SOLUTIONS MANUAL FOR. Advanced Engineering Mathematics with MATLAB Third Edition. Dean G. Duffy
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1 SOLUTIONS MANUAL FOR Advanced Engineering Mathematics with MATLAB Third Edition by Dean G. Duffy
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3 SOLUTIONS MANUAL FOR Advanced Engineering Mathematics with MATLAB Third Edition by Dean G. Duffy Taylor & Francis Group, an informa business
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5 hapter Solution Manual Section.... 5i + i = 5i + i i i = 5 + i 4 + = + i 5 + 5i 4i i = 5 + 5i 4i + 4i + 4i i 4 i 4 i = + 7i i 5 = i + i 4i + i = + i 5i 4i + 4i + 4i + i 5i + i = + i = 5i 5i ( i 4 = ( i ( i = ( i = 4 i( i ( + i = i( i + i + = + i r = + ( = θ = tan ( / = π/ or π/ Because i lies below the real axis, θ = π/ and z = e πi/. 7. r = 4 + = 4 and θ = tan (/ 4 = or π Because 4 lies in the left side of the complex plane, θ = π and z = 4e πi. 8. r = 4 + = 4
6 Advanced Engineering Mathematics with MATLAB θ = tan ( / = tan ( = π/ or 4π/ Because + i lies in the first quadrant, z = 4e πi/. 9. r = = 5 θ = tan 5/( 5 = tan ( = π/4 or 7π/4 Because 5 + 5i lies in the second quadrant, z = 5 e πi/4.. r = = θ = tan ( / = tan ( = π/4 or 7π/4 Because i lies in the fourth quadrant, z = e 7πi/4.. r = + = θ = tan /( = tan ( = π/ or 5π/ Because + i lies in the second quadrant, z = e πi/.. e (α+βi = e αi e βi cos(α + β + isin(α + β = cos(α + isin(αcos(β + isin(β = cos(α cos(β sin(α sin(β + icos(αsin(β + isin(αcos(β Taking the real and imaginary parts, cos(α + β = cos(αcos(β sin(αsin(β and sin(α + β = cos(αsin(β + sin(αcos(β.. N e int = n= expi(n + t exp(it expi(n + t/ exp i(n + t/ = exp(int/ exp(it/ exp( it/ int/ sin(n + t/ = e. sin(t/
7 Worked Solutions To obtain the final answer, take the real and imaginary parts of the last equation. 4. (a ǫ n e int = ǫexp(it = ǫcos(t iǫsin(t n= = ǫcos(t + iǫsin(t + ǫ. ǫcos(t To obtain the final answer, take the real and imaginary parts of the last equation. (b e na e a sin(t sin(nt = + e a e a cos(t = sin(t cosh(a cos(t. n= Now multiply both sides of the equation by. Section.. Therefore, or 8 = e i+kπi. z k = e kπi/, k =,,,,4,5 z =, z = ( π ( π cos + isin = + z = cos z 4 = cos ( π + isin ( 4π ( π = i + z = e πi =, ( 4π + isin = i i,, and z 5 = cos ( 5π + isin ( 5π = i.. = e πi+kπi.
