζ 3 ζ z(0 if z 3.2πi if z 3Furthermore, we deduce that the third integral is zero sinceI1ddζ 2dz ζ 3 (ζ z)Combining the results, we have"I#1dζ 0. ζ 3 ζ zf ′ (z) (2πi)(6z 7)if z 3. We observe that 1 i is inside z 3 since 1 i 2 3.Therefore, we obtainf ′(1 i) 2πi [6(1 i) 7] 12π 26πi.57
ExampleEvaluatee2zdz, where C is the circle z 3.4(z 1)CISolutionLet f (z) e2z and a 1 in the Cauchy integral formulaIf (z)n!(n)dz.f (a) n 12πi C (z a)If n 3, then f ′′′(z) 8e2z and f ′′′( 1) 8e 2. Hence,3!8e 2 2πiIe2zdz4(z 1)from which we see the required integral has the value 8πie 2/3.58
Cauchy inequalitySuppose f (z) is analytic on and inside the disc z z0 r, 0 r ,and letM (r) max f (z) , z z0 rthen f (k) (z) M (r) ,kk!rk 0, 1, 2, . . . .This inequality follows from the generalized Cauchy integral formula.59
ExampleSuppose f (z) is analytic inside the unit circle z 1 andshow that1, f (z) 1 z n1. f (n)(0) (n 1)! 1 n Solution(ζ)n , where f (ζ) is analyticWe integrate fn 1around the circle ζ n 1ζon and inside the circle. Using the generalized Cauchy integralformula, we haveIn!f (ζ)(n)f (0) dζnn 12πi ζ n 1 ζ60
niθ Z 2πf n 1 en!n eiθ i dθ n 1n2πi 0n 1i(n 1)θen 1 n Z1n! 2πnf 1 eiθ e inθ dθ.n2π 0n 1The modulus f (n) (0) is bounded by f (n) (0) Z1 n n! 2πn1 feiθ dθn2π 0n 1 Z1 n n! 2π1 1 dθnn2π 0 1 n 1 1 n n! 1 [(n 1) 2π]n 2π 1 n (n 1)! 1 .n 61
Gauss’ mean value theoremIf f (z) is analytic on and inside the disc Cr : z z0 r, thenZ1 2πf (z0) f (z0 reiθ ) dθ.2π 0ProofFrom the Cauchy integral formula, we havef (z0) Write u(z) Re f (z), itI1f (z)dz2πi Cr z z0Z 2π1f (z0 reiθ )ireiθdθiθ2πiZ 0re1 2πf (z0 reiθ ) dθ.2π 0is known that u is harmonic. We have1 2πu(z0) u(z0 reiθ ) dθ.2π 0Z62
ExampleFind the mean value of x2 y 2 x on the circle z i 2.SolutionFirst, we observe that x2 y 2 x Re(z 2 z). The mean valueis defined byZ1 2πu(i 2eiθ ) dθ,2π 0where u(z) Re(z 2 z). By Gauss’ mean value theorem,Z1 2πu(i 2eiθ ) dθ Re(z 2 z)2π 0z i Re( 1 i) 1.63
Maximum modulus theoremIf f (z) is analytic on and inside a domain D bounded by a simpleclosed curve C, then the maximum value of f (z) occurs on C,unless f (z) is a constant.ExampleFind the maximum value of z 2 3z 1 in the disk z 1.SolutionThe triangle inequality gives z 2 3z 1 z 2 3 z 1 5,for z 1.64
However, the maximum value is actually smaller than this, as thefollowing analysis shows.The maximum of z 2 3z 1 must occur on the boundary of the disk( z 1). The latter can be parameterized as z eit , 0 t 2π;whence z 2 3z 1 2 (ei2t 3eit 1)(e i2t 3e it 1).Expanding and gathering terms reduces this to (11 2 cos 2t), whosemaximum value is 13. The maximum value is obtained by takingt π/2 or t 3π/2.Thus the maximum value of z 2 3z 1 is 13, which occurs atz i.65
ExampleLet R denote the rectangular region:0 x π,0 y 1,the modulus of the entire functionf (z) sin zhas a maximum value in R that occurs on the boundary.To verify the claim, consider f (z) qsin2 x sinh2 y,the term sin2 x is greatest at x π/2 and the increasing functionsinh2 y is greatest when y 1. The maximum value of f (z) in Rπoccurs at the boundary point, 1 and at no other point in R.266
Proof of the Maximum Modulus TheoremProof by contradiction. Suppose f (z) attains its maximum at α D. Using the Gauss Mean Value Theorem:1 2πf (α reiθ ) dθf (α) 2π 0where the neighborhood N (α; r) lies inside D. By the modulus inequality,Z1 2π f (α) f (α reiθ ) dθ.2π 0ZSince f (α) is a maximum, then f (α reiθ ) f (α) for all θ, giving1 2π1 2πiθ f (α re ) dθ f (α) dθ f (α) .2π 02π 0Combining the results, we obtainZZZ 2π0[ f (α) f (α reiθ ) ] dθ 0. {znon-negative}67
