(a) To show f(z) is analytic, explicitly evaluate partials,, etc. and show. = 0. To find v, integrate u = v to get v = dy u =

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1 Homework -5 Solutions Problems (a) z = + 0i, (b) z = i 2 f(z) = u(x, y) + iv(x, y) with u(x, y) = e 2y cos(2x) and v(x, y) = e 2y sin(2x) u (a) To show f(z) is analytic, explicitly evaluate partials,, etc and show x u that they satisfy C-R equations: = v, u = v x y y x (b) f (z) = u x + i v x = e2y 2 sin 2x ie 2y 2 cos 2x = 2if(z) (c) f(z) = e 2iz 3 To prove u = sin x cosh y + 2 cos x sinh y + x 2 y 2 + 4xy is harmonic, just show that 2 u + 2 u = 0 To find v, integrate u = v to get v = dy u = x 2 y 2 x y x cos x sinh y 2 sin x cosh y + 2xy + 2y 2 + φ(x) To determine the function φ(x), plug u and v into the second C-R equation u = v, which gives y x φ (x) = 4x so φ(x) = 2x 2 + const Finally, a little algebra can be used to show f(z) = u + iv = ( 2i) [sin z + ] + i const 4 u = ax 3 + bx 2 y + cxy 2 + dy 3 (a) Harmonic equation 2 u if c = 3a, b = 3d + 2 u x 2 y 2 = (6a+2c)x+(2b+6d)y = 0 This is satisfied (b) Write u = ax 3 3dx 2 y 3axy 2 + dy 3 Then using C-R equations as in problem 3, you can find v = 3ax 2 y ay 3 3dxy 2 + dx 3 + const (c) A little algebra gives f(z) = (a + id)z 3 + i const 5 We want to show that static Maxwell equations in free space imply that E (z) = E x ie y is analytic as a function of the two dimensions x and y with z = x+iy Writing E (z) = f(z) = u + iv in the standard notation, gives u = E x, v = E y Free space means ρ = 0, J = 0, and D = E, H = B Static means timeindependent ( E B = 0, = 0) Thus, the relevant Maxwell equations t t become E = 0 and E = 0 In 2-dimensions, E = 0 implies Ex + Ey x y implies Ex y Ey x = u y + v x = 0 = u x v y = 0 and E = 0

2 The two equations u = v u and = v are the Cauchy-Riemann equations Assuming continuity of partial derivatives, leads to analyticity of x y y x E (z) 6 Find Cauchy-Riemann equations in polar coordinates (z = re iθ ) There are several ways to do this Here s one f = u+i v and z = e iθ r +re iθ i θ We can take the derivative along a path with r constant or θ constant, then take the limit in the other variable If we keep θ constant, θ = 0 Then df If we keep r constant, r = 0 Then df = u + i v dz e iθ r e iθ r = u + i v dz re iθ i θ re iθ i θ Multiplying through by e iθ and equating real and imaginary parts of the above expressions gives u = v, v = u r r θ r r θ 7 The principle branch of the complex logarithm Log(z) (a) The cut z plane associated with Log(z) There is a Branch point singularity at the origin There is a branch cut along the negative real axis The phase is θ = π on the negative real axis and approaching it from above The phase is θ = π when approaching the negative real axis from below (b) The fact that the derivative of analytic function f(z) can be expressed as f (z) = e iθ ( u + i v ) follows directly from the solution to problem 6 r r (c) dlog(z) = e iθ = dz r z 8 Evaluate the following (a) Log( 2i)=Log(2e iπ/2 )=log 2 i π 2 (b) Tan ( 2i)= i [Log( i) Log(3i)]= i 2 2 [Log(e iπ/2 ) Log(3e iπ/2 )] = i [ i π (log 3 + i π)] = π i log (c) sin( 2i) = ei( 2i) e i( 2i) 2i = e2 e 2 2i = i sinh 2 9 The integral, z dz, does not make sense Since z is not an analytic function, this integral will depend on the path Since no path is given, the integral is ambiguous Since sin 2z is an analytic function, the other integral has one unambiguous answer, independent of path It is i 0 sin 2z dz = cos 2 2z i 0 = (cos 2i ) = ( cosh 2) Use Cauchy s integral formula to do the following contour integrals: (a) I = C ez dz, where C = boundary of a square with sides x = ±2, y = ±2 z πi/2 The square encompasses the pole at z = πi/2, so we get I = 2πi[e z ] z=πi/2 = 2πie πi/2 = 2πi

