SOLUTIONS MANUAL FOR. Advanced Engineering Mathematics with MATLAB Third Edition. Dean G. Duffy

Size: px
Start display at page:

Download "SOLUTIONS MANUAL FOR. Advanced Engineering Mathematics with MATLAB Third Edition. Dean G. Duffy"

Transcription

1 SOLUTIONS MANUAL FOR Advanced Engineering Mathematics with MATLAB Third Edition by Dean G. Duffy

2

3 SOLUTIONS MANUAL FOR Advanced Engineering Mathematics with MATLAB Third Edition by Dean G. Duffy Taylor & Francis Group, an informa business

4

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.

Complex Function. Chapter Complex Number. Contents

Complex Function. Chapter Complex Number. Contents Chapter 6 Complex Function Contents 6. Complex Number 3 6.2 Elementary Functions 6.3 Function of Complex Variables, Limit and Derivatives 3 6.4 Analytic Functions and Their Derivatives 8 6.5 Line Integral

More information

Residues and Contour Integration Problems

Residues and Contour Integration Problems Residues and ontour Integration Problems lassify the singularity of fz at the indicated point.. fz = cotz at z =. Ans. Simple pole. Solution. The test for a simple pole at z = is that lim z z cotz exists

More information

18.04 Practice problems exam 1, Spring 2018 Solutions

18.04 Practice problems exam 1, Spring 2018 Solutions 8.4 Practice problems exam, Spring 8 Solutions Problem. omplex arithmetic (a) Find the real and imaginary part of z + z. (b) Solve z 4 i =. (c) Find all possible values of i. (d) Express cos(4x) in terms

More information

18.04 Practice problems exam 2, Spring 2018 Solutions

18.04 Practice problems exam 2, Spring 2018 Solutions 8.04 Practice problems exam, Spring 08 Solutions Problem. Harmonic functions (a) Show u(x, y) = x 3 3xy + 3x 3y is harmonic and find a harmonic conjugate. It s easy to compute: u x = 3x 3y + 6x, u xx =

More information

Math 312 Fall 2013 Final Exam Solutions (2 + i)(i + 1) = (i 1)(i + 1) = 2i i2 + i. i 2 1

Math 312 Fall 2013 Final Exam Solutions (2 + i)(i + 1) = (i 1)(i + 1) = 2i i2 + i. i 2 1 . (a) We have 2 + i i Math 32 Fall 203 Final Exam Solutions (2 + i)(i + ) (i )(i + ) 2i + 2 + i2 + i i 2 3i + 2 2 3 2 i.. (b) Note that + i 2e iπ/4 so that Arg( + i) π/4. This implies 2 log 2 + π 4 i..

More information

Suggested Homework Solutions

Suggested Homework Solutions Suggested Homework Solutions Chapter Fourteen Section #9: Real and Imaginary parts of /z: z = x + iy = x + iy x iy ( ) x iy = x #9: Real and Imaginary parts of ln z: + i ( y ) ln z = ln(re iθ ) = ln r

More information

n } is convergent, lim n in+1

n } is convergent, lim n in+1 hapter 3 Series y residuos redit: This notes are 00% from chapter 6 of the book entitled A First ourse in omplex Analysis with Applications of Dennis G. Zill and Patrick D. Shanahan (2003) [2]. auchy s

More information

MA 201 Complex Analysis Lecture 6: Elementary functions

MA 201 Complex Analysis Lecture 6: Elementary functions MA 201 Complex Analysis : The Exponential Function Recall: Euler s Formula: For y R, e iy = cos y + i sin y and for any x, y R, e x+y = e x e y. Definition: If z = x + iy, then e z or exp(z) is defined

More information

Syllabus: for Complex variables

Syllabus: for Complex variables EE-2020, Spring 2009 p. 1/42 Syllabus: for omplex variables 1. Midterm, (4/27). 2. Introduction to Numerical PDE (4/30): [Ref.num]. 3. omplex variables: [Textbook]h.13-h.18. omplex numbers and functions,

More information

(1) Let f(z) be the principal branch of z 4i. (a) Find f(i). Solution. f(i) = exp(4i Log(i)) = exp(4i(π/2)) = e 2π. (b) Show that

(1) Let f(z) be the principal branch of z 4i. (a) Find f(i). Solution. f(i) = exp(4i Log(i)) = exp(4i(π/2)) = e 2π. (b) Show that Let fz be the principal branch of z 4i. a Find fi. Solution. fi = exp4i Logi = exp4iπ/2 = e 2π. b Show that fz fz 2 fz z 2 fz fz 2 = λfz z 2 for all z, z 2 0, where λ =, e 8π or e 8π. Proof. We have =

More information

Math Spring 2014 Solutions to Assignment # 8 Completion Date: Friday May 30, 2014

Math Spring 2014 Solutions to Assignment # 8 Completion Date: Friday May 30, 2014 Math 3 - Spring 4 Solutions to Assignment # 8 ompletion Date: Friday May 3, 4 Question. [p 49, #] By finding an antiderivative, evaluate each of these integrals, where the path is any contour between the

More information

BTL What is the value of m if the vector is solenoidal. BTL What is the value of a, b, c if the vector may be irrotational.