8 4 Advanced Engineering Mathematics with MATLAB Therefore, or and z k = e (π+kπi/, k =,, ( π ( π z = e πi/ = cos + isin = + i, z = e πi = z = e 5πi/ = cos ( ( 5π 5π + isin = i.. Therefore, or and i = e πi/+kπi. z k = e πi/+kπi/, k =,, z = e πi/ = i, ( ( 7π 7π z = e 7πi/6 = cos + isin = 6 6 i ( ( π π z = e πi/6 = cos + isin = 6 6 i. 4. Therefore, or and 7i = e πi/+kπi. z k = e πi/4+kπi/, k =,,,,4,5 z = ( π ( π cos + isin,z = cos 4 4 z = cos z = cos z 4 = cos z 5 = cos ( π ( 5π 4 ( 9π + isin + isin + isin ( π + isin ( 7π + isin ( π ( 5π 4 ( 9π,,, ( π. ( 7π,
9 Worked Solutions 5 5. If we find one root w, then the other root is w. Let z = re θi, where r = a + b and θ = tan ( b/a. The value of θ lies between π/ and π because z lies in the fourth quadrant. We know that w = re θi/ = r cos(θ/ + isin(θ/. But cos(θ/ = ± + cos(θ/ and sin(θ/ = ± cos(θ/, where cos(θ = a/ a + b. Therefore, a a w = (a + b /4 + b + a + b + i a a + b a + b = ( a a + b + a + i + b + a. Our choice of signs for cos(θ/ and sin(θ/ is determined by the fact that π/4 < θ/ < π. 6. From the quadratic formula, z = (i ± 9 + 8/ = i,i. Therefore, z, = ± e πi/4 = ±( + i and z,4 = ±e πi/4 = ±( + i. 7. From the quadratic formula, Therefore, z = 6i ± 6 64 = 8i,i. z, = ± e πi/4 = ±( + i and z,4 = ± e πi/4 = ±( i. Section.. w = u + iv = i(x + iy + = y + xi Therefore, u = y and v = x. Then, u x =, v y =, v x = and u y =. Thus, u x = v y and v x = u y for all x and y. Therefore, iz + is an entire function because the derivatives are continuous.. w = u + iv = e (x+iy = e x cos(y isin(y. Therefore, u = e x cos(y and v = e x sin(y. Then, v x = e x sin(y, u x = e x cos(y, v y = e x cos(y and u y = e x sin(y. Thus, u x = v y and v x = u y for all
10 6 Advanced Engineering Mathematics with MATLAB x and y. Therefore, e z is an entire function because the derivatives are continuous.. w = u + iv = (x + iy = x xy + i(x y y Therefore, u = x xy and v = x y y. Then, u x = x y, v y = x y, v x = 6xy and u y = 6xy. Thus, u x = v y and v x = u y for all x and y. Therefore, z is an entire function because the derivatives are continuous. 4. w = u + iv = cosh(x + iy = cosh(xcos(y + isinh(xsin(y. Therefore, u = cosh(xcos(y and v = sinh(xsin(y. Then, u x = sinh(xcos(y, v y = sinh(xcos(y, v x = cosh(xsin(y and u y = cosh(xsin(y. Thus, u x = v y and v x = u y for all x and y. Therefore, cosh(z is an entire function because the derivatives are continuous. 5. f (z = ( + z / (z = z( + z / 6. f (z = (z + z/ / ( + z / 7. f (z = ( + 4iz 8. f (z = z + i z i (z + i = 5i (z + i 9. f (z = i(iz 4. z iz lim z i z 4 + z + = lim z i z i 4z + 4z = lim z i z + 4 = 4. z sin(z cos(z sin(z cos(z lim z z = lim z z = lim = lim = z 6z z 6 6
11 Worked Solutions 7. Because u = x and v = y, we have u x = and v y =. Therefore, u x v y for any x and y and f(z is not differentiable anywhere on the complex plane.. Because u xx = and u yy =, we have u xx + u yy = and u(x,y is harmonic. To find v(x,y, we use the auchy-riemann equations: v y = u x = x or v(x,y = xy + f(x. To find f(x we use v x = y + f (x = u y = y or f (x =. Therefore, the final answer is v(x,y = xy + constant. 4. Because u xx = x y and u yy = x + y, we have that u xx + u yy = and u(x,y is harmonic. To find v(x,y, we use the auchy-riemann equations: v y = u x = 4x xy + or v(x,y = 4x y 4xy +y +f(x. To find f(x we use v x = x y 4y +f (x = u y = x y 4y or f (x =. Therefore, the final answer is v(x,y = 4x y 4xy + y + constant. 