Math Spring 2014 Solutions to Assignment # 6 Completion Date: Friday May 23, 2014

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1 Math 11 - Spring 014 Solutions to Assignment # 6 Completion Date: Friday May, 014 Question 1. [p 109, #9] With the aid of expressions 15) 16) in Sec. 4 for sin z cos z, namely, sin z = sin x + sinh y cos z = cos x + sinh y, show that a) sinh y sin z cosh y; b) sinh y cos z cosh y. a) Note that sin z = sinx + i y) = sin x cosiy) + cos x siniy) = sin x cosh y + i cos x sinh y = sin x cosh y + cos x sinh y sin x cosh y + cos x cosh y = cosh y, since sinh y cosh y for all y R, sin z cosh y. Also, sin z = sin x cosh y + sinh y cos x sin x sinh y + cos x sinh y = sinh y, sinh y sin z. b) Note that cos z = cosx + i y) = cos x cosiy) sin x siniy) = cos x cosh y i sin x sinh y = cos x cosh y + sin x sinh y cos x cosh y + sin x cosh y = cosh y, cos z cosh y.

2 Also, cos z = cos x cosh y + sin x sinh y cos x sinh y + sin x sinh y = sinh y cos z sinh y. Question. [p 109, #14] Show that a) cosi z) = cos i z) for all z; b) sini z) = sin i z) if only if z = nπi n = 0, ±1, ±,... ). a) If z = x + i y C, then cos i z) = cosy + i x) = cosh x cos y i sin y sinh x cosi z) = cos y + i x) = cosh x cos y) i sin y) sinh x cosi z) = cosh x cos y + i sin y sinh x, cosi z) = cos i z) for all z C. b) Since sin i z) = sin y cosh x + i cos y sinh x sini z) = sin y cosh x i cos y sinh x then sin i z) = sini z) if only if sin y cosh x = 0 cos y sinh x = 0. Now, since cosh x 1, the first of these equations holds if only if sin y = 0, then in the second equation since cos y 0, we must have sinh x = 0, therefore these two equations hold if only if x = 0 y = nπ, for n = 0, ±1, ±,... so sin i z) = sini z) if only if z = nπi, Question. [p 109, #15] Find all roots of the equation sin z = cosh 4 by equating real imaginary parts of sin z cosh 4. Ans: π + nπ ) ± 4i n = 0, ±1, ±,... ).

3 Note that if only if sin z = sinx + i y) = sin x cosh y + i cos x sinh y = cosh 4 sin x cosh y = cosh 4 sinh y cos x = 0. Now, if sinh y = 0, then cosh y = 1, the first equation implies that sin x = cosh 4 > 1 which is a contradiction. Therefore we must have sinh y 0, the second equation implies that cos x = 0, so that n + 1)π x = For these values of x we have sin x = ±1, since cosh 4 > 0, cosh y > 0, then we must have sin x = +1, cosh y = cosh 4, so that y = ±4 x = Therefore, the solutions to the equation sin z = cosh 4 are z = 4n + 1)π 4n + 1)π ± 4i, Question 4. [p 111, #6] Show that sinh x cosh z cosh x by using a) identity 1), Sec. 5, namely cosh z = sinh x + cos y; b) the inequalities sinh y cos z cosh y, obtained in Exercise 9b), Sec.4. a) We have cosh z = sinh x + cos y sinh x 1) cosh z = cosh x cos y + sinh x sin y cosh x cos y + cosh x sin y, cosh z cosh x ) combining 1) ) we get sinh x cosh z cosh x. b) Starting from the inequality sinh y cos z cosh y, we replace z by iz, then since iz = y + i x cosiz) = cosh z, we have sinh Imiz)) cosiz) cosh Imiz)), sinh x cosh z cosh x.

4 Question 5. [p 11, #9] Using the results proved in Exercise 8, locate all zeros singularities of the hyperbolic tangent function. Note that tanh z = sinh z cosh z = 0 if only if sinh z = 0 if only if ez = e z if only if e z = 1, tanh z = 0 if only if e x e iy = 1 if only if e x = 1 y = πn Therefore tanh z = 0 if only if z = nπi, n = 0, ±1, ±,.... Note that the singularities of tanh z are precisely the points z C for which cosh z = 0, cosh z = 0 if only if e z = e z if only if e z = 1, cosh z = 0 if only if e x e iy = 1 = e πi if only if e x = 1 y = π + πn Therefore cosh z = 0 if only if z = πi + nπi = n + 1 ) πi Question 6. [p 11, #16] Find all roots of the equation cosh z =. Compare this exercise with Exercise 16, Sec 4.) Ans : ± ln + ) + n + 1)πi n = 0, ±1, ±,... ). Note that if only if cosh z = cosh x cos y + i sinh x sin y = cosh x cos y = sinh x sin y = 0 Now, if sinh x = 0, then x = 0 this implies that cosh x = 1, then the first equation implies that cos y = which is a contradiction. Therefore, sinh x 0, from the second equation we must have sin y = 0. Thus, y is a multiple of π, since cosh x 1, then we must have cos y = 1, cosh x =. Therefore, cosh z = if only if for n = 0, ±1, ±,..., if only if x = cosh 1 ), y = n + 1)π z = cosh 1 ) + n + 1)πi In order to simplify the expression for cosh 1 ), note that x = cosh 1 ) if only if cosh x = ex + e x =,

5 if only if e x 4e x + 1 = 0, solving this quadratic equation, we get two real roots, e x = ± or x = ln ± ). However, note that ln ) ) + ) ) = ln + we have cosh z = if only if z = ± ln ) 1 = ln + = ln + ), + ) + n + 1)πi Not Assigned: Excersise 16 in Sec. 4 asked for all roots of the equation cos z =. Note that cos z = if only if cos z = cosx + iy) = cos x cosh y i sin x sinh y =, if only if cos x cosh y = sin x sinh y = 0. Now if these equations are satisfied sinh y = 0, then y = 0 so cosh y = 1, the first equation implies that cos x =, which is a contradiction. Thus, we must have sin x = 0, then x = nπ for some integer n, so that cos x = ±1, since cosh y 1 > 0, we must have cos x = +1 cosh y =. Therefore, cos z = if only if z = nπ + i cosh 1 ), for n = 0, ±1, ±,..., as we saw before, since cosh 1 ) = ± ln + ), then cos z = if only if z = nπ ± i ln + ) Question 7. [p 110, #] Solve the equation sin z = for z by a) equating real imaginary parts in that equation; b) Using expression ), Sec. 5, for sin 1 z, namely sin 1 z = i log [ i z + 1 z ) 1/]. a) We have sin z = sinx + i y) = sin x cosh y + i cos x sinh y =

6 if only if sin x cosh y = cos x sinh y = 0. If these equations hold sinh y = 0, then y = 0 so cosh y = 1, from the first equation this implies that sin x =, which is a contradiction. Therefore, sinh y 0, from the second equation n + 1)π we must have cos x = 0, so that x =, Also, since cosh y 1 > 0, then sin x = +1 cosh y =. Therefore, sin z = if only if x = if only if n + 1)π, where n is an even integer, y = ± ln + ), z = 4n + 1) π ± i ln + ) b) Using sin 1 z = i log [iz + 1 z ) 1/] with z =, we get by definition of the logarithm, Since log i + ) = log [ sin 1 ) = i log i + ) 1/] = i log[ i ± i ] = i log[ i ± ) ] [ + )e iπ/] = ln + [ π ] ) + i + πn = 1 +, the roots of the equation sin z = are given by z = sin 1 = i ± ln + π )) ) + i + πn, for n = 0, ±1, ±,..., as before. z = sin 1 = 4n + 1)π ± i ln + )