BTL What is the value of m if the vector is solenoidal. BTL What is the value of a, b, c if the vector may be irrotational. VALLIAMMAI ENGINEERING OLLEGE SRM NAGAR, KATTANDKULATHUR Department of Mathematics MA65 - MATHEMATIS II QUESTION BANK - 6 UNIT - I VETOR ALULUS Part - A. Find, if at (, -, ). BTL-. Find the Directional

More information

Solution for Final Review Problems 1

Solution for Final Review Problems 1 Solution for Final Review Problems Final time and location: Dec. Gymnasium, Rows 23, 25 5, 2, Wednesday, 9-2am, Main ) Let fz) be the principal branch of z i. a) Find f + i). b) Show that fz )fz 2 ) λfz

More information

Exercises for Part 1

Exercises for Part 1 MATH200 Complex Analysis. Exercises for Part Exercises for Part The following exercises are provided for you to revise complex numbers. Exercise. Write the following expressions in the form x+iy, x,y R:

More information

Exercises for Part 1

Exercises for Part 1 MATH200 Complex Analysis. Exercises for Part Exercises for Part The following exercises are provided for you to revise complex numbers. Exercise. Write the following expressions in the form x + iy, x,y

More information

(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 =

(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 = Homework -5 Solutions Problems (a) z = + 0i, (b) z = 7 + 24i 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,,

More information

Analyse 3 NA, FINAL EXAM. * Monday, January 8, 2018, *

Analyse 3 NA, FINAL EXAM. * Monday, January 8, 2018, * Analyse 3 NA, FINAL EXAM * Monday, January 8, 08, 4.00 7.00 * Motivate each answer with a computation or explanation. The maximum amount of points for this exam is 00. No calculators!. (Holomorphic functions)

More information

Math 185 Fall 2015, Sample Final Exam Solutions

Math 185 Fall 2015, Sample Final Exam Solutions Math 185 Fall 2015, Sample Final Exam Solutions Nikhil Srivastava December 12, 2015 1. True or false: (a) If f is analytic in the annulus A = {z : 1 < z < 2} then there exist functions g and h such that

More information

Complex Variables...Review Problems (Residue Calculus Comments)...Fall Initial Draft

Complex Variables...Review Problems (Residue Calculus Comments)...Fall Initial Draft Complex Variables........Review Problems Residue Calculus Comments)........Fall 22 Initial Draft ) Show that the singular point of fz) is a pole; determine its order m and its residue B: a) e 2z )/z 4,

More information

Math 411, Complex Analysis Definitions, Formulas and Theorems Winter y = sinα

Math 411, Complex Analysis Definitions, Formulas and Theorems Winter y = sinα Math 411, Complex Analysis Definitions, Formulas and Theorems Winter 014 Trigonometric Functions of Special Angles α, degrees α, radians sin α cos α tan α 0 0 0 1 0 30 π 6 45 π 4 1 3 1 3 1 y = sinα π 90,

More information

The Calculus of Residues

The Calculus of Residues hapter 7 The alculus of Residues If fz) has a pole of order m at z = z, it can be written as Eq. 6.7), or fz) = φz) = a z z ) + a z z ) +... + a m z z ) m, 7.) where φz) is analytic in the neighborhood

More information

MTH 3102 Complex Variables Final Exam May 1, :30pm-5:30pm, Skurla Hall, Room 106

MTH 3102 Complex Variables Final Exam May 1, :30pm-5:30pm, Skurla Hall, Room 106 Name (Last name, First name): MTH 02 omplex Variables Final Exam May, 207 :0pm-5:0pm, Skurla Hall, Room 06 Exam Instructions: You have hour & 50 minutes to complete the exam. There are a total of problems.

More information

Exercises involving elementary functions

Exercises involving elementary functions 017:11:0:16:4:09 c M K Warby MA3614 Complex variable methods and applications 1 Exercises involving elementary functions 1 This question was in the class test in 016/7 and was worth 5 marks a) Let z +

More information

Chapter 9. Analytic Continuation. 9.1 Analytic Continuation. For every complex problem, there is a solution that is simple, neat, and wrong.

