Notes on Complex Analysis in Physics

1 Notes on Complex Analysis in PhysicsJim NapolitanoMarch 9, 2013 These Notes are meant to accompany a graduate level Physics course, to provide a basicintroduction to the necessary concepts in Complex Analysis . They are not complete, nor areany of the proofs considered rigorous. The immediate goal is to carry through enough of thework needed to explain the Cauchy Residue Theorem. Other material may be added Numbers and Complex FunctionsA Complex numberzcan be written asz=x+iyorz=rei withr 0wherei= 1, andx,y,r, and are real numbers. Clearly,x=rcos andy=rsin leading to a description in terms of the Complex plane. The Complex conjugate ofzisz =x iyorz =re i The modulus ofzis|z| z z=r= x2+y2and is often called the phase Complex functionf(z) typically returns a Complex number. Generically, we writef(z) =u(x,y) +iv(x,y)(1)for purposes of proofs or illustrations. The behavior of the (real) functionsu(x,y) andv(x,y)are critical for classifying Complex functions, as seen when we consider taking and AnalyticityWe define the derivativef (z) =df/dzof a Complex functionf(z) in the same was as we dofor the derivatives of real functions.

2 That is, forz0 x0+iy0,f (z0) =dfdz z=z0= limz z0f(z) f(z0)z z01 However, there is clearly an ambiguity, depending on whether we approachz0along the liney=y0or alongx=x0. (Of course, we could also say the ambiguity is along any line ofconstant = 0, but it is sufficient to consider just two orthogonal directions.) That is,f (z0) = limx x0u(x,y0) u(x0,y0)x x0+ilimx x0v(x,y0) v(x0,y0)x x0= u x+i v xorf (z0) = limy y0u(x0,y) u(x0,y0)iy iy0+ilimy y0v(x0,y) v(x0,y0)iy iy0= i u y+ v yTherefore, in order to remove the ambiguity and have a consistent definition of the derivative, u x= v yand u y= v x(2)These are called the Cauchy-Riemann conditions. A functionf(z) which satisfies theserather restrictive conditions is calledanalytic. Indeed, analytic functions have very manyapplications in Physics , and we will merely scratch the surface example, the functionf(z) =ez=ex(cosy+isiny) is analytic. This is easy to (x,y) =excosyandv(x,y) =exsiny, u x=excosy= v yand u y= exsiny= v xso the Cauchy-Riemann conditions (2) are is simple to show thatf(z) =azis analytic, whereais a Complex constant.

3 It is alsonot hard to show that the product of two analytic functions is analytic, so any function ofthe formf(z) =anzn, wherenis a non-negative integer, is also analytic. Of course, anysum of analytic functions is analytic, so we see that any polynomial inzis analytic in theentire Complex examples beg the question: If a functionf(z) can be written explicitly in terms ofz, is it analytic? The answer is Yes. To see this, realize that instead ofxandy, we couldalways write a Complex function in terms ofzandz usingx= (z+z )/2 andy= (z z ) consider f z = f x x z + f y y z =( u x+i v x)(12)+( u y+i v y)( 12i)=12( u x v y)+i2( u y+ v x)= 0so long as the Cauchy-Riemann conditions (2) are satisfied. That is, if the expression forf(z) contains onlyz(and notz ) then the function is is some common terminology. A functionf(z) need not be analytic in the entirecomplex plane. (If it is, we called the function entire.)

4 If it is analytic at a pointz0then we call that a regular point. Otherwise,z0is called a singular point. Much of ourdiscussion of Complex integration will focus on the notion of singular and Series ExpansionSimilarly to differentiation, we approach integration of Complex functions the same way aswith real functions, but we need to be aware that there is now an arbitrariness of the path of integration. Withdz=dx+idyand using using (1), we have z2z1f(z)dz= z2z1(u dx v dy) +i z2z1(v dx+u dy) = z2z1A dx+i z2z1B dx(3)whereA=u x v yandB=v x+u y. So, we can now think of the two integrals on theright as real integrals of vector functions over curves in thexyplane. However, if we invokeStokes Theorem, these become integrals of the curls, and using (2), we find A=( v x u y)= 0and B=( u x v y)= 0(4)and each of the two integrals on the right in (3) are path-independent. Hence, the integral ofananalyticcomplex functionf(z) is path-independent and can be unambiguously here on, we assume all functions to be analytic unless explicitly is obvious from (3) that, when integrating around a closed pathC, Cf(z)dz= 0which is known as the Cauchy-Goursat Theorem.

