The Real And Complex Number Systems

Proof: Since
tan z = sin z sin (x + iy) sin x cosh y + i cos x sinh y
= =
cos z cos (x + iy) cos x cosh y − i sin x sinh y
(sin x cosh y + i cos x sinh y) (cos x cosh y + i sin x sinh y)
=
(cos x cosh y − i sin x sinh y) (cos x cosh y + i sin x sinh y)
(
sin x cos x cosh 2 y − sin x cos x sinh 2 y ) + i ( sin 2 x cosh y sinh y + cos 2 x cosh y sinh y )
=
(cos x cosh y) 2 − (i sin x sinh y) 2
= sin x cos x ( cosh 2 y − sinh 2 y ) + i (cosh y sinh y)
cos 2 x cosh 2 y + sin 2 x sinh 2 since sin 2 x + cos 2 x = 1
y
(sin x cos x) + i (cosh y sinh y)
=
cos 2 x + sinh 2 since cosh 2 y = 1 + sinh 2 y
y
1
2
=
sin 2x + i sinh 2y
2
cos 2 x + sinh 2 since 2 cosh y sinh y = sinh 2y and 2 sin x cos x = sin 2x
y
sin 2x + i sinh 2y
=
2 cos 2 x + 2 sinh 2 y
sin 2x + i sinh 2y
=
2 cos 2 x − 1 + 2 sinh 2 y + 1
sin 2x + i sinh 2y
=
cos 2x + cosh 2y since cos 2x = 2 cos2 x − 1 and 2 sinh 2 y + 1 = cosh 2y.
1.47 Let w be a given complex number. If w ≠ ±1, show that there exists
two values of z = x + iy satisfying the conditions cos z = w and −π < x ≤ π.
Find these values when w = i and when w = 2.
Proof: Since cos z = eiz +e −iz
, if we let e iz = u, then cos z = w implies
2
that
w = u2 + 1
⇒ u 2 − 2wu + 1 = 0
2u
which implies that
So, by Theorem 1.51,
(u − w) 2 = w 2 − 1 ≠ 0 since w ≠ ±1.
e iz = u = w + ∣ w 2 − 1 ∣ 1/2 e iφ k
, where φ k = arg (w2 − 1)
2
(
)
= w ± ∣ w 2 − 1 ∣ 1/2 e i arg(w 2 −1)
2
+ 2πk , k = 0, 1.
2
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