Chapter 9. Analytic Continuation. 9.1 Analytic Continuation. For every complex problem, there is a solution that is simple, neat, and wrong. Chapter 9 Analytic Continuation For every complex problem, there is a solution that is simple, neat, and wrong. - H. L. Mencken 9.1 Analytic Continuation Suppose there is a function, f 1 (z) that is analytic

More information

Physics 2400 Midterm I Sample March 2017

Physics 2400 Midterm I Sample March 2017 Physics 4 Midterm I Sample March 17 Question: 1 3 4 5 Total Points: 1 1 1 1 6 Gamma function. Leibniz s rule. 1. (1 points) Find positive x that minimizes the value of the following integral I(x) = x+1

More information

Conformal maps. Lent 2019 COMPLEX METHODS G. Taylor. A star means optional and not necessarily harder.

Conformal maps. Lent 2019 COMPLEX METHODS G. Taylor. A star means optional and not necessarily harder. Lent 29 COMPLEX METHODS G. Taylor A star means optional and not necessarily harder. Conformal maps. (i) Let f(z) = az + b, with ad bc. Where in C is f conformal? cz + d (ii) Let f(z) = z +. What are the

More information

8. Solutions to Part 1

8. Solutions to Part 1 MATH Complex Analysis 8. Solutions to Part 8. Solutions to Part Solution. (i) (3+4i) 9+4i 6 7+4i. (ii) (iii) (iv) (v) +3i 3 4i +3i 3+4i 3 4i3+4i (+3i)(3+4i) 5 5i +3i 8 5 +i 5. i i+ i i+ i. +i i i. 6 5

More information

MATH243 First Semester 2013/14. Exercises 1

MATH243 First Semester 2013/14. Exercises 1 Complex Functions Dr Anna Pratoussevitch MATH43 First Semester 013/14 Exercises 1 Submit your solutions to questions marked with [HW] in the lecture on Monday 30/09/013 Questions or parts of questions

More information

Poles, Residues, and All That

Poles, Residues, and All That 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

More information

North MaharashtraUniversity ; Jalgaon.

North MaharashtraUniversity ; Jalgaon. North MaharashtraUniversity ; Jalgaon. Question Bank S.Y.B.Sc. Mathematics (Sem II) MTH. Functions of a omplex Variable. Authors ; Prof. M.D.Suryawanshi (o-ordinator) Head, Department of Mathematics, S.S.V.P.S.

More information

6. Residue calculus. where C is any simple closed contour around z 0 and inside N ε.

6. Residue calculus. where C is any simple closed contour around z 0 and inside N ε. 6. Residue calculus Let z 0 be an isolated singularity of f(z), then there exists a certain deleted neighborhood N ε = {z : 0 < z z 0 < ε} such that f is analytic everywhere inside N ε. We define Res(f,

More information

NPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India

NPTEL web course on Complex Analysis. A. Swaminathan I.I.T. Roorkee, India. and. V.K. Katiyar I.I.T. Roorkee, India NPTEL web course on omplex Analysis A. Swaminathan I.I.T. Roorkee, India and V.K. Katiyar I.I.T. Roorkee, India A.Swaminathan and V.K.Katiyar (NPTEL) omplex Analysis 1 / 18 omplex Analysis Module: 6: Residue

More information

MATH 106 HOMEWORK 4 SOLUTIONS. sin(2z) = 2 sin z cos z. (e zi + e zi ) 2. = 2 (ezi e zi )

MATH 106 HOMEWORK 4 SOLUTIONS. sin(2z) = 2 sin z cos z. (e zi + e zi ) 2. = 2 (ezi e zi ) MATH 16 HOMEWORK 4 SOLUTIONS 1 Show directly from the definition that sin(z) = ezi e zi i sin(z) = sin z cos z = (ezi e zi ) i (e zi + e zi ) = sin z cos z Write the following complex numbers in standard

More information

Math 423/823 Exam 1 Topics Covered

Math 423/823 Exam 1 Topics Covered Math 423/823 Exam 1 Topics Covered Complex numbers: C: z = x+yi, where i 2 = 1; addition and multiplication behaves like reals. Formally, x + yi (x,y), with (x,y) + (a,b) = (x + a,y + b) and (x,y)(a,b)

More information

EE2007 Tutorial 7 Complex Numbers, Complex Functions, Limits and Continuity

EE2007 Tutorial 7 Complex Numbers, Complex Functions, Limits and Continuity EE27 Tutorial 7 omplex Numbers, omplex Functions, Limits and ontinuity Exercise 1. These are elementary exercises designed as a self-test for you to determine if you if have the necessary pre-requisite

More information

MATH 452. SAMPLE 3 SOLUTIONS May 3, (10 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic.

MATH 452. SAMPLE 3 SOLUTIONS May 3, (10 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic. MATH 45 SAMPLE 3 SOLUTIONS May 3, 06. (0 pts) Let f(x + iy) = u(x, y) + iv(x, y) be an analytic function. Show that u(x, y) is harmonic. Because f is holomorphic, u and v satisfy the Cauchy-Riemann equations:

More information

1 z n = 1. 9.(Problem) Evaluate each of the following, that is, express each in standard Cartesian form x + iy. (2 i) 3. ( 1 + i. 2 i.

