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 + 7 f 1 z) = z 7z + 10. At what points does this function have poles? [1 mark] Express the function f 1 z) in partial fraction form and state the residue at any pole. [9 marks] b) Determine the residue at z = 1 of the following function. c) Let f z) = z9 + 1 z 7 1. f 3 z) = z + 1) z + ) 3. [7 marks] Determine the partial fraction representation and give the residue at z =. [8 marks] a) f 1 z) has simple poles at and at 5. z 7z + 10 = z )z 5). f 1 z) = z + 7 z )z 5). The partial fraction representation has the form f 1 z) = A z + B z 5. z + 7 A = limz )f 1 z) = lim z z z 5 = 9 3 = 3. z + 7 B = limz 5)f 1 z) = lim z 5 z 5 z = 1 3 = 4. A = 3 is the residue at z = and B = 4 is the residue at z = 5. 1 mark 9 marks

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications b) The residue is lim z 1)f z) = limz 9 z 1 + 1) lim z 1 z 1 z 1 z 7 1 = lim 1 z 1 7z 6 = 7. 7 marks c) As there is one pole at we can get the partial fraction representation using the Taylor polynomial of the numerator about. z +1) = z +4) 3) = 9 6z +4)+z +4) = 9 1z +)+4z +). The residue at is 4. f 3 z) = 4 z + 1 z + ) + 9 z + ) 3. 8 marks. This question was in the class test in 015/6 and was worth 4 marks. a) Express the function f 1 z) defined below in partial fraction form and state the residue at any pole. 3z f 1 z) = z 1)z + ). [9 marks] b) Determine the residue at of the following rational function. f z) = z8 z 4. [7 marks] c) Determine the residue at 1 of the following rational function. f 3 z) = zz + 3) z + 1) 3. [8 marks] a) f 1 z) = 3z z 1)z + ) = A z 1 + B z +. A is the residue at z = 1 and B is the residue at z =. A = lim z 1 z 1)f 1 z) = 3z z + = 1 z=1

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 3 and B = lim z z + )f 1z) = 3z z 1 = 6 z= 3 =. 9 marks b) z 4 = z )z + ). f z) = z8 z 4 = polynomial of degree 6) + A z + B z +. A is the residue at z = and B is the residue at z =. A = limz )f z) = z8 z z + = 56 z= 4 = 64. c) Let The derivatives are Hence gz) = zz + 3) = z + 3z. g z) = z + 3, g z) =. 7 marks gz) = g 1) + g 1)z + 1) + g 1) z + 1) = + z + 1) + z + 1). Thus f 3 z) = and the residue at 1 is 1. zz + 3) z + 1) 3 = z + 1) 3 + 1 z + 1) + 1 z + 1) 8 marks 3. This question was in the class test in 013/4 and was worth 18 marks. a) Determine the partial fraction representation of the function fz) = and give all the residues of the function. z + 3 z + 1)z + ) b) Determine the residue at z = of the following function. gz) = z 6 z + 1)z + ). [10 marks]

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 4 [8 marks] a) fz) = z + 3 z + 1)z + ) = A z + 1 + B z + Az + ) + Bz + 1) =. z + 1)z + ) A = lim z 1 z + 1)fz) = 1 1 = 1, 1 B = lim z + )fz) = z 1 = 1. A is the residue at z = 1 and B is the residue at z =. 10 marks b) fz) = The residue at is z 6 z + 1)z + ) B = lim z + )fz) = z6 z A = degree 4 polynomial) + z + 1 + B z +. z + 1 = )6 z= 1 = 64. 4. The following was in the class test in 014/015 and was worth 5 marks. 8 marks Express the functions f 1 and f defined below in partial fraction form and state the residue of any pole. a) b) f 1 z) = z 1. f z) = Determine the residue at z = of the following function [10 marks] z. [8 marks] z + 1) f 3 z) = z 10 z z. [7 marks]