3 (b) I = C dz, where C = circle of radius, center i Factoring the denominator of the integral, I = dz +2 C (z+ 2i)(z, reveals that the integrand has 2i) two poles at z = ± 2i Only the pole at z = + 2i is inside the contour, so we get I = 2πi [ ] z+ 2i z= = 2πi 2i 2 = π/ 2 2i Do the following integrals: (a) I = C z4 +2z+ (z z 0 dz, where C is any closed contour containing z ) 4 0 Using the derivatives of Cauchy s integral formula gives I = 2πi [ d 3 (z 4 + 2z + ) ] = 2πi z 3! dz 3 z=z 0 3! 0 = 8πiz 0 (b) I = C cosh z dz, where C is any closed contour containing the origin Using z n+ the derivatives of Cauchy s integral formula gives I = 2πi [ d n cosh z ] = 2πi [ d n n! dz n z=0 n! dz n 2 (ez + e z ) ] z=0 = 2πi [ n! 2 (ez + ( ) n e z ) ] = 2πi [+( ) n ] z=0 n! 2 2 Find the order of the pole at the origin, the residue there, and the integral around a path C enclosing the origin, but no other singularities, for each of the following: (a) I like to use Taylor expansions to find these cot z = cos z = 2! z2 + = sin z z 2! z2 + = + 3! z3 + z( 3! z2 +) z The pole is order The Residue at the origin is The integral around the origin is 2πi (b) csc 2 z log ( z) = (c) log ( z) sin 2 z = z 2 z2 (z 3! z3 +) 2 = z(+ 2 z+) ( 3! z2 +) 2 = z + The pole is order The Residue at the origin is The integral around the origin is 2πi z = sin z tan z z cos z = z( 2! z2 +) = ( sin z(cos z ) (z 3! z3 +)( 2! z2 + 2! z2 +) 4! z4 + ) ( 3! z2 +)( 2! z2 + 4! z4 +) ( 2! z2 +) 2! z2 ( 3! z2 +)( 2! = = 2! ! z2 +) z The pole is order 2 The Residue at the origin is 0 The integral around the origin is 0 (Note that this function is even in z, so there are no odd powers in the Laurent expansion) 3 We want to evaluate I = dx 0 by considering the contour integral x 3 +a 3 C dz where the contour C consists of: C = a ray along θ = 0 from r = 0 to r = R, C 2 = an arc at r = R from θ = 0 to θ = 2π/3, C 3 = a ray along θ = 2π/3 from r = R to r = 0, taking R The various contributions to the contour integral give z 3 +a 3,

4 C : Let z = x, dz = dx Then C dz z 3 +a 3 C 2 : Let z = Re iθ, dz = Re iθ idθ Then C 2 = R 0 dx I as R x 3 +a 3 dz R/R 3 0 as R z 3 +a 3 C 3 : Let z = xe 2πi/3, dz = dxe 2πi/3 Then dz C 3 = 0 e 2πi/3 dx z 3 +a 3 R (e 2πi/3 x) 3 +a e2πi/3 I, where we used e 2πi = 3 Thus, the contour integral is C dz = ( e 2πi/3 )I We can evaluate the z 3 +a 3 contour integral using the residue theorem There are poles at a, ae iπ/3 and ae iπ/3 Only ae iπ/3 is inside the contour, so C dz = ( e 2πi/3 )I = 2πiRes [ ] = 2πi [ ] = 2πi [ ] z 3 +a 3 z 3 +a 3 z=ae πi/3 3 z=ae πi/3 3a 2 e 2πi/3 (I used the trick, Res [ ] P (z) = P (z 0) Q(z) z 0 Q (z 0 if the polynomial Q(z) has only a simple ) zero at z = z 0 ) Solving for I gives I = 2πi = 3a 2 e 2πi/3 ( e 2πi/3 ) 2π 3 3a 2 4 To evaluate I = 2π dθ 0, let z = e iθ and dz = e iθ idθ With this substitutionq, it becomes a contour integral around the