1 z n = 1. 9.(Problem) Evaluate each of the following, that is, express each in standard Cartesian form x + iy. (2 i) 3. ( 1 + i. 2 i. . 5(b). (Problem) Show that z n = z n and z n = z n for n =,,... (b) Use polar form, i.e. let z = re iθ, then z n = r n = z n. Note e iθ = cos θ + i sin θ =. 9.(Problem) Evaluate each of the following,

More information

Complex Homework Summer 2014

Complex Homework Summer 2014 omplex Homework Summer 24 Based on Brown hurchill 7th Edition June 2, 24 ontents hw, omplex Arithmetic, onjugates, Polar Form 2 2 hw2 nth roots, Domains, Functions 2 3 hw3 Images, Transformations 3 4 hw4

More information

Exercises involving elementary functions

Exercises involving elementary functions 017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 1 Exercises involving elementary functions 1. This question was in the class test in 016/7 and was worth 5 marks. a) Let

More information

Math Final Exam.

Math Final Exam. Math 106 - Final Exam. This is a closed book exam. No calculators are allowed. The exam consists of 8 questions worth 100 points. Good luck! Name: Acknowledgment and acceptance of honor code: Signature:

More information

MATH FINAL SOLUTION

MATH FINAL SOLUTION MATH 185-4 FINAL SOLUTION 1. (8 points) Determine whether the following statements are true of false, no justification is required. (1) (1 point) Let D be a domain and let u,v : D R be two harmonic functions,

More information

MATH 417 Homework 4 Instructor: D. Cabrera Due July 7. z c = e c log z (1 i) i = e i log(1 i) i log(1 i) = 4 + 2kπ + i ln ) cosz = eiz + e iz

MATH 417 Homework 4 Instructor: D. Cabrera Due July 7. z c = e c log z (1 i) i = e i log(1 i) i log(1 i) = 4 + 2kπ + i ln ) cosz = eiz + e iz MATH 47 Homework 4 Instructor: D. abrera Due July 7. Find all values of each expression below. a) i) i b) cos i) c) sin ) Solution: a) Here we use the formula z c = e c log z i) i = e i log i) The modulus

More information

Part IB. Complex Analysis. Year

Part IB. Complex Analysis. Year Part IB Complex Analysis Year 2018 2017 2016 2015 2014 2013 2012 2011 2010 2009 2008 2007 2006 2005 2018 Paper 1, Section I 2A Complex Analysis or Complex Methods 7 (a) Show that w = log(z) is a conformal

More information

Section 7.2. The Calculus of Complex Functions

Section 7.2. The Calculus of Complex Functions Section 7.2 The Calculus of Complex Functions In this section we will iscuss limits, continuity, ifferentiation, Taylor series in the context of functions which take on complex values. Moreover, we will

More information

2.5 (x + iy)(a + ib) = xa yb + i(xb + ya) = (az by) + i(bx + ay) = (a + ib)(x + iy). The middle = uses commutativity of real numbers.

2.5 (x + iy)(a + ib) = xa yb + i(xb + ya) = (az by) + i(bx + ay) = (a + ib)(x + iy). The middle = uses commutativity of real numbers. Complex Analysis Sketches of Solutions to Selected Exercises Homework 2..a ( 2 i) i( 2i) = 2 i i + i 2 2 = 2 i i 2 = 2i 2..b (2, 3)( 2, ) = (2( 2) ( 3), 2() + ( 3)( 2)) = (, 8) 2.2.a Re(iz) = Re(i(x +

More information

David A. Stephens Department of Mathematics and Statistics McGill University. October 28, 2006

David A. Stephens Department of Mathematics and Statistics McGill University. October 28, 2006 556: MATHEMATICAL STATISTICS I COMPUTING THE HYPEBOLIC SECANT DISTIBUTION CHAACTEISTIC FUNCTION David A. Stephens Department of Mathematics and Statistics McGill University October 8, 6 Abstract We give

More information

x y x 2 2 x y x x y x U x y x y

x y x 2 2 x y x x y x U x y x y 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

More information

1 Discussion on multi-valued functions

1 Discussion on multi-valued functions Week 3 notes, Math 7651 1 Discussion on multi-valued functions Log function : Note that if z is written in its polar representation: z = r e iθ, where r = z and θ = arg z, then log z log r + i θ + 2inπ

More information

MATH 311: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE

MATH 311: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE MATH 3: COMPLEX ANALYSIS CONTOUR INTEGRALS LECTURE Recall the Residue Theorem: Let be a simple closed loop, traversed counterclockwise. Let f be a function that is analytic on and meromorphic inside. Then

More information

Synopsis of Complex Analysis. Ryan D. Reece

Synopsis of Complex Analysis. Ryan D. Reece Synopsis of Complex Analysis Ryan D. Reece December 7, 2006 Chapter Complex Numbers. The Parts of a Complex Number A complex number, z, is an ordered pair of real numbers similar to the points in the real

More information

Man will occasionally stumble over the truth, but most of the time he will pick himself up and continue on.