One then infer that f (α) f (α reiθ ) . However, it may bepossible to have f (α reiθ ) f (α) at isolated points. We arguethat this is not possible due to continuity of f (z).If f (α reiθ ) f (α) at a single point, then f (α reiθ ) f (α) for a finite arc on the circle, giving1 2π f (α reiθ ) dθ f (α) ,2π 0a contradiction. We can then deduce thatZ f (α) f (α reiθ ) for all points on the circle. Since r can be any value, f (z) is constant in any neighborhoodof α lying inside D.68
Finally, we need to show that f (z) is constant at any point in D.Take any z D, we can join α to z by a curve lying completelyinside D. Taking a sequence of points z0 α, z1, · · · , zn z suchthat each of these points is the center of a disc (plus its boundary)lying completely inside D and zk is contained in the disk centered atzk 1, k 1, 2, · · · , n.We then have f (z1) f (α) . Also z2 is contained inside the disccentered at z1, so f (z2) f (z1) , · · · , and deductively f (z) f (α) .Lastly, we use the result that if f (z) constant throughout D,then f (z) constant throughout D.69
Liouville’s TheoremIf f is entire and bounded in the complex plane, then f (z) is constantthroughout the plane.Prooff (ζ)(ζ z)2around CR : ζ z R. By the generalized Cauchy integral formulaIt suffices to show that f ′ (z) 0 for all z C. We integrateI1f (ζ)′f (z) dξ,2πi CR (ζ z)2which remains valid for any sufficiently large R since f (z) is entire.Since f (z) is bounded, so f (z) B for all z C,1 f ′(z) 2πIf (ζ)1 BBdζ 2πR .222π RRCR (ζ z)70
Now, B is independent of R and R can be arbitrarily large. Theinequality can hold for arbitrarily large values of R only if f ′ (z) 0.Since the choice of z is arbitrary, this means that f ′(z) 0 everywhere in the complex plane. Consequently, f is a constant function.RemarkNon-constant entire functions must be unbounded. For example,sin z and cos z are unbounded, unlike their real counterparts.71
4. Complex integration: Cauchy integral theorem and Cauchy integral formulas Definite integral of a complex-valued function of a real variable Consider a complex valued function f(t) of a real variable t: f(t) u(t) iv(t), which is assumed to be a piecewise continuous
Lecture 35. Cauchy, Who Set the Foundation of Analysis Figure 34.1 Cauchy was living in a cottage in Arcueil Augustin Cauchy Augustin Louis Cauchy (1789-1857) was a French mathematician and is one of the greatest modern mathematicians. Cauchy pioneered the elds of analysis, both real and
Topic 4 Notes Jeremy Orlo 4 Cauchy’s integral formula 4.1 Introduction Cauchy’s theorem is a big theorem which we will use almost daily from here on out. Right away it will reveal a number of interesting and useful properties of analytic functions. More will follow as the course progresses.
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work and language games. - reading texts extending the main story-line. - translation exercises. - some songs and poems for learners. At the end of the book you will find the texts of the listening exercises, a summary of the grammar covered in the course, a pronunciation guide and Russian to English and English to Russian dictionaries.