unit circle, z = (a+cos θ) 2 Using cos θ = (e iθ + e iθ )/2 = (z + z )/2, we get I = dz iz(a+(z+z )/2) 2 = 4 i z dz ( +2az+) 2 = 4 i z dz (z z + ) 2 (z z, ) 2 where z ± = a ± a 2 Assuming a >, then only z + is inside the unit circle We can now use the Residue theorem (or the derivative of Cauchy s integral formula) We find I = 42πi [ ] d z i dz (z z = 8π (z ++z ) ) 2 z=z + (z + z = 2πa ) 2 (a 2 ) 3/2 5 We want to evaluate I = eax dx, where 0 < a <, by considering the +e x contour integral eaz C dz along the closed contour C connecting ρ+0i, ρ+0i, +e z ρ + 2πi, ρ + 2πi, and then letting ρ Along the segment ρ + 0i to ρ + 0i: Let z = x, dz = dx Then dz eaz = ρ eax +e z ρ dx I as ρ +e x Along the segment ρ + 0i to ρ + 2πi: Let z = ρ + yi, dz = idy Then dz eaz = 2π +e z 0 idy ea(ρ+iy) e (a )ρ 0 as ρ +e (ρ+iy) Along the segment ρ + 2πi to ρ + 2πi: Let z = x + 2πi, dz = dx Then dz eaz = ρ +e z ρ dx eax e 2πai e 2πai I as ρ +e x e 2πi Along the segment ρ + 2πi to ρ + 0i: Let z = ρ + yi, dz = idy Then dz eaz = 0 ea( ρ+iy) +e π idy e aρ 0 as ρ +e ( ρ+iy) Thus, we get eaz C dz = ( e 2πai )I The contour integral has only one pole +e z inside the contour, at z = πi The residue can be found by Taylor expanding around the pole: + Thus, we get C e az = eaπi e a(z πi) +e z e (z πi) eaπi (z πi) dz eaz +e z = ( e 2πai )I = 2πi [ e aπi ] Solving for I gives I = 2πieaπi e 2πai = π sin πa

5 6 We want to evaluate I = 0 dx log x by considering the contour integral x 2 +b 2 C dz log2 (z) ( +b 2 ) around the contour shown in Fig 622 in the text The contour has four contributions: A straight line just above the real axis, with phase θ = 0 Let z = x, dz = dx Then dz log2 (z) = R ( +b 2 ) ρ dx log2 (x) (x 2 +b 2 ) 0 dx log2 (x) as ρ 0 and (x 2 +b 2 ) R A large circle at radius R: Let z = Re iθ, dz = Re iθ idθ Then dz log2 (z) R log2 R 0 as R ( +b 2 ) R 2 A straight line just below the real axis, with phase θ = 2π Let z = xe 2πi, dz = dxe 2πi Then dz log2 (z) = ρ ( +b 2 ) R dx (log(x)+2πi)2 (x 2 +b 2 ) 0 dx (log(x)+2πi)2 (x 2 +b 2 ) as ρ 0 and R A small circle at radius ρ: Let z = ρe iθ, dz = ρe iθ idθ Then dz log2 (z) ρ log2 ρ 0 as ρ 0 ( +b 2 ) b 2 Combining these contributions, we obtain C dz log2 (z) = ( +b 2 ) 0 dx log2 (x) (log(x)+2πi) 2 = (x 2 +b 2 ) Solving for the integral I that we want, we get 0 dx (2π)2 (x 2 +b 2 ) 0 dx 4πi log(x) (x 2 +b 2 ) { I = 2π2 4πi dx (x 2 +b 2 ) C dz } log2 (z) ( +b 2 ), where I ve used the symmetry of the first integral to divide by 2 and change the lower limit on the integration to The poles are at z = ±ib The first integral can be done by completing in the upper half-plane (hence only picking up the residue at z = +ib), while the second integral uses the contour of Fig 622 and { picks up the residue at both poles We get I = (2πi) Res [ ] 2π 2 Res [ ] log 2 (z) Res [ ] } log 2 (z) 4πi ( +b 2 ) z=ib ( +b 2 ) z=ib ( +b 2 ) z= ib { = 2π 2 } (log(b)+iπ/2)2 (log(b)+3iπ/2)2 2 2ib 2ib 2ib = π log b 2b (A comment about phases: The choice of branch cut requires that θ range between 0 and 2π Thus, +ib = be πi/2 and ib = be 3πi/2 ) 7 We want to evaluate the integral I = P cos(ax) cos(bx) dx by examining the x 2 contour integral C dz eiaz e ibz along the real axis and on a large semi-circle in the upper half plane (radius R), with the origin excluded using a small semi-circle (radius ρ) Assume that both a and b are positive The contribution from the large semi-circle goes to zero as R The contribution of the segments along the real axis give the principle value integral P dx eiax e ibx as R and ρ 0 x 2 The contribution of the small semi-circle gives πires [ e iaz e ibz 0 ] z=0 as ρ

