Poles, Residues, and All That

Transcription

1 hapter Ten Poles, Residues, and All That 0.. Residues. A point z 0 is a singular point of a function f if f not analytic at z 0, but is analytic at some point of each neighborhood of z 0. A singular point z 0 of f is said to be isolated if there is a neighborhood of z 0 which contains no singular points of f sav 0.In other words, f is analytic on some region 0 z z 0. Examples The function f given by fz zz 2 4 has isolated singular points at z 0, z 2i, and z 2i. Every point on the negative real axis and the origin is a singular point of Log z, but there are no isolated singular points. Suppose now that z 0 is an isolated singular point of f. Then there is a Laurent series fz c j z z 0 j j valid for 0 z z 0 R, for some positive R. The coefficient c of z z 0 is called the residue of f at z 0, and is frequently written Res f. zz 0 Now, why do we care enough about c to give it a special name? Well, observe that if is any positively oriented simple closed curve in 0 z z 0 R and which contains z 0 inside, then c 2i fzdz. 0.

2 This provides the key to evaluating many complex integrals. Example We shall evaluate the integral e /z dz where is the circle z with the usual positive orientation. Observe that the integrand has an isolated singularity at z 0. We know then that the value of the integral is simply 2i times the residue of e /z at 0. Let s find the Laurent series about 0. We already know that j0 j! zj for all z. Thus, e /z j0 j! zj z 2! z 2 The residue c, and so the value of the integral is simply 2i. Now suppose we have a function f which is analytic everywhere except for isolated singularities, and let be a simple closed curve (positively oriented) on which f is analytic. Then there will be only a finite number of singularities of f inside (why?). all them z, z 2,, z n. For each k,2,,n, let k be a positively oriented circle centered at z k and with radius small enough to insure that it is inside and has no other singular points inside it. 0.2

3 Then, fzdz fzdz 2 fzdz n fzdz 2i Res f 2i Res f 2i Res f zz zz2 zzn n 2i k Res f. zz k This is the celebrated Residue Theorem. It says that the integral of f is simply 2i times the sum of the residues at the singular points enclosed by the contour. Exercises Evaluate the integrals. In each case, is the positively oriented circle z 2.. e /z2 dz. 2. sin z dz. 3. cos z dz z sin z dz. z cos z dz. 0.3

4 0.2. Poles and other singularities. In order for the Residue Theorem to be of much help in evaluating integrals, there needs to be some better way of computing the residue finding the Laurent expansion about each isolated singular point is a chore. We shall now see that in the case of a special but commonly occurring type of singularity the residue is easy to find. Suppos 0 is an isolated singularity of f and suppose that the Laurent series of f at z 0 contains only a finite number of terms involving negative powers of z z 0. Thus, fz c n z z 0 n c n z z 0 n c z z 0 c 0 c z z 0. Multiply this expression by z z 0 n : z z z 0 n fz c n c n z z 0 c z z 0 n. What we see is the Taylor series at z 0 for the function z z z 0 n fz. The coefficient of z z 0 n is what we seek, and we know that this is. n z 0 n! The sought after residue c is thus c Res f n z 0 zz0 n!, wher z z 0 n fz. Example We shall find all the residues of the function fz z 2 z 2. First, observe that f has isolated singularities at 0,and i. Let s see about the residue at 0. Here we have 0.4

5 z z 2 fz z 2. The residue is simply 0 : z z2 2z z 2 2. Hence, Res f 0. z0 Next, let s see what we have at z i: z z ifz z 2 z i, and so Res fz i ei zi 2i. In the same way, we see that Let s find the integral Res f ei zi 2i. dz, where is the contour pictured: z 2 z 2 0.5

6 This is now easy. The contour is positive oriented and encloses two singularities of f; viz, i and i. Hence, dz 2i z 2 z 2 zi Res f Res f zi 2i 2i ei ei 2i 2isin. Miraculously easy! There is some jargon that goes with all this. An isolated singular point z 0 of f such that the Laurent series at z 0 includes only a finite number of terms involving negative powers of z z 0 is called a pole. Thus, if z 0 is a pole, there is an integer n so that z z z 0 n fz is analytic at z 0, and fz 0 0. The number n is called the order of the pole. Thus, in the preceding example, 0 is a pole of order 2, while i and i are poles of order. (A pole of order is frequently called a simple pole.) We must hedge just a bit here. If z 0 is an isolated singularity of f and there are no Laurent series terms involving negative powers of z z 0, then we say z 0 is a removable singularity. Example Let fz sinz z ; then the singularity z 0 is a removable singularity: fz z sinz z3 z z 3! z5 5! 3! z2 z4 5! and we see that in some sense f is really analytic at z 0 if we would just define it to be the right thing there. A singularity that is neither a pole or removable is called an essential singularity. Let s look at one more labor-saving trick or technique, if you prefer. Suppose f is a function: 0.6

7 fz pz qz, where p and q are analytic at z 0, and we have qz 0 0, while q z 0 0, and pz 0 0. Then and so fz pz qz pz 0 p z 0 z z 0 q z 0 z z 0 q z 0 z z 2 0 2, z z z 0 fz pz 0 p z 0 z z 0 q z 0 q z 0 z z 2 0. Thus z 0 is a simple pole and Res f z 0 zz 0 pz 0 q z 0. Example Find the integral dz, where is the rectangle with sides x, y, and y 3. The singularities of the integrand are all the places at which, or in other words, the points z 0,2i,4i,. The singularities enclosed by are 0 and 2i. Thus, dz 2i z0 Res f Res f, z2i where fz. 0.7

8 Observe this is precisely the situation just discussed: fz pz, where p and q are qz analytic, etc.,etc. Now, pz q z. Thus, Res f cos0 z0 Res f cos2i z2i e 2i,and e2 e 2 2 cosh2. Finally, dz 2i z0 Res f Res f z2i 2i cosh2 Exercises 6. Suppose f has an isolated singularity at z 0. Then, of course, the derivative f also has an isolated singularity at z 0. Find the residue Res f. zz0 7. Given an example of a function f with a simple pole at z 0 such that Res f 0, or explain zz0 carefully why there is no such function. 8. Given an example of a function f with a pole of order 2 at z 0 such that Res f 0, or zz0 explain carefully why there is no such function. 9. Suppose g is analytic and has a zero of order n at z 0 (That is, gz z z 0 n hz, where hz 0 0.). Show that the function f given by fz gz 0.8

9 has a pole of order n at z 0. What is Res f? zz0 0. Suppose g is analytic and has a zero of order n at z 0. Show that the function f given by fz g z gz has a simple pole at z 0, and Res f n. zz0. Find z 2 4 dz, where is the positively oriented circle z Find tan zdz, where is the positively oriented circle z Find z 2 z dz, where is the positively oriented circle z