Lecture 7 Appendi B: Some sample problems from Boas Here are some solutions to the sample problems assigned for hapter 4 4: 8 Solution: We want to learn about the analyticity properties of the function i f z z re r cos i r sin y cos where cos y y cos sin Sign cos sin Sign y y y y U y y y y y V y y y Recall that the square root function has sheets connected by a branch cut running from the origin to infinity In polar coordinates (see the first line above) the phase must vary from 0 to 4 before z returns to its starting point In the last epressions the signs account for this sheet structure We proceed by considering the auchy- Riemann conditions and perform some simple manipulations to find Physics 8 Lecture 7 Appendi B Winter 009
U y y y y y y y y y y V y y y y y y V y y y y y y y U y y y y y y y y y U V V U : z 0 y y The square root is analytic (on a single sheet) everywhere in the comple plane ecept the origin which is a branch point Physics 8 Lecture 7 Appendi B Winter 009
4: 9 Solution: We want to learn about the analyticity properties of the logarithm function f z ln z ln i ln y i tan y U y y V y ln tan y Again there is a branch cut from the origin to infinity but now there are an infinite number of branches (due to the multi-valued arctan function) Net consider the auchy-riemann conditions U V y y y V y U y y y y U V V U : z 0 y y The logarithm is analytic (on a single sheet) everywhere in the (finite) comple plane ecept the origin which is a branch point 4: 3 Solution: We want to learn about the analyticity properties of the function f z iy z y z z y U y V y y y So we consider the auchy-riemann conditions Physics 8 Lecture 7 Appendi B 3 Winter 009
U y V y y y y y y y y y V y U y y y y U V V U : z 0 y y Thus this function is analytic everywhere in the comple plane ecept the origin where the function has a simple pole In fact that is all this function is just the pole /z 4: 34 Solution: We consider the function ln z epanded about the origin We know from last quarter that this function has a series epansion n n 3 z z z ln z z n 3 which we further know converges for z Let s consider this result in our new language The auchy-riemann conditions tell us y ln z ln y itan y U y ln y V y tan U V y y y V y U y y y y U V V U : z 0 y y Physics 8 Lecture 7 Appendi B 4 Winter 009
This logarithm is analytic (on a single sheet) everywhere in the (finite) comple plane ecept the point (on the real ais) = y = 0 which is a branch point This singular point eplains why the power series epansion about the origin cannot converge for z 4: 48 Solution: In this eercise we want to think about the auchy-riemann conditions in i polar coordinates In Boas notation we have z re i and thus z r e z z and z iz Hence a unique derivative requires (equating real and imaginary parts in the last step and finding -R in polar coordinates) f df z df z f df z df iz r dz r dz z dz dz df z f z U V i f i U V i i dz z r z r r z z U V V U r r r r Now we can apply this form to the square root function (see 4:8 above) We find i f z z re r cos i r sin U r r cos V r r sin U V cos cos r r r r V U sin sin r r r r U V V U : r 0 r r r r Physics 8 Lecture 7 Appendi B 5 Winter 009
So we obtain our previous result that the square root is analytic in the entire comple plane ecept the origin where there is a branch point However the calculation is clearly simpler in polar coordinates 43: 7 Solution: We want to consider the following (closed) contour integrals sin z sin z a) z dz dz : z z b) z To proceed we must recognize two features of the integrand sin z z First the sine function is an entire function analytic in the entire comple plane (recall from last quarter that we know that its power series epansion is convergent for all z) Second we see that there is a simple pole at z 5 y 0 0 0 0 Now we are ready to apply auchy s integral theorem For the first contour the pole is outside of the contour and the integral vanishes while the second contour encircles the pole and we find a b sin z dz 0 z sin z dz i sin i z 43: 3 Solution: Now consider a contour integral around the square with vertices at z i of the following integrand 3z 3ln 3 zln e e e 4 4 zln zln n0 n4 n 3 z ln dz dz 8 dz n! auchy tells us that we are only interested in the simple pole term corresponding n 3in the sum Thus we have Physics 8 Lecture 7 Appendi B 6 Winter 009