6 (I have left out details here, because the arguments are exactly as we did in class) Thus, we get P dx eiax e ibx = x 2 C dz eiaz e ibz + πires [ ] e iaz e ibz z=0 The only singularities of the integrand are at z = 0, which is excluded from the contour C Thus, C dz eiaz e ibz = 0 The residue at zero can be found by e Taylor expansion: iaz e ibz = +iaz+ (+ibz+) = i(a b) + Thus, we get z P dx eiax e ibx = πires [ ] e iaz e ibz = πi[i(a b)] = π(b a) Taking the x 2 z=0 real part of both sides of the equation gives I = π(b a) 8 We want to evaluate I = 0 dx x2a with 0 < a < Consider the contour b 2 +x 2 integral z2a C dz, where the contour C is the same as in problem 6 There b 2 + are four contributions: A straight line just above the real axis, with phase θ = 0 Let z = x, dz = dx Then dz z2a = R x2a b 2 + ρ dx I as ρ 0 and R b 2 +x 2 A large circle at radius R: Let z = Re iθ, dz = Re iθ idθ Then dz z2a R2a 0 as R b 2 + R 2 A straight line just below the real axis, with phase θ = 2π Let z = xe 2πi, dz = dxe 2πi Then dz z2a = ρ b 2 + R dx (xe2πi ) 2a e 2πi(2a ) I as ρ 0 b 2 +x 2 and R A small circle at radius ρ: Let z = ρe iθ, dz = ρe iθ idθ Then dz z2a ρ2a 0 as ρ 0 b 2 + b 2 Combining these contributions, we obtain z2a C dz = ( e 2πi(2a ) )I b 2 + The poles of the complex function are at +ib = be πi/2 and ib = be 3πi/2, where we are careful to define the phase with 0 θ < 2π So I = 2πi [ ] (be πi/2 ) 2a + (be3πi/2 ) 2a e 2πi(2a ) 2ib 2ib After some algebra this can be reduced to I = π 2 (b2 ) a sin (πa) 9 We are given χ I (ω) = (ω ω 0 ) (a) We want to find χ R (ω) Since we cannot assume χ(t) is real, we cannot use the Kramers-Kronig formula We must use the Hilbert transform formula, which makes no reality assumptions Thus, we have χ R (ω) = π P dz χ I(z) z ω = π P dz z ω (z ω 0 ) = P π dz z ω (z ω 0 +i)(z ω 0 i) There are three poles The one at z = ω is on the real axis and is treated by the Principle-value prescription The pole at z = ω 0 + i is in the upper half-plane The pole at z = ω 0 i is in the lower half-plane Using the method for finding Principle-value integrals from class we obtain

7 { 2πiRes [ z ω ] + πires [ (z ω 0 +i)(z ω 0 i) z=ω 0 +i z ω ] } (z ω 0 +i)(z ω 0 i) z=ω χ R (ω) = π = ω 0 ω (ω ω 0 ) Note that only the residue in the upper half plane is included in the first term in brackets (b) We can combine χ I (ω) and χ R (ω) into a single complex function of a complex variable, χ(ω) = χ R (ω) + i χ I (ω) = ω 0 ω+i (ω ω 0 = ) ω (ω 0 i) This function has a pole in the lower half-plane Thus, it is () analytic in the upper half-plane, and (2) its magnitude goes to zero as ω Thus, the use of the Hilbert transform is justified (c) Include a sketch of χ I (ω) and χ R (ω) for = /0 and ω 0 = in some units here 20 Find the shortest distance between two points in polar coordinates The differential line element is ds 2 = dr 2 + r 2 dθ 2 One can choose r = r(θ) or θ = θ(r) Either will work as easily as the other I will choose r = r(θ) Then the distance of a path between the two points is