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 5 a) z 1 = z + 1)z 1). The partial fraction representation has the form f 1 z) = z + 1)z 1) = A z + 1 + B z 1. A = lim z 1 z + 1)f 1z) = 1 1 = 1. B = lim z 1 z 1)f 1 z) = 1 + 1 = 1. 8 marks A = 1 is the residue at 1 and B = 1 is the residue at 1. b) The Taylor expansion of the numerator about 1 is z = 1 + z + 1) and hence the partial fraction representation is f z) = 1 z + 1) + 1 z + 1. marks The residue at 1 is 1. 7 marks 1 mark z z = z )z + 1). The function has a representation of the form The residue at z = is f 3 z) = polynomial of degree 8) + A = lim z z )f 3 z) = A z + B z + 1. z10 z + 1 = 104 z= 3. 7 marks

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 6 5. Let Rz) = pz) z ζ 1 ) r 1 z ζ ) r z ζ n ) rn denote a rational function in which ζ 1,..., ζ n are distinct points, where each r k 1 is an integer and where pz) is a polynomial which is non-zero at these n points. What can you say about the order of the poles of R z) and R z) and what can you say about the residues of the function R z)? To cater for a numerator of any degree we have a representation of the form n Ak1 Rz) = polynomial) + + + A ) kr k z ζ k z ζ k ) r. k Differentiating gives R z) = deriv of polynomial) k=1 n k=1 Ak1 z ζ k ) + + r ) ) ka krk z ζ k ) r. k+1 The order of the pole of R z) at ζ k is one more than the order of the pole of Rz) at ζ k and in particular this implies that R z) has no simple pole terms and hence all the residues are 0. Similarly R z) has poles which are of a order which is two more that of Rz) and it has no residues. 6. Let qz) = z ζ 1 ) r 1 z ζ ) r z ζ n ) rn where ζ 1,..., ζ n are distinct points. What can you say about the multiplicity of the zeros of q z) at the points ζ 1,..., ζ n? Using a derivation based on partial fractions show that q z) qz) = r 1 z ζ 1 + r z ζ + + r n z ζ n. Observe that the result is consistent with the result in an exericise of the previous exercise sheet which involved a proof by induction.) qz) has a zero at ζ k of multiplicity r k and the derivative has a zero at ζ k of multiplicity r k 1 for k = 1,,..., n. The ratio qz)/q z) thus has a simple zero at ζ k and the ratio q z)/qz) has a simple pole at ζ k. From this it follows that the partial fraction representation is hence of the form and q z) qz) = n k=1 A k z ζ k z ζ k )q z) A k = lim. z ζk qz)

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 7 Now to get A k observe that qz) = z ζ k ) r k gz), with gζ k ) 0, q z) = r k z ζ k ) rk 1 gz) + z ζ k ) r k g z), z ζ k )q z) = r k z ζ k ) r k gz) + z ζ k ) rk+1 g z) and for the ratio z ζ k )q z) qz) = r k + z ζ k ) g z) gz) r k as z ζ k and we have the required result. 7. This question was in the class test in 016/7 and was worth 6 marks. The definition of sinh z is sinh z = ez e z. With x, y R express sinhx + iy) in terms of cosh x, sinh x, cos y and sin y. 8. By using the definitions show that sinhx + iy) = ex e iy e x e iy sin z = 1 i = ex cos y + i sin y) e x cos y i sin y) e x e x ) e x + e x ) = cos y + i sin y = sinh x cos y + i cosh x sin y. e iz e iz), cos z = 1 e iz + e iz) sinz 1 + z ) = sin z 1 cos z + cos z 1 sin z, cosz 1 + z ) = cos z 1 cos z sin z 1 sin z. Further show that ) z1 + z sin z 1 sin z = cos sin z1 z ).