5 We will be exploring circumstances wherethe integrand is explicitly singular at one or more the first example, we prove the Cauchy Integral Formula, namelyf(z0) =12 i Cf(z)z z0dz(5)whereCis a closed contour in the Complex plane that contains the pointz0and traversed inthe counterclockwise direction. We can break up a contourCinto something that looks likewhereC0is a tinycircularcontour around the singular point, but in the clockwise is, we replaceCwith limC0 0(C+C0). However, forC06= 0, the new contourCdoesnot include the singular point, so by (4) we write (5) asf(z0) = 12 i C0f(z)z z0dz(6)The shrinking contourC0is parameterized asz z0=rei forr 0 and = 2 0, so 12 i C0f(z)z z0dz= 12 if(z0) 02 1rei irei d = 12 if(z0)i( 2 ) =f(z0)proving the Cauchy Integral Formula (5). A trivial, but suggestive, rewriting of (5) givesf(z) =12 i Cf( ) zd (7)which leads to a convenient way to write the derivatives of a Complex function, namelyf(n)(z) =dnfdzn=n!

6 2 i Cf( )( z)n+1d (8)Now consider the series expansion of an analytic functionf(z). We would naturally writef(z) =f(z0) +f (z0)(z z0) + = n=0an(z z0)n(9)wherean 1n!f(n)(z0) =12 i Cf( )( z0)n+1d (10)Such a Taylor Series expansion works out as expected, but the curveCspecifies regions inwhich the series idea can be expanded to include n , still using the right side of (10) todefinean, and with modified regions of convergence. Such an expansion is called a LaurentSeries. It clearly is not, in general, an analytic function because of poles that appear forn <0. These, however, lead us to one of the most important theorems of Complex Analysis ,so far as mathematical Physics is Cauchy Residue TheoremLetg(z) have an isolated singularity atz=z0. If the Laurent expansion can be written asg(z) = n= an(z z0)n=b1z z0+ n=0an(z z0)n(11)4then we say thatg(z) has a simple pole atz=z0. Higher order poles are possible, butwe re not going to consider them a contourCwithin the radius of convergence ofg(z).

7 Separate the integral ofg(z) around this contour into two terms, one for each of the two terms on the right in (11).The second term is a polynomial inz; therefore it is analytic and the integral is zero. Recallthat we reduced the contour to a small circle around the pole in order to prove the CauchyIntegral Formula. We can do the same thing here, and Cg(z)dz=b1 C1z z0= 2 i b1(12)We refer tob1as the residue ofg(z0), sometimes written as Res[g(z0)]. We haveRes[g(z0)] = limz z0(z z0)g(z)for a simple pole there are more than one simple pole within the contourC, this result is easy togeneralize. Instead of redrawing the contour with a small loop about the single pole, do itfor allNpoles within the contour. The result is clearly Cg(z)dz= 2 iN k=1 Res[g(zk)](13)We refer to this as the Cauchy Residue Theorem. It is widely used in mathematical usefulness of the Residue Theorem can be illustrated in many ways, but here is oneimportant example.

8 It is a warm-up to evaluating the integral ( ) inModern QuantumMechanics, 2nd exercise is to evaluate the integralI= eikaq2 k2k dk= lim 0 eikaq2 k2+i k dk(14)wherek,a,q, and >0 are all real variables. We use the second version above because thismoves the singularities atk= qoff the real axis. To be sure, we could have moved off thereal axis by using i instead of +i , and in fact, this would give us a different answer. Aphysical rationale is needed to justify one sign or the other. Leave that for a Physics can evaluate (14) using contour integration by first allowingkto be Complex andthen noting thateikx 0 as Im(k) + . Therefore (14) can be rewritten as an integralover a semicircular contourCthat runs (counter clockwise) along the Re(k) axis and closesas a semicircle in the Im(k)>0 plane. Then for 0, the integrand in (14) has poles atk= q2 i = q(1 i q2) q i 5where we redefine (withq >0) so that it is still small and has the same pole atk=k0 +q i does not matter to us, since it is outside the integrationcontour.

9 However, the pole atk= k0= q+i is inside, so we use the Residue Theoremto writeI= lim 0 Ceika(k k0)(k+k0)k dk= lim 02 ieikak k0k k= k0= ilim 0e ik0a= ie iqaOther TopicsThere is of course much more to Complex Analysis than what is covered here, even leavingaside the question of rigor. Two upper level texts I generally recommend are Arfken, Methods for Physicists, 4th ed., Academic Press, 1995; andByron, and of Classical and Quantum Physics , Dover, other important topics include the following: Convergence criteria Branch cuts in the Complex plane The Cauchy Principal Value Conformal mapping6