Transcription of 4 Cauchy’s integral formula - MIT Mathematics
1 Topic 4 NotesJeremy Orloff4 cauchy s integral IntroductionCauchy s theorem is a big theorem which we will use almost daily from here on out. Rightaway it will reveal a number of interesting and useful properties of analytic functions. Morewill follow as the course you learn just one theorem this week it should be cauchy s integral formula !We start with a statement of the theorem for functions. After some examples, we ll give ageneralization to all derivatives of a function. After some more examples we will prove thetheorems. After that we will see some remarkable consequences that follow fairly directlyfrom the cauchy s cauchy s integral for functionsTheorem ( cauchy s integral formula ) SupposeCis a simple closed curve and thefunctionf(z) is analytic on a region containingCand its interior.
2 We assumeCis orientedcounterclockwise. Then for anyz0insideC:f(z0) =12 i Cf(z)z z0dz(1)Re(z)Im(z)z0 CACauchy s integral formula : simple closed curveC,f(z) analytic on and is remarkable: it says that knowing the values offon the boundary curveCmeans weknow everything aboutfinsideC!! This is probably unlike anything you ve encounteredwith functions of real a slight change of notation (zbecomeswandz0becomesz) we often writethe formula asf(z) =12 i Cf(w)w zdw(2)14 cauchy S integral FORMULA2 Aside re not being entirely fair to functions of real variables. We will see that forf=u+ivthe real and imaginary partsuandvhave many similar remarkable called conjugate harmonic ExamplesExample cez2z 2dz, whereCis the curve (z)Im(z)C2 Solution:Letf(z) = (z) is entire.
3 SinceCis a simple closed curve (counterclockwise)andz= 2 is insideC, cauchy s integral formula says that the integral is 2 if(2) = 2 the same integral as the previous example withCthe curve (z)Im(z)C2 Solution:Sincef(z) = ez2/(z 2) is analytic on and insideC, cauchy s theorem says thatthe integral is the same integral as the previous examples withCthe curve (z)Im(z)C2 Solution:This one is trickier. Letf(z) = ez2. The curveCgoes around 2 twice in theclockwisedirection, so we breakCintoC1+C2as shown in the next cauchy S integral FORMULA3Re(z)Im(z)C1C1C22 These are both simple closed curves, so we can apply the cauchy integral formula to eachseparately. (The negative signs are because they go clockwise aroundz= 2.)
4 Cf(z)z 2dz= C1f(z)z 2dz+ C2f(z)z 2dz= 2 if(2) 2 if(2) = 4 if(2). cauchy s integral formula for derivativesCauchy s integral formula is worth repeating several times. So, now we give it for allderivativesf(n)(z) off. This will include the formula for functions as a special s integral formula for derivatives. Iff(z) andCsatisfy the samehypotheses as for cauchy s integral formula then, for allzinsideCwe havef(n)(z) =n!2 i Cf(w)(w z)n+1dw, n= 0,1,2,..(3)where,Cis a simple closed curve, oriented counterclockwise,zis insideCandf(w) isanalytic on and Ce2zz4dzwhereC:|z|= :With cauchy s formula for derivatives this is easy. Letf(z) = e2z. Then,I= Cf(z)z4dz=2 i3!
5 F (0) =83 LetCbe the contour shown below and evaluate the same integral asin the previous :Again this is easy: the integral is the same as the previous example, cauchy S integral Another approach to some basic examplesSupposeCis a simple closed curve around 0. We have seen that C1zdz= 2 cauchy integral formula gives the same result. That is, letf(z) = 1, then the formulasays12 i Cf(z)z 0dz=f(0) = cauchy s formula for derivatives shows C1(z)ndz= Cf(z)zn+1dz=f(n)(0) = 0,for integersn > More examplesExample Ccos(z)z(z2+ 8)dzover the contour (z)Im(z)2i 2iCSolution:Letf(z) = cos(z)/(z2+ 8).f(z) is analytic on and inside the curveC. That is,the roots ofz2+ 8 are outside the curve.