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 8 We consider parts of the terms on the right hand side as follows. sin z 1 cos z = 1 4i eiz 1 e iz 1 )e iz + e iz ) = 1 e iz 1 +z ) e iz 1+z ) + e iz 1 z ) e ) iz z 1 ), 4i cos z 1 sin z = 1 4i eiz 1 + e iz 1 )e iz e iz ) = 1 e iz 1 +z ) e iz 1+z ) e iz 1 z ) + e ) iz z 1 ). 4i When we add the last two terms cancel and the first two term combine to give sin z 1 cos z + cos z 1 sin z = 1 e iz 1 +z ) e ) iz 1+z ) = sin z 1 + z ). i Similarly for the expression involving cosine we have cos z 1 cos z = 1 4 eiz 1 + e iz 1 )e iz + e iz ) = 1 e iz 1 +z ) + e iz 1+z ) + e iz 1 z ) + e ) iz z 1 ), 4 sin z 1 sin z = 1 4 eiz 1 e iz 1 )e iz e iz ) = 1 e iz 1 +z ) + e iz 1+z ) e iz 1 z ) e ) iz z 1 ). 4 Combining gives cos z 1 cos z sin z 1 sin z = 1 e iz 1 +z ) + e ) iz 1+z ) = cos z 1 + z ). The last part can be shown in a similar way to the real case. If z 1 = a + b and z = a b then and the result follows. sina + b) sina b) = cosa) sinb). a = z 1 + z, b = z 1 z 9. The following was in the class test in 014/015 and was worth 14 marks. The complex sine function is defined by sin z = 1 i e iz e iz). Describe in Cartesian form the image of each of the following 3 line segments under the mapping w = fz) = sin z.

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 9 a) Γ 1 = { π/ iy : y 0}. b) Γ = [ π/, π/]. c) Γ 3 = {π/ iy : y 0}. State, as concisely as possible, what the image of Γ 1 Γ Γ 3 is under the mapping f. When z = π/ iy, e iz = e iπ/ e i iy) = ie y, e iz = e iπ/ e i iy) = ie y. Thus sin z = 1 e y e y) = cosh y. As cosh y 1 the image of Γ 1 is the real interval, 1]. 5 marks When z = π/ iy, e iz = e iπ/ e i iy) = ie y, e iz = e iπ/ e i iy) = ie y. Thus sin z = 1 e y + e y) = cosh y. The image of Γ 3 is the real interval [1, ). 4 marks The image of Γ is the real interval [ 1, 1]. marks The image of Γ 1 Γ Γ 3 is the real axis. 3 marks 10. Let z = x + iy with x, y R. What is the image of the rectangle {z = x + iy : 1 x 1, 0 y π} when we have the function w = expz) = e z. From the definition we have e z = e x cos y + i sin y).

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 10 When x is fixed and y varies we get a circular arc with centre 0 and radius e x. When y is fixed and x varies we get a radial line. The sides x = 1 and x = 1 map to the half circles in the upper half plane of radius e 1 and e. The side y = 0 maps to the real axis between x = e 1 to x = e and the side y = π maps to the real axis between x = e and x = e 1. The image of the rectangle is half an annulus as shown below. y e 1/e 0 1/e e x 11. This question was in the class test in 016/7 and was worth 10 marks. The complex cosine function is such that for x, y R. cosx + iy) = cos x cosh y i sin x sinh y Let b > 0 and let z 1 = ib, z = 0, z 3 = π and z 4 = π + ib. Describe as concisely as possible the image of the polygonal path z 1 to z, z to z 3 and z 3 to z 4 under the mapping w = cosz). [8 marks] State, as concisely as possible, the image of the line segment joining π/ and π/+ib under the mapping w = cosz). [ marks] When x = 0, cosiy) = cosh y 1. The image of z 1 to z is the line segment from cosh b to 1. When y = 0, cosz) = cosx) and when x = 0 this is 1 and when x = π this is 1. The image of z to z 3 is the line segment from 1 to 1. When x = π, cosπ +iy) = cosh y 1. The image of z 3 to z 4 is the line segment from 1 to cosh b. The image of the polygonal path is the line segment from cosh b to cosh b on the real axis.

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 11 8 marks If x = π/ then cosπ/ + iy) = i sinh y. The image of the line segment joining π/ and π/ + ib under the mapping w = cosz) is the line segment joining 0 and i sinh b on the imaginary axis. marks 1. This question was in the class test in 013/4 and was worth 10 marks. The complex cosine is given by cosx + iy) = cosx) coshy) i sinx) sinhy) for x, y R. Describe the lines in the complex z-plane when cosx + iy) is real and describe the part of the complex plane when cosx + iy) is both real and cosx + iy) > 1. The complex cosine is real when sinx) sinhy) = 0. sinhy) = 0 only when y = 0 and this gives the real axis. sinx) = 0 when x = kπ where k Z which are vertical lines parallel to the imaginary axis. When sinx) = 0 we have cosx) = 1 and coskπ + iy) = coshy) > 1 when y 0. The part of the complex plane when the cosine is real and greater than 1 in magnitude is {kπ + iy : k Z, y 0}. 13. Show that if y R then tan π 4 + iy ) = 1. As tan z = sin z cos z we first consider the numerator and denominator when z = π/4 + iy. π ) i sin 4 + iy = exp i π )) 4 + iy exp i π 4 + iy )) π ) cos 4 + iy = e iπ/4 e y e iπ/4 e y = e iπ/4 e y 1 e iπ/ e y), = e iπ/4 e y + e iπ/4 e y = e iπ/4 e y 1 + e iπ/ e y).