5. Because u xx = sin(x xcos(xe y + ye y sin(x and u yy = xcos(xe y + e y sin(x ye y sin(x, we have u xx + u yy = and u(x,y is harmonic. To find v(x,y, we use the auchy-riemann equations: v y = u x = cos(x xsin(xe y ye y cos(x or v(x,y = xsin(xe y + ye y cos(x + f(x. To find f(x we use v x = sin(xe y + xcos(xe y sin(xye y + f (x = u y = xcos(xe y + sin(xe y ye y or f (x =. Therefore, the final answer is v(x,y = xsin(xe y + ye y cos(x + constant. 6. Because u xx = cos(ye x +4xcos(ye x +(x y cos(ye x 4y sin(ye x xy cos(ye x and u yy = cos(ye x +4y sin(ye x (x y cos(ye x 4xcos(ye x +xy sin(ye x,
12 8 Advanced Engineering Mathematics with MATLAB we have that u xx + u yy = and u(x,y is harmonic. To find v(x,y, we use the auchy-riemann equations: or v y = u x = xcos(ye x + (x y cos(ye x y sin(ye x xy sin(ye x To find f(x we use v(x,y = xy cos(ye x + (x y sin(ye x + f(x. v x = y cos(ye x + xy cos(ye x + xsin(ye x + (x y sin(ye x + f (x = u y = y cos(ye x + (x y sin(ye x + xsin(ye x + xy cos(ye x or f (x =. Therefore, the final answer is v(x,y = xy cos(ye x + (x y sin(ye x + constant. Section.4. Because z = e θi, z = e θi and dz = ie θi dθ. Then (z dz = π e θi ie θi dθ = π π e θi idθ = e θi =.. Then z dz = z dz = Finally, z dz = (x + y (dx + idy = z dz + z dz + z dz + z dz 4 (x + dx =, z dz = (x + dx =, 4 z dz = z dz = + i. ( + y idy = i( +,. We have z = e θi and dz = ie θi dθ with π/ < θ < π/. Then z dz = π/ π/ ie θi dθ = e θi π/ π/ = i. ( + y idy = i.
13 Worked Solutions 9 4. Because y = x, z = x + ix and dz = dx + idx. Then e z dz = e (+ix ( + idx = e (+ix = e +i e i = sinh(cos( + icosh(sin(. 5. Because y = x, z = x + x i and dz = ( + ixdx. Then (z dz = = = (x ix ( + ixdx (x x 4 ix ( + ixdx (x + x 4 ix 5 dx = 4 5 i. 6. For (a, z = e θi and dz = ie θi dθ. Then For (b dz π iexp(θi = z exp(θi/ dθ = π ie θi/ dθ = e θi/ π = + i. dz = ie θi/ dθ = e θi/ = + i. z π π Note the jump in e θi/ as we move across the negative real axis. Just above the negative real axis, e θi/ = i while below this axis we have e θi/ = i. This jump occurs because of the presence of the branch cut along the negative real axis associated with our multivalued, complex square root function. Section.5. Because u = e x cos(y and v = e x sin(y, u x = e x cos(y = v y and v x = e x sin(y = u y for all x and y. Therefore, any integration between two points is path independent. Thus, +πi πi e z dz = e z +πi πi = e 4 6πi e +πi = e e 4.. Because u = e x cos(y cos(xcosh(y and v = e x sin(y + sin(xsinh(y, u x = e x cos(y + sin(xcosh(y = v y and v x = e x sin(y + cos(xsinh(y =
14 Advanced Engineering Mathematics with MATLAB u y for all x and y. Therefore, any integration between two points is path independent. Thus, π e z cos(zdz = e z sin(z π = e π.. Because u = cos(xcosh(y and v = sin(xsinh(y, u x = sin(xcosh(y = v y and v x = cos(xsinh(y = u y for all x and y. Therefore, any integration between two points is path independent. Thus, π sin (zdz = π cos(zdz = z sin(z π = π/. 4. Because u = x + and v = y, u x = = v y and v x = = u y for all x and y. Therefore, any integration between two points is path independent. Thus, i i (z + dz = ( z + z i i = + i. Section.6... z = z = z = sin 6 (z (z π/ z =/ sin 6 (z z π/6 dz = πisin6 (π/6 = πi πi dz =! d sin 6 (z dz z=π/6 = πi sin 4 (π/6cos (π/6 6sin 6 (π/6 = πi/6 dz z(z + 4 = /(z + 4 dz = πi z = z z = tan(z z dz = πitan(z z= = dz (z (z = /(z z =/ z z + 4 = πi z= dz = πi z = πi z=