Man will occasionally stumble over the truth, but most of the time he will pick himself up and continue on. hapter 3 The Residue Theorem Man will occasionally stumble over the truth, but most of the time he will pick himself up and continue on. - Winston hurchill 3. The Residue Theorem We will find that many

More information

1. DO NOT LIFT THIS COVER PAGE UNTIL INSTRUCTED TO DO SO. Write your student number and name at the top of this page. This test has SIX pages.

1. DO NOT LIFT THIS COVER PAGE UNTIL INSTRUCTED TO DO SO. Write your student number and name at the top of this page. This test has SIX pages. Student Number Name (Printed in INK Mathematics 54 July th, 007 SIMON FRASER UNIVERSITY Department of Mathematics Faculty of Science Midterm Instructor: S. Pimentel 1. DO NOT LIFT THIS COVER PAGE UNTIL

More information

MTH 3102 Complex Variables Solutions: Practice Exam 2 Mar. 26, 2017

MTH 3102 Complex Variables Solutions: Practice Exam 2 Mar. 26, 2017 Name Last name, First name): MTH 31 omplex Variables Solutions: Practice Exam Mar. 6, 17 Exam Instructions: You have 1 hour & 1 minutes to complete the exam. There are a total of 7 problems. You must show

More information

Solutions to Exercises 1.1

Solutions to Exercises 1.1 Section 1.1 Complex Numbers 1 Solutions to Exercises 1.1 1. We have So a 0 and b 1. 5. We have So a 3 and b 4. 9. We have i 0+ 1i. i +i because i i +i 1 {}}{ 4+4i + i 3+4i. 1 + i 3 7 i 1 3 3 + i 14 1 1

More information

Complex Analysis I Miniquiz Collection July 17, 2017

Complex Analysis I Miniquiz Collection July 17, 2017 Complex Analysis I Miniquiz Collection July 7, 207. Which of the two numbers is greater? (a) 7 or 0.7 (b) 3 8 or 0.3 2. What is the area A of a circular disk with radius? A = 3. Fill out the following

More information

secθ 1 cosθ The pythagorean identities can also be expressed as radicals

secθ 1 cosθ The pythagorean identities can also be expressed as radicals Basic Identities Section Objectives: Students will know how to use fundamental trigonometric identities to evaluate trigonometric functions and simplify trigonometric expressions. We use trig. identities

More information

Qualifying Exam Complex Analysis (Math 530) January 2019

Qualifying Exam Complex Analysis (Math 530) January 2019 Qualifying Exam Complex Analysis (Math 53) January 219 1. Let D be a domain. A function f : D C is antiholomorphic if for every z D the limit f(z + h) f(z) lim h h exists. Write f(z) = f(x + iy) = u(x,

More information

Ma 416: Complex Variables Solutions to Homework Assignment 6

Ma 416: Complex Variables Solutions to Homework Assignment 6 Ma 46: omplex Variables Solutions to Homework Assignment 6 Prof. Wickerhauser Due Thursday, October th, 2 Read R. P. Boas, nvitation to omplex Analysis, hapter 2, sections 9A.. Evaluate the definite integral

More information

Math Homework 2

Math Homework 2 Math 73 Homework Due: September 8, 6 Suppose that f is holomorphic in a region Ω, ie an open connected set Prove that in any of the following cases (a) R(f) is constant; (b) I(f) is constant; (c) f is

More information

Second Midterm Exam Name: Practice Problems March 10, 2015

Second Midterm Exam Name: Practice Problems March 10, 2015 Math 160 1. Treibergs Second Midterm Exam Name: Practice Problems March 10, 015 1. Determine the singular points of the function and state why the function is analytic everywhere else: z 1 fz) = z + 1)z

More information

Math 421 Midterm 2 review questions

Math 421 Midterm 2 review questions Math 42 Midterm 2 review questions Paul Hacking November 7, 205 () Let U be an open set and f : U a continuous function. Let be a smooth curve contained in U, with endpoints α and β, oriented from α to

More information

Math 417 Midterm Exam Solutions Friday, July 9, 2010

Math 417 Midterm Exam Solutions Friday, July 9, 2010 Math 417 Midterm Exam Solutions Friday, July 9, 010 Solve any 4 of Problems 1 6 and 1 of Problems 7 8. Write your solutions in the booklet provided. If you attempt more than 5 problems, you must clearly