6 So, we rewrite the integral as Ccos(z)/(z2+ 8)zdz= Cf(z)zdz= 2 if(0) = 2 i18= C1(z2+ 4)2dzover the contour cauchy S integral FORMULA5Re(z)Im(z)Ci2i i 2iSolution:We factor the denominator as1(z2+ 4)2=1(z 2i)2(z+ 2i) (z) =1(z+ 2i)2. Clearlyf(z) is analytic insideC. So, by cauchy s formula for derivatives: C1(z2+ 4)2dz= Cf(z)(z 2i)2= 2 if (2i) = 2 i[ 2(z+ 2i)3]z=2i=4 i64i= 16 Example Czz2+ 4dzover the curveCshown (z)Im(z)Ci2i i 2iSolution:The integrand has singularities at 2iand the curveCencloses them both. Thesolution to the previous solution won t work because we can t find an appropriatef(z) thatis analytic on the whole interior ofC. Our solution is to split the curve into two thatC3is traversed both forward and cauchy S integral FORMULA6Re(z)Im(z)C1C1C3C2C2 C3i2i i 2iSplit the original curveCinto 2 pieces that each surround just one havezz2+ 4=z(z 2i)(z+ 2i).
7 We letf1(z) =zz+ 2iandf2(z) =zz ,zz2+ 4=f1(z)z 2i=f2(z)z+ integral , can be written out as Czz2+ 4dz= C1+C3 C3+C2zz2+ 4dz= C1+C3f1(z)z 2idz+ C2 C3f2(z)z+ 2idzSincef1is analytic inside the simple closed curveC1+C3andf2is analytic inside thesimple closed curveC2 C3, cauchy s formula applies to both integrals. The total integralequals2 i(f1(2i) +f2( 2i)) = 2 i(1/2 + 1/2) = 2 We could also have done this problem using partial fractions:z(z 2i)(z+ 2i)=Az 2i+Bz+ will turn out thatA=f1(2i) andB=f2( 2i). It is easy to apply the cauchy integralformula to both an upcoming topic we will formulate the cauchy residue will allow us to compute the integrals in Examples in an easier and less adhoc cauchy S integral The triangle inequality for integralsWe discussed the triangle inequality in the Topic 1 notes.
8 It says that|z1+z2| |z1|+|z2|,(4a)with equality if and only ifz1andz2lie on the same ray from the useful variant of this statement is|z1| |z2| |z1 z2|.(4b)This follows because Equation 4a implies|z1|=|(z1 z2) +z2| |z1 z2|+|z2|.Now subtractingz2from both sides give Equation 4bSince an integral is basically a sum, this translates to the triangle inequality for ll state it in two ways that will be useful to (Triangle inequality for integrals) Supposeg(t) is a complex valued func-tion of a real variable, defined ona t b. Then bag(t)dt ba|g(t))|dt,with equality if and only if the values ofg(t) all lie on the same ray from the follows by approximating the integral as a Riemann sum.
9 Bag(t)dt g(tk) t |g(tk)| t ba|g(t)| middle inequality is just the standard triangle inequality for sums of complex num-bers. Theorem (Triangle inequality for integrals II) For any functionf(z) and any curve , we have f(z)dz |f(z)||dz|.Heredz= (t)dtand|dz|=| (t)| follows immediately from the previous theorem: f(z)dz = baf( (t)) (t)dt ba|f( (t))|| (t)|dt= |f(z)||dz|. |f(z)|< MonCthen Cf(z)dz M (length ofC).4 cauchy S integral (t), witha t b, be a parametrization ofC. Using the triangle inequality Cf(z)dz C|f(z)||dz|= ba|f( (t))|| (t)|dt baM| (t)|dt=M (length ofC).Here we have used that| (t)|dt= (x )2+ (y )2dt=ds,the arclength element.
10 Example the real integralI= 1(x2+ 1)2dxSolution:The trick is to integratef(z) = 1/(z2+ 1)2over the closed contourC1+CRshown, and then show that the contribution ofCRto this integral vanishes asRgoes to .Re(z)Im(z)CRCRC1R RiThe only singularity off(z) =1(z+i)2(z i)2inside the contour is atz=i. Letg(z) =1(z+i) analytic on and inside the contour, cauchy s formula gives C1+CRf(z)dz= C1+CRg(z)(z i)2dz= 2 ig (i) = 2 i 2(2i)3= parametrizeC1by (x) =x,with R x , C1f(z)dz= R R1(x2+ 1) goes toI(the value we want to compute) asR .Next, we parametrizeCRby ( ) =Rei ,with 0 .4 cauchy S integral FORMULA9So, CRf(z)dz= 01(R2e2i + 1)2iRei d By the triangle inequality for integrals, ifR >1 CRf(z)dz 0 1(R2e2i + 1)2iRei d.