15 Worked Solutions 6. z =5 exp(z z dz = πi! d (e z dz = πi z= ( 4z e z + e z z= = πi 7. z = z + z dz = (z + /(z + z = z dz = πi z + z + = πi z= z = z = z = z (z + i z (z πi dz =! cos(z πi dz = zn+ (n! πi dz = 4! d (z dz d (z dz d n cos(z dz n = z= = 6πiz z=i = 6π z=i = πi z= (n! ( n MATLAB code used in the complex variables project MATLAB ode for omplex Variables Project initialize parameters clear n = ; R = ; x = ; y = ; load in values for Gaussian-Legendre quadrature x gauss( = ; A( = ; x gauss( = ; A( = ; x gauss( =.; A( = ; x gauss(4 = ; A(4 = ; x gauss(5 = ; A(5 = ; compute n!
16 Advanced Engineering Mathematics with MATLAB factorial = ; radius = ; answer ave = ; for j = :n factorial = j * factorial; radius = R * radius; end test out various resolutions for m = :85 M = m+5; m plot(m = M; dtheta = * pi / M; answer = ; answer = ; now do the integration the circle for j = :M a = (j-*dtheta; b = j*dtheta; h =.5*(b-a; ave =.5*(b+a; for k = :5 theta = ave + h * x gauss(k; x = x + R * cos(theta; y = y + R * sin(theta; z = x + i*y; f = 8 * z / (z * z+4; u = real(f; v = imag(f; cosine = cos(n * theta; sine = sin(n * theta; integrand = u * cosine + v * sine; integrand = v * cosine - u * sine; answer = answer + h * A(k * integrand; answer = answer + h * A(k * integrand; end end nth derivative = answer + i * answer format long answer = factorial * answer / ( * pi * radius; answer = factorial * answer / ( * pi * radius; derivative plot(m = answer; answer ave = answer ave + derivative plot(m; end
17 Worked Solutions plot difference in the answer as a function of order of scheme answer ave = answer ave / m; plot(m plot,derivative plot-answer ave xlabel( number of Gaussian intervals, Fontsize, ylabel( the eleventh derivative of f(z, Fontsize, Section.7. ( z = n= f (n ( z n n! Because f (n (z = (n +!/( z n+, f (n ( = (n +! Then ( z = n= (n +! z n = n! (n + z n. n=. f(z = e(z e (z = e(z + z (z (z + + +!!! = e(z + e(z + e(z + 6 e(z 4 +. ( f(z = z z + z 6z +!z + = z z 9 + z8 6 z7 +!z + We have an essential singularity and the residue equals /! 4. f(z = z sin (z = z cos(z ( = z + (z! (z4 4! + (z6 6! = z z + 5 z 4! 6! We have a simple pole with a residue that equals. 5. f(z = cosh(z z = ( z + z! + z4 4! + z6 6! + =! + z 4! + z4 6! +
18 4 Advanced Engineering Mathematics with MATLAB We have a removable singularity where the value of the residue equals zero. 6. f(z = z + (z + = (z + + z + We have a second-order pole where the residue equals one. 7. e z + e z = + z + z + 6 z + z + z 6 z + = z ( + z + z + 6 z + ( + z + z + = z 7 6 z z We have a simple pole and the residue equals. 8. e iz z + b = e i(z bi e b (z bibi + (z bi = e b bi = e b bi z bi (z bi + bi (bi z bi + i(z bi + i (z bi + z bi + e b e b ( + b 4b 8b i ( + b + b (z bi + We have a simple pole and the residue equals e b /(bi. 9. f(z = + (z (z = z (z + z 4 = z 4 + (z 8 We have a simple pole and the residue equals /.. f(z = ( z 4 + z + z4! + z6! + = z 4 + z + + z 6 + We have a fourth-order pole and the residue equals zero. Section.8
19 Worked Solutions 5. because z = ( z + z + z 4 dz = πi z z 4 z ; = πi 4, ( z + z 4 z ; = lim z = lim z d! dz z z + z (z (z + z + (z = 8.. z = (z + 4 (z + 4 z 4 + 5z dz = πi + 6z z 4 + 5z + 6z ; = 6πi 9 because (z + 4 z 4 + 5z + 6z ; d = lim z (z + 4 z dz z (z + 5z + 6 = lim z (z + 4 z + 5z + 6 (z + 4 (z + 5 (z + 5z + 6 = Because e z = z z = z = ( z, z ( + z + we have a simple pole at z = with =. Therefore, z = dz = πi. ez 4. because z = z 4 z 4 dz = πi (z 4 (z 4 ; = z 4 (z 4 ; =! lim d z dz (z 4 z 4 (z 4 =.