More information

A REVIEW OF RESIDUES AND INTEGRATION A PROCEDURAL APPROACH

A REVIEW OF RESIDUES AND INTEGRATION A PROCEDURAL APPROACH A REVIEW OF RESIDUES AND INTEGRATION A PROEDURAL APPROAH ANDREW ARHIBALD 1. Introduction When working with complex functions, it is best to understand exactly how they work. Of course, complex functions

More information

Chapter II. Complex Variables

Chapter II. Complex Variables hapter II. omplex Variables Dates: October 2, 4, 7, 2002. These three lectures will cover the following sections of the text book by Keener. 6.1. omplex valued functions and branch cuts; 6.2.1. Differentiation

More information

Complex Numbers and Exponentials

Complex Numbers and Exponentials Complex Numbers and Exponentials Definition and Basic Operations A complexnumber is nothing morethan a point in the xy plane. The first component, x, ofthe complex number (x,y) is called its real part

More information

6.1: Verifying Trigonometric Identities Date: Pre-Calculus

6.1: Verifying Trigonometric Identities Date: Pre-Calculus 6.1: Verifying Trigonometric Identities Date: Pre-Calculus Using Fundamental Identities to Verify Other Identities: To verify an identity, we show that side of the identity can be simplified so that it

More information

Summary for Vector Calculus and Complex Calculus (Math 321) By Lei Li

Summary for Vector Calculus and Complex Calculus (Math 321) By Lei Li Summary for Vector alculus and omplex alculus (Math 321) By Lei Li 1 Vector alculus 1.1 Parametrization urves, surfaces, or volumes can be parametrized. Below, I ll talk about 3D case. Suppose we use e

More information

Here are brief notes about topics covered in class on complex numbers, focusing on what is not covered in the textbook.

Here are brief notes about topics covered in class on complex numbers, focusing on what is not covered in the textbook. Phys374, Spring 2008, Prof. Ted Jacobson Department of Physics, University of Maryland Complex numbers version 5/21/08 Here are brief notes about topics covered in class on complex numbers, focusing on

More information

MA 412 Complex Analysis Final Exam

MA 412 Complex Analysis Final Exam MA 4 Complex Analysis Final Exam Summer II Session, August 9, 00.. Find all the values of ( 8i) /3. Sketch the solutions. Answer: We start by writing 8i in polar form and then we ll compute the cubic root:

More information

lim when the limit on the right exists, the improper integral is said to converge to that limit.

lim when the limit on the right exists, the improper integral is said to converge to that limit. hapter 7 Applications of esidues - evaluation of definite and improper integrals occurring in real analysis and applied math - finding inverse Laplace transform by the methods of summing residues. 6. Evaluation

More information

Introduction to Complex Analysis

Introduction to Complex Analysis Introduction to Complex Analysis George Voutsadakis 1 1 Mathematics and Computer Science Lake Superior State University LSSU Math 413 George Voutsadakis (LSSU) Complex Analysis October 2014 1 / 67 Outline

More information

1 + z 1 x (2x y)e x2 xy. xe x2 xy. x x3 e x, lim x log(x). (3 + i) 2 17i + 1. = 1 2e + e 2 = cosh(1) 1 + i, 2 + 3i, 13 exp i arctan

1 + z 1 x (2x y)e x2 xy. xe x2 xy. x x3 e x, lim x log(x). (3 + i) 2 17i + 1. = 1 2e + e 2 = cosh(1) 1 + i, 2 + 3i, 13 exp i arctan Complex Analysis I MT333P Problems/Homework Recommended Reading: Bak Newman: Complex Analysis Springer Conway: Functions of One Complex Variable Springer Ahlfors: Complex Analysis McGraw-Hill Jaenich:

More information

Complex Derivative and Integral

Complex Derivative and Integral hapter 19 omplex Derivative and Integral So far we have concerned ourselves with the algebra of the complex numbers. The subject of complex analysis is extremely rich and important. The scope and the level

More information

INTRODUCTION TO COMPLEX ANALYSIS W W L CHEN

INTRODUCTION TO COMPLEX ANALYSIS W W L CHEN INTRODUTION TO OMPLEX ANALYSIS W W L HEN c W W L hen, 986, 2008. This chapter originates from material used by the author at Imperial ollege, University of London, between 98 and 990. It is available free

More information

EE2 Mathematics : Complex Variables

EE2 Mathematics : Complex Variables EE Mathematics : omplex Variables J. D. Gibbon (Professor J. D Gibbon 1, Dept of Mathematics) j.d.gibbon@ic.ac.uk http://www.imperial.ac.uk/ jdg These notes are not identical word-for-word with my lectures

More information

Complex Variables. Instructions Solve any eight of the following ten problems. Explain your reasoning in complete sentences to maximize credit.

