If n is a positive integer, show that : P + i Q 1 n + P - i Q 1 n = 2 P 2 + Q 2 1 2 n · cos 1 n tan - 1 Q P

To prove: P + i Q 1 n + P - i Q 1 n = 2 P 2 + Q 2 1 2 n · cos 1 n tan - 1 Q P Let P + i Q = r { cos θ + i sin θ } , here r cos θ = P , r sin θ = Q ⇒ r 2 = P 2 + Q 2 ∴ r = P 2 + Q 2 and similarly P - i Q = r { cos θ - i sin θ } Thus, cos θ = P P 2 + Q 2 and sin θ = Q P 2 + Q 2 , therefore tan θ = Q P ⇒ θ = tan - 1 Q P . . . . i Therefore, L.H.S. = ( P + i Q ) 1 n + ( P - i Q ) 1 n = { r ( cos θ + i sin θ ) } 1 n + { r ( cos θ - i sin θ ) } 1 n = r 1 n cos θ n + i sin θ n + cos θ n - i sin θ n ∵ ( cos θ + i sin θ ) n = cos n θ + i sin n θ , n ∈ Z = P 2 + Q 2 1 n 2 cos 1 n θ = 2 P 2 + Q 2 1 2 n cos 1 n tan - 1 Q P [Using i ] = R.H.S Hence, proved.

If n is a positive integer, prove that 1 + sin θ − i cos θ 1 + sin θ + i cos θ n = cos n π 2 − θ − i sin n π 2 − θ .

To prove 1 + sin θ − i cos θ 1 + sin θ + i cos θ n = cos n π 2 − θ − i sin n π 2 − θ Let Z = sin θ - i cos θ . So Z = 1 Now, 1 Z = 1 sin θ - i cos θ = sin θ + i cos θ sin θ - i cos θ sin θ + i cos θ = sin θ + i cos θ sin 2 θ + cos 2 θ = sin θ + i cos θ LHS = 1 + sin θ - i cos θ 1 + sin θ + i cos θ n = 1 + Z 1 + 1 Z n = Z Z + 1 1 + Z n = Z n And Z n = sin θ - i cos θ n = cos π 2 − θ − i sin π 2 − θ n = cos n π 2 − θ − i sin n π 2 − θ = RHS [By De Moivre's Theorem, cos θ + i sin θ n = cos n θ + i sin n θ for all n ∈ Z ] Hence, proved.

If n is an integer and Z = cis ⁡ θ , θ ≠ ( 2 n + 1 ) π 2 then show that Z 2 n − 1 Z 2 n + 1 = i tan ⁡ n θ .

Given that : Z = cis θ = cos θ + i sin θ , θ ≠ ( 2 n + 1 ) π 2 , where n is an integer. To prove: Z 2 n - 1 Z 2 n + 1 = i tan n θ From, LHS, we have, = Z 2 n - 1 Z 2 n + 1 = ( cos θ + i sin θ ) 2 n - 1 ( cos θ + i sin θ ) 2 n + 1 = cos 2 n θ + i sin 2 n θ - 1 cos 2 n θ + i sin 2 n θ + 1 ∵ ( cos A + i sin A