20 6 Advanced Engineering Mathematics with MATLAB 5. The singularities are located where z 4 = or z n = ± and ±i. The corresponding residues are and ( z z 4 ; (z z ( ( = lim z z 4 = lim z z ( z z 4 ; ( z z 4 ;i (z + z = lim z z 4 = lim z i (z iz ( = lim z z lim z ( ( ( z 4 = lim z lim z i z i 4z = 4, lim z 4z 4z = 4, = 4 ( z z 4 ; i (z + iz ( ( = lim z i z 4 = lim z i z lim z i 4z = 4. Therefore, z = z z 4 dz = πi. 6. Because the Laurent expansion of the integrand is f(z = z n + z n + + n+ +, (n +! z we have an essential singularity and the residue equals n+ /(n+! Therefore, z n e /z n dz = 4πi (n +!. z = 7. Because e /z cos(/z = e (+i/z + e ( i/z = + z + z +, we have an essential singularity and the residue equals one. Then e /z cos(/zdz = πi. z = 8. z = + 4cos(πz z(z { + 4cos(πz dz = πi z(z ; } + 4cos(πz + z(z ; = 6πi
21 Worked Solutions 7 because and + 4cos(πz z(z ; = lim z + 4cos(πz z z(z = 6 + 4cos(πz d z(z ; = lim (z + 4cos(πz z dz z(z =. Section.9. dx x 4 + = dx x 4 + = dz z 4 +, where denotes a semicircle of infinite radius in the upper half-plane. This contour encloses two simple poles at z = e πi/4 and z = e πi/4. Therefore, dz z 4 = πi + z 4 + ;eπi/4 = πi lim z e πi/4 z e πi/4 = πi + πi z πi e πi/4 + e 9πi/4 = π. z 4 + ;eπi/4 lim z e πi/4 z e πi/4 z 4 + The final result follows by substitution into the first equation.. dx (x + 4x + 5 = dz (z + 4z + 5, where denotes a semicircle of infinite radius in the upper half-plane. This contour encloses a second-order pole at z = + i. Therefore, dz (z = πi + 4z + 5 (z + 4z + 5 ; + i d = πi lim z +i dz (z + i (z + i (z + + i = πi lim z= +i (z + + i = π. The final result follows by substitution into the first equation.. x dx (x + (x + x + = z dz (z + (z + z +,
22 8 Advanced Engineering Mathematics with MATLAB where denotes a semicircle of infinite radius in the upper half-plane. This contour encloses simple poles at z = i and z = + i. Therefore, z dz (z + (z + z + = πi z (z + (z + z + ;i z + πi (z + (z + z + ; + i (z iz = πi lim z i (z i(z + i(z + z + + πi lim = πi z +i (z + iz (z + (z + i(z + + i + i = π (i( i 5. i (i( + i + The final result follows by substitution into the first equation. 4. x dx x 6 + = x dx x 6 + = z dz z 6 +, where denotes a semicircle of infinite radius in the upper half-plane. This contour encloses three simple poles at z = e πi/6, z = e πi/ and z = e 5πi/6. Therefore, z ( ( dz z z z 6 = πi + z 6 + ;eπi/6 + πi z 6 + ;eπi/ ( z + πi z 6 + ;e5πi/6 (z e πi/6 z (z e πi/ z = πi lim z e πi/6 z 6 + πi lim + z e πi/ z 6 + (z e 5πi/6 z + πi lim z e 5πi/6 z 6 + = πi 6i + 6i + = π 6i. The final result follows by substitution into the first equation. 5. dx (x + = dx (x + = dz (z +, where denotes a semicircle of infinite radius in the upper half-plane. This