Complex Variables. Instructions Solve any eight of the following ten problems. Explain your reasoning in complete sentences to maximize credit. Instructions Solve any eight of the following ten problems. Explain your reasoning in complete sentences to maximize credit. 1. The TI-89 calculator says, reasonably enough, that x 1) 1/3 1 ) 3 = 8. lim

More information

EE2007: Engineering Mathematics II Complex Analysis

EE2007: Engineering Mathematics II Complex Analysis EE2007: Engineering Mathematics II omplex Analysis Ling KV School of EEE, NTU ekvling@ntu.edu.sg V4.2: Ling KV, August 6, 2006 V4.1: Ling KV, Jul 2005 EE2007 V4.0: Ling KV, Jan 2005, EE2007 V3.1: Ling

More information

MTH4101 CALCULUS II REVISION NOTES. 1. COMPLEX NUMBERS (Thomas Appendix 7 + lecture notes) ax 2 + bx + c = 0. x = b ± b 2 4ac 2a. i = 1.

MTH4101 CALCULUS II REVISION NOTES. 1. COMPLEX NUMBERS (Thomas Appendix 7 + lecture notes) ax 2 + bx + c = 0. x = b ± b 2 4ac 2a. i = 1. MTH4101 CALCULUS II REVISION NOTES 1. COMPLEX NUMBERS (Thomas Appendix 7 + lecture notes) 1.1 Introduction Types of numbers (natural, integers, rationals, reals) The need to solve quadratic equations:

More information

Definite integrals. We shall study line integrals of f (z). In order to do this we shall need some preliminary definitions.

Definite integrals. We shall study line integrals of f (z). In order to do this we shall need some preliminary definitions. 5. OMPLEX INTEGRATION (A) Definite integrals Integrals are extremely important in the study of functions of a complex variable. The theory is elegant, and the proofs generally simple. The theory is put

More information

Math 715 Homework 1 Solutions

Math 715 Homework 1 Solutions . [arrier, Krook and Pearson Section 2- Exercise ] Show that no purely real function can be analytic, unless it is a constant. onsider a function f(z) = u(x, y) + iv(x, y) where z = x + iy and where u

More information

Solutions to practice problems for the final

Solutions to practice problems for the final Solutions to practice problems for the final Holomorphicity, Cauchy-Riemann equations, and Cauchy-Goursat theorem 1. (a) Show that there is a holomorphic function on Ω = {z z > 2} whose derivative is z

More information

Mathematics 104 Fall Term 2006 Solutions to Final Exam. sin(ln t) dt = e x sin(x) dx.

Mathematics 104 Fall Term 2006 Solutions to Final Exam. sin(ln t) dt = e x sin(x) dx. Mathematics 14 Fall Term 26 Solutions to Final Exam 1. Evaluate sin(ln t) dt. Solution. We first make the substitution t = e x, for which dt = e x. This gives sin(ln t) dt = e x sin(x). To evaluate the

More information

Topic 4 Notes Jeremy Orloff

Topic 4 Notes Jeremy Orloff Topic 4 Notes Jeremy Orloff 4 auchy s integral formula 4. Introduction auchy s theorem is a big theorem which we will use almost daily from here on out. Right away it will reveal a number of interesting

More information

Midterm Examination #2

Midterm Examination #2 Anthony Austin MATH 47 April, 9 AMDG Midterm Examination #. The integrand has poles of order whenever 4, which occurs when is equal to i, i,, or. Since a R, a >, the only one of these poles that lies inside

More information

Theorem [Mean Value Theorem for Harmonic Functions] Let u be harmonic on D(z 0, R). Then for any r (0, R), u(z 0 ) = 1 z z 0 r

Theorem [Mean Value Theorem for Harmonic Functions] Let u be harmonic on D(z 0, R). Then for any r (0, R), u(z 0 ) = 1 z z 0 r 2. A harmonic conjugate always exists locally: if u is a harmonic function in an open set U, then for any disk D(z 0, r) U, there is f, which is analytic in D(z 0, r) and satisfies that Re f u. Since such

More information

McGill University April Calculus 3. Tuesday April 29, 2014 Solutions

McGill University April Calculus 3. Tuesday April 29, 2014 Solutions McGill University April 4 Faculty of Science Final Examination Calculus 3 Math Tuesday April 9, 4 Solutions Problem (6 points) Let r(t) = (t, cos t, sin t). i. Find the velocity r (t) and the acceleration

More information

Introduction to Complex Analysis

Introduction to Complex Analysis Introduction to Complex Analysis George Voutsadakis Mathematics and Computer Science Lake Superior State University LSSU Math 43 George Voutsadakis (LSSU) Complex Analysis October 204 / 58 Outline Consequences

More information

Types of Real Integrals

Types of Real Integrals Math B: Complex Variables Types of Real Integrals p(x) I. Integrals of the form P.V. dx where p(x) and q(x) are polynomials and q(x) q(x) has no eros (for < x < ) and evaluate its integral along the fol-

More information

MATH MIDTERM 1 SOLUTION. 1. (5 points) Determine whether the following statements are true of false, no justification is required.