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 1 By taking the absolute value of the ratio of these terms and noting that e iπ/ = i and e iπ/ = i we get π ) tan 4 + iy 1 ie y = 1 + ie y. 1 ie y is the complex conjugate of 1 + ie y with both values having the same magnitude and hence tan π 4 + iy ) = 1. 14. Give the definition of the principal value of z α and show that d dz zα = α z α 1. The principal value of z α is defined as Differentiating using the chain rule gives z α = expαlog z). d dz zα = expαlog z) d αlog z) dz α ) = expαlog z) z expαlog z) = α explog z) = α expα 1)Log z) = αz α 1 where the last part is a consequence of the properties of the exponential. 15. a) What are the real and imaginary parts of the principal value of 1 + i ) i. b) Determine all the values of 1 1/π. a) The principal value of z α is defined as z α = expαlog z) = z α expiαarg z).

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 13 In this case z = ) 1 + i, z = 1, and Arg z = π 4 giving Log z = i π 4, and with α = i αlog z = π 4. Thus the required value is exp π/4) and hence the real part is exp π/4) and the imaginary part is 0. b) For the multi-valued logarithm all values of log 1 means kπi for k = 0, ±1, ±,.... Thus 1 log 1 means ki for k = 0, ±1, ±,... π and all the values of 1 1/π are e ki, k = 0, ±1, ±,....

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 14 16. This was in the class test in 013/4 and was worth 1 marks. State the definition of the principal valued logarithm function and also state the definition of the principal value of z α for any α C and any z 0. [4 marks] With e denoting the usual mathematical constant exp1) =.7188... determine the principal value of the following. a) b) Logie). Reie) i ). [4 marks] [4 marks]

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 15 a) ie = e, Argie) = π. Logie) = Log e + i π = 1 + iπ. 4 marks b) ie) i = expi Log ie)) and ilogie) = π + i. Hence Reie) i ) = exp π/) cos1). 4 marks 17. This was in the class test in 016/7 and was worth 13 marks. State the definition of the principal valued logarithm function. State the definition of the principal value of z α for any α C and any z 0. a) Determine the principal value of 1) i. [4 marks] [4 marks] b) Let π < α π. For all possible α in this interval give the value of Logie iα ) in cartesian form where Log denotes the principal valued logarithm. [5 marks] Log z = Log z + iarg z, where Arg z is the principal value of the argument. The principal value of z α is z α = expα Log z). 4 marks

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 16 a) In polars 1 = cosπ) + i sinπ) and Log 1) = iπ. ilog 1) = π. Thus the principal value of 1) i is e π. 4 marks b) i = e iπ/, ie iα = e iπ/+α). π/ + α is in the range π, π] when α is in the range π, π/]. ) i π + α, α π, π/], Logie iα ) = i π + π ) + α, α π/, π]. 5 marks 18. This was in the class test in 015/6 and was worth 10 marks. State the definition of the principal valued logarithm function. State the definition of the principal value of z α for any α C and any z 0. [4 marks] If z = α = e iθ, θ π, π] then explain why the magnitude of the principal value of z α is e θ sin θ. [6 marks] Log z = Log z + iarg z, where Arg z is the principal value of the argument. The principal value of z α is z α = expα Log z). 4 marks With z = α = e iθ, θ π, π] Log z = iθ,

017:11:0:16:4:09 c M. K. Warby MA3614 Complex variable methods and applications 17 and α Log z = iθe iθ = iθcos θ + i sin θ) = θ sin θ + iθ cos θ. With the exponential function the magnitude just depends on the real part and thus the magnitude of z α is e θ sin θ. 6 marks