23 Worked Solutions 9 contour encloses a second-order pole at z = i. Therefore, dz (z = πi + (z + ;i d (z i = πi lim z i dz (z i (z + i = πi lim z i (z + i = π. The final result follows by substitution into the first equation. 6. dx (x + (x + 4 = dx (x + (x + 4 = dz (z + (z + 4, where denotes a semicircle of infinite radius in the upper half-plane. This contour encloses simple poles at z = i and a second-order pole at z = i. Therefore, dz (z + (z = πi + 4 (z + (z + 4 ;i + πi (z + (z + 4 ;i = πi lim z i d + πi lim z i dz z i (z + i(z i(z + 4 (z i (z + (z i (z + i = πi (i(9 + (πi( (4i ( + (πi( (4i (4i ( = 5π 44. The final result follows by substitution into the first equation. 7. x (x + a (x + b dx = z (z + a (z + b dz, where denotes a semicircle of infinite radius in the upper half-plane. This contour encloses a simple pole at z = ai and a second-order pole at z = bi. Therefore, z z (z + a (z + b dz = πi (z + a (z + b ;ai + πi z (z + a (z + b ;bi.
24 Advanced Engineering Mathematics with MATLAB At z = ai, z (z + a (z + b ;ai = lim z ai = (z ai (z ai(z + ai lim z ai ai (a b. z (z + b At z = bi, z (z + a (z + b ;bi d z = lim z bi dz (z + a (z + bi i = b(a b bi (a b + i 4b(a b. Therefore, x (x + a (x + b dx = πi ai (a b i b(a b bi (a b + i 4b(a b π = b(a + b. 8. t (t + t (a/h + + (a/h dt = (t + t (a/h + + (a/h dt z = ( + a/h (z + (z + b dz where b = ( h/a/( + h/a. We have a simple pole at z = i and z = bi. Then z z (z + (z + b dz = πi (z + (z + b ;i + πi t z (z + (z + b ;bi = πi lim z i (z iz (z i(z + i(z + b + πi lim z bi (z biz (z + (z bi(z + bi = πi (i(b + πi b (bi( b = π( b b
25 Worked Solutions and ( z (z + (z + b dz = π h/a. ( h/a/( + h/a + h/a Substitution into the first equation yields the desired result. 9. We begin by converting the real integral into a closed contour integration: π/ dθ a + sin (θ = π 4 dθ a + sin (θ = i z = z (z 4az dz, where z = e θi. The integrand has four poles: z = ± a ± + a. Only two are located inside the contour: z = a + + a and z = a + a. The corresponding residues are and z (z 4az ;z z (z 4az ;z (z z z = lim z z (z 4az = 8 a + a (z z z = lim z z (z 4az ;z = 8 a + a. Employing the residue theorem and substituting into the top line, we have that π/ dθ a + sin (θ = π a + a.. We begin by converting the real integral into a closed contour integration: π/ dθ a cos (θ + b sin (θ = π dθ 4 a cos (θ + b sin (θ z = i a (z + b (z dz, z = where z = e θi. The integrand has four simple poles located at z + = b + a b a, and z = b a b + a.
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