given by the integral L = θ B θ A dθ Letting f = r 2 + (r ) 2, we have f r 2 + (r ) 2 = r f r and = r r Since there 2 +(r ) 2 r r 2 +(r ) 2 is no explicit dependence of f on θ, we can use the integrated Euler-Lagrange equation r f r solving for r gives r = dr r f = c, where c is a constant Plugging into this equation and r = r r 2 c 2 We can integrate this by quadrature: dθ c d r r = θ d θ = θ A, where A is an integration constant The inte- 2 /c 2 gration over r can be done by change-of-variables I will do this in two steps First, let r = c/x, d r = [c/x 2 ]dx This gives θ A = c/r dx x 2 Now let x = sin φ, dx = cos φ dφ This gives θ A = sin c/r dφ = sin c/r We now have the equation r sin(θ A) = c, which is a straight line in polar coordinates This can be seen explicitly by rewriting it as r sin θ cos A r cos θ sin A = c = y cos A x sin A = c 2 Fermat s principle says to extremize I = ds = x 2 +(y u x dx ) 2 u(y) (a) We will use this to derive Snell s law The function f = + (y ) 2 /u(y) does not depend explicitly on x, so we can use the integrated Euler- Lagrange equation, y f we need f y = y u(y) +(y ) f = c, where c is a constant For this y f Plugging f and into the integrated Euler- 2 Lagrange equation, a little algebra gives u(y) +(y = c ) 2 Now we just need to use geometry Referring to Fig 26 in the textbook, y

8 we see that sin φ = dx ds = dx dx +(y = ) +(y 2 ) 2 Combining this with the result from the Euler-Lagrange equation gives sin φ/u(y) = c = constant (b) Suppose the speed is proportional to y Write u(y) = Ay, where A is some constant The Euler-Lagrange equation from part (a) gives +(y = ) 2 Ay +(y ) 2 = a constant Let the constant be /(ca) We can solve to find y = c 2 /y 2 We can integrate this by quadrature: y ỹdỹ = x d x = x x 0 c 2 ỹ 2 Use change of variables, letting c 2 ỹ 2 = w 2, 2ỹdỹ = 2wdw This gives c 2 y 2 dw = c 2 y 2 = x x 0 Squaring both sides gives (x x 0 ) 2 + y 2 = c 2, which is a circle centered at (x 0, 0) 22 Consider the cycloid that connects the origin (0, 0) and the point (πa, 2a) (a) Calculate the time required for a mass to slide from the origin to (πa, 2a) The equations for the cycloid, from class are x = a(θ sin θ) and y = a( cos θ) So θ goes from 0 to π The time to the bottom is T = (πa, 2a) (0,0) We can do the integral in terms of θ, using ds = dθ (dx/dθ) 2 + (dy/dθ) 2 = dθ a 2( cos θ) and v = 2gy Plugging these into the equation for the total time gives T = π 0 dθ a 2( cos θ) = π a/g 2ga( cos θ) (b) Suppose the mass is released from rest at an arbitrary point (x A, y A ) (other than the origin) on this cycloid We want to show that the time is unchanged from the result in part (a) Since it is released from rest at a different point, conservation of energy gives mv 2 /2 = mg(y A y), so we get v = 2g(y A y) We can now write the integral for the total time in terms of θ as before, except we start at an arbitrary initial θ A and the velocity formula is replaced with the new one Thus, we have T = (πa, 2a) ds (x A,y A ) = π a 2( cos θ) v θ A dθ 2g[a( cos θ) a( cos θ A )] = a/g π sin θ A dθ 2 (θ/2) [cos 2 (θ A /2) cos 2 (θ/2)] Letting u = cos(θ/2) sin(θ/2)dθ cos(θ A, du = gives T = /2) 2 cos(θ A a/g /2) 0 du 2 u 2 This integral can now be done by the substitution u = sin φ, du = cos φdφ, giving T = a/g π/2 0 2dφ = π a/g, exactly as before u(y) ds v