MATH MIDTERM 1 SOLUTION. 1. (5 points) Determine whether the following statements are true of false, no justification is required. MATH 185-4 MIDTERM 1 SOLUTION 1. (5 points Determine whether the following statements are true of false, no justification is required. (1 (1pointTheprincipalbranchoflogarithmfunctionf(z = Logz iscontinuous

More information

1. The COMPLEX PLANE AND ELEMENTARY FUNCTIONS: Complex numbers; stereographic projection; simple and multiple connectivity, elementary functions.

1. The COMPLEX PLANE AND ELEMENTARY FUNCTIONS: Complex numbers; stereographic projection; simple and multiple connectivity, elementary functions. Complex Analysis Qualifying Examination 1 The COMPLEX PLANE AND ELEMENTARY FUNCTIONS: Complex numbers; stereographic projection; simple and multiple connectivity, elementary functions 2 ANALYTIC FUNCTIONS:

More information

III.2. Analytic Functions

III.2. Analytic Functions III.2. Analytic Functions 1 III.2. Analytic Functions Recall. When you hear analytic function, think power series representation! Definition. If G is an open set in C and f : G C, then f is differentiable

More information

Power Series. x n. Using the ratio test. n n + 1. x n+1 n 3. = lim x. lim n + 1. = 1 < x < 1. Then r = 1 and I = ( 1, 1) ( 1) n 1 x n.

Power Series. x n. Using the ratio test. n n + 1. x n+1 n 3. = lim x. lim n + 1. = 1 < x < 1. Then r = 1 and I = ( 1, 1) ( 1) n 1 x n. .8 Power Series. n x n x n n Using the ratio test. lim x n+ n n + lim x n n + so r and I (, ). By the ratio test. n Then r and I (, ). n x < ( ) n x n < x < n lim x n+ n (n + ) x n lim xn n (n + ) x

More information

Math Spring 2014 Solutions to Assignment # 12 Completion Date: Thursday June 12, 2014

Math Spring 2014 Solutions to Assignment # 12 Completion Date: Thursday June 12, 2014 Math 3 - Spring 4 Solutions to Assignment # Completion Date: Thursday June, 4 Question. [p 67, #] Use residues to evaluate the improper integral x + ). Ans: π/4. Solution: Let fz) = below. + z ), and for

More information

FINAL EXAM { SOLUTION

FINAL EXAM { SOLUTION United Arab Emirates University ollege of Sciences Department of Mathematical Sciences FINAL EXAM { SOLUTION omplex Analysis I MATH 5 SETION 0 RN 56 9:0 { 0:45 on Monday & Wednesday Date: Wednesday, January

More information

Assignment 6 Solution. Please do not copy and paste my answer. You will get similar questions but with different numbers!

Assignment 6 Solution. Please do not copy and paste my answer. You will get similar questions but with different numbers! Assignment 6 Solution Please do not copy and paste my answer. You will get similar questions but with different numbers! This question tests you the following points: Integration by Parts: Let u = x, dv

More information

Assignment 2 - Complex Analysis

Assignment 2 - Complex Analysis Assignment 2 - Complex Analysis MATH 440/508 M.P. Lamoureux Sketch of solutions. Γ z dz = Γ (x iy)(dx + idy) = (xdx + ydy) + i Γ Γ ( ydx + xdy) = (/2)(x 2 + y 2 ) endpoints + i [( / y) y ( / x)x]dxdy interiorγ

More information

PROBLEM SET 3 FYS3140

PROBLEM SET 3 FYS3140 PROBLEM SET FYS40 Problem. (Cauchy s theorem and integral formula) Cauchy s integral formula f(a) = πi z a dz πi f(a) a in z a dz = 0 otherwise we solve the following problems by comparing the integrals

More information

Evaluation of integrals

Evaluation of integrals Evaluation of certain contour integrals: Type I Type I: Integrals of the form 2π F (cos θ, sin θ) dθ If we take z = e iθ, then cos θ = 1 (z + 1 ), sin θ = 1 (z 1 dz ) and dθ = 2 z 2i z iz. Substituting

More information

R- and C-Differentiability

R- and C-Differentiability Lecture 2 R- and C-Differentiability Let z = x + iy = (x,y ) be a point in C and f a function defined on a neighbourhood of z (e.g., on an open disk (z,r) for some r > ) with values in C. Write f (z) =

More information

MAT389 Fall 2016, Problem Set 11

MAT389 Fall 2016, Problem Set 11 MAT389 Fall 216, Problem Set 11 Improper integrals 11.1 In each of the following cases, establish the convergence of the given integral and calculate its value. i) x 2 x 2 + 1) 2 ii) x x 2 + 1)x 2 + 2x

More information