3z 4 z ln e z 3 3 ln 8 dz 8 dz i 7 7 i 3! 6 Note that for this nonzero result it is essential that ln 0693 (so that the pole is inside of the contour) 46: 3 4 Solution: We want to epand the function sin z z in a Laurent series around the singular point at the origin This is just the familiar series for the sine function 4 divided by z We have sin z n n n3 z z!! 4 4 z z n0 n n0 n n Hence the residue (from the simple pole term n ) is then n n n R! 3! 6 46: 5 Solution: Now consider terms of z we find z e z about the point z Epressing everything in z z n m e ee e z m z! z z z z n0 n m0 e z z z z z e z z 4 Physics 8 Lecture 7 Appendi B 7 Winter 009
Thus the residue (the first term) is just R e 47: Solution: Finally we want to practice evaluating integrals using auchy onsider first an integral around a circle which we can think of as an integral in the comple i plane using the polar representation z e i dz ide izd Thus we can write d dz dz 5 3 z i z dz i 5z z 5i 5 3 5sin iz 5z 6iz 5 0 z z z Thus the one simple pole inside the contour is the one at z i 5 while the other pole (at z 5i) is outside of the contour So finally we obtain d dz 4 i 3 5sin i i 4 6 5 z z 5i 5 5i 5 5 0 z 47: 3 Solution: Now consider an integral along ½ of the real ais which we can etend to the full ais by using the even symmetry of the integrand I d d 0 4 9 i i 3i 3i We see that there are 4 simple poles that we can use via auchy to evaluate the integral above the real ais (+i +3i) and below (-i -3i) As usual we want to Physics 8 Lecture 7 Appendi B 8 Winter 009
add an arc at infinity which we are allowed to do (by Jordan) since the integral along the arc behaves like i i R e id Re i lim lim ie d 0 i i R R R e 4R e 9 R R arc at 0 If we close the contour in the upper half plane (counter-clockwise) we encircle the poles at +i and +3i to find I z dz 3 3 losed Above z i z iz iz i i 4 9 3 4i4 9 9 46i 5 0 0 If we choose to close the contour in the lower half plane (clockwise) we encircle the poles at -i and -3i with the opposite residues from above but with an etra minus sign due to the opposite direction Thus we obtain the same answer either way 47: 9 Solution: This integral is similar to the one we just did but now with a cosine function in the numerator Again we can use the fact that the integrand is symmetric to etend it along the entire real ais and then due to the anti-symmetry of the sine function we can for free replace the cosine function with the comple eponential We have I i cosd cosd e d 3 3 3 3 0 4 9 i i 3 i i Now we have to worry about the way the eponential behaves on the arc We have Physics 8 Lecture 7 Appendi B 9 Winter 009
z Re i iz ir cosi sin Rsin ircos e e e e Thus the arc will be (eponentially) damped iff we choose sin 0 0 So in this case we choose to close in the upper half plane and find I iz e dz 3 losed Above 3 3 z i z i The final detail is that we must epand the integrand in order to find the appropriate residue at z = 3i/ (simple poles) To this end we note that iz 3 i z3i e e e 3 3 3 3 3 z i z i 3 z i 3i z i 3 e 3 3 i z i z i 3 3i 88z i 3 e 8 3 i z i 3 3 88z i Hence the desired residue and integral are given by 3 3 e 8i e R i 88 3 08 3 I ir e 54 Physics 8 Lecture 7 Appendi B 0 Winter 009
47: 33 Solution: Here we want to consider an integrand with a branch cut as in eample 5 in Boas We start with the integral d I 0 where we choose to put the branch cut of the square root along the positive real ais Net we consider a corresponding comple path integral around a contour as illustrated in Fig 74 in Boas z z J dz dz z z i z i = Fig 74 = Fig 74 lim i i i i Re ide R re ide r R r0 i i i R e i r e arc zr e arc zr e R r i d e d I r R where we have used the fact that neither arc makes a finite contribution and due to the phase change of the square root the line integrals along the positive real ais in opposite directions on the sides of the branch cut contribute equally Finally we use auchy to evaluate the closed contour integral J The contour encircles both poles z i in a counter-clockwise direction and we find i 4 i3 4 i 4 i 4 i i e e i e e J i i ie i i i i i sin 4 J I To obtain the correct result we need to carefully evaluate the residues especially noting that the phases of the square root at the two poles z i are determined by our choice of where to put the branch cut ie 0 on the sheet of interest Physics 8 Lecture 7 Appendi B Winter 009