598 ANSWERS(^3) · t an z -_ 2( cos (^2) x cosh^2 cosxsl (^2) y+slnn x (^2) x sinh (^2) y) + i • 2 (cos• x cosh2cos 2 h 11sinhy y+sin2 x sinh" 11 )' Th e numer-
a.tors simplify to sin 2x and sinh 2y, respectively. Show that the denominator
equals cos 2x + cosh 2y by using the identities cos 2x = cos^2 x - sin^2 x and
cosh^2 y -sinh^2 y = 1.
5a. This follows immediately from sin (z1 + z2) = sin z1 cos z2 +cos z1 sin Z2·5c. This follows immediately from sinhz = sinh xcosy +icoshxsiny, where we
replace z = x + iy with z = x + iy + i1r = x + i (y + 11').5e. This follows immediately from sin z = sinxcoshy + icosxsinh y, where we
replace z = x + iy with i z = - y + ix.7a. -:;\cos ( ~), valid for z f O.
7c. 2 zsecz^2 tanz^2 , valid for z f (k + 4) 11', where k is an integer.
7e. z cosh z + sinh z, valid for all z.9a. Use the same methods as in Exercise lla of Section 5.1.
11. By identity (5-33), sinz = 0, if and only ifsinxcoshy+icosxsinhy = 0.
Equate real and imaginary parts to show this occurs iff x = k11', where k is
an integer.13. Combining (5-36) and (5- 37 ) , and letting z = x + iy, we get lsinzl^2 +
I cos zi
2= sin^2 x + sinh^2 y + cos^2 x + sinh^2 y = 1 + 2 sinh^2 y. This quan-
tity equals 1 iffy = 0 ( when z is a real number) and is greater than 1
otherwise.15a. Consider the real part of Identity (5-33), and appeal to Theore m 3.8.
15c. Consider the imaginary part of sin (iz), and appeal to Theorem 3.8.17. z = 10 + lOi.
Section 5.5. Inverse Trigonometric and Hyperbolic Functions: page
192la. (4 + 2n) n ± i ln2, where n is an integer.le. (4 +2n)n±iln(3+ 2v'2), wherenis an integer.
le. - (4 + n) 11' + i In v'3, where n is an integer.lg. i U + 2n) 11', where n is an integer.
li. ln (v'z + 1) + i (~ + 2n) n and ln ( v'z-1) + i (-~ + 2n) n, where n is an
integer.