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


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1 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 indicated limits of integration: (a) i/ i e πz dz; (b) π+i ( z cos dz; ) (c) 3 (z ) 3 dz. Ans: (a) ( + i)/π; (b) e + (/e) ; (c). Solution: (a) Since d dz ( ) π eπz = e πz for all z, then F (z) = π eπz is an antiderivative of f(z) = e πz, i/ i e πz dz = i/ π eπz = π i ( e iπ/ e iπ) = ( + i). π (b) Since d ( sin (z/)) = cos (z/) for all z, then the function F (z) = sin (z/) is an antiderivative dz of f(z) = cos (z/), and π+i cos (z/) dz = sin (z/) π+i = [sin (π/ + i) sin()] = sin (π/ + i) = cos(i) = e + e. (c) Since d ( ) (z )4 = (z ) 3 for all z, then the function F (z) = dz 4 4 (z )4 is an antiderivative of f(z) = (z ) 3, and 3 (z ) 3 dz = 4 (z )4 3 = 4 [ (3 ) 4 ( ) 4] =. Question. [p 49, #5] Show that where z i denotes the principal branch z i dz = + e ( i), z i = exp(i Log z) ( z >, < Arg z < π) and where the path of integration is any contour from z = to z = that, except for its end points, lies above the real axis. (ompare with Exercise 7, Sec. 4.)
2 Suggestion: Use an antiderivative of the branch of the same power function. z i = exp(i log z) ( z >, π < arg z < 3π ) Solution: Let f(z) = z i, z >, < Arg z < π, y x then f(z) = z i = exp(i Log z) is not defined at z =, but if we consider the branch f (z) = z i = exp(i log z), where log z = ln z + iθ, π < θ < 3π, then f (z) is defined and analytic at each point of and its values coincide with the values of f(z) except at z =. y x An antiderivative of f (z) is given by F (z) = + i zi+, Therefore, and F (z) = + i exp((i + ) log z), z >, π < θ < 3π. z i dz = z i dz = + i f (z) dz = F (z) dz = F () F ( ), [ ( ) i+ ] = [ e ((+i)[ln +iπ])], + i z i dz = [ e (+i)πi] = ( e iπ e ) = i ( + e ). + i + i Question 3. [p 6, # (a)] Apply the auchygoursat theorem to show that either direction, and when f(z) = z z 3. f(z) dz = when the contour is the circle z =, in
3 Solution: Since f(z) = z is analytic inside and on the contour z =, then z 3 z z 3 dz = if the contour is traversed in either direction. z = Question 4. [p 6, # (c)] Apply the auchygoursat theorem to show that either direction, and when f(z) = Solution: Since f(z) = then z + z +. f(z) dz = when the contour is the circle z =, in z + z + = is analytic inside and on the contour z =, (z + i)(z + + i) z + z + dz = z = if the contour is traversed in either direction. Question 5. [p 6, # (f)] Apply the auchygoursat theorem to show that either direction, and when f(z) = Log(z + ). f(z) dz = when the contour is the circle z =, in Solution: Since the branch cut for f(z) = Log(z + ) extends from the point z = along the negative real axis, then f(z) is analytic inside and on the contour z =, Log(z + ) dz = if the contour is traversed in either direction. z = Question 6. [p 7, # (a)] Let denote the positively oriented boundary of the square whose sides lie along the lines x = ± and y = ±. Evaluate the integral e z z (πi/) dz. Ans: π. Solution: Let f(z) = e z, then f is analytic inside and on, and since z = πi auchy s integral formula e z πi z (πi/) dz = f (πi/) = ei/ = i, e z dz = πi( i) = π. z (πi/) is interior to, then by
4 Question 7. [p 7, # (b)] Let denote the positively oriented boundary of the square whose sides lie along the lines x = ± and y = ±. Evaluate the integral cos z z(z + 8) dz. Ans: πi/4. Solution: Let f(z) = cos z z + 8, then f is analytic inside and on, and since z = is interior to, then by auchy s integral formula πi cos z cos() z(z dz = f() = = + 8) 8 8, cos z πi z(z dz = + 8) 4. Question 8. [p 7, # (e)] Let denote the positively oriented boundary of the square whose sides lie along the lines x = ± and y = ±. Evaluate the integral tan(z/) (z x ) dz ( < x < ). Ans: iπ sec (x /). Solution: Let f(z) ( = tan(z/), then the singularities of f(z) are the zeros of cos(z/) and these occur at the points z = n + ) π, n =, ±, ±,..., all of which are outside the square. Therefore, f(z) is analytic inside and on, and since x is interior to, from auchy s integral formula we have f(x ) = πi tan(z/) (z x ) dz, and f (x ) =! πi tan(z/) (z x ) dz. Now, f (x ) = sec (x /), tan(z/) πi (z x ) dz = sec (x /), tan(z/) (z x ) dz = iπ sec (x /). Question 9. [p 7, #] Find the value of the integral of g(z) around the circle z i = in the positive sense when (a) g(z) = z + 4 ; (b) g(z) = (z + 4). Ans: (a) π/; (b) π/6.
5 Solution: (a) Let g(z) = z and f(z) = + 4 z + i, then f is analytic inside and on, and since z = i is interior to z i =, then from auchy s integral formula we have πi z i = z + 4 dz = πi z i = f(z) z i dz = f(i) = 4i, z i = z + 4 dz = π. (b) Let g(z) = (z + 4), and let f(z) = (z + i), then f is analytic inside and on, and since z = i is interior to z i =, then from auchy s integral formula we have! g(z) dz = f(z) πi z i = πi z i = (z i) dz = f (i) = (i + i) 3 = 64( i), z i = πi (z dz = + 4) 3i = π 6. Question. [p 7, #3] Let be the circle z = 3, described in the positive sense. Show that if g(w) = z z z w then g() = 8πi. What is the value of g(w) when w > 3? dz ( w 3), Solution: Let f(z) = z z, then f is analytic inside and on, and from the auchy integral formula we have z z g() = dz = πif() = πi(8 ) = 8πi. z z =3 If w > 3, then h(z) = z z z w for w > 3. is analytic inside and on, and from the auchygourat theorem, z =3 z z z w dz = Question. [p 7, #7] Let be the unit circle z = e iθ ( θ π). First show that, for any real constant a, e az dz = πi. z Then write this integral in terms of θ to derive the integration formula π e a cos θ cos(a sin θ) dθ = π.
6 Solution: Let f(z) = e az, then f is analytic inside and on, and from the auchy integral formula, we have e az πi z dz = e =, e az z Now, on, we have z = e iθ and dz = ie iθ dθ, and therefore dz = πi. e az π a(cos θ+i sin θ) z dz = e π e iθ i e iθ dθ = i e a cos θ e ai sin θ dθ, πi = π e az z Equating real and imaginary parts, we have π dz = i e a cos θ [cos(a sin θ) + i sin(a sin θ)] dθ, π e a cos θ cos(a sin θ) dθ + i e a cos θ sin(a sin θ) dθ = π. π e a cos θ cos(a sin θ) dθ = π and π e a cos θ sin(a sin θ) dθ =, and since the function h(θ) = e a cos θ cos(a sin θ) is even, then π e a cos θ cos(a sin θ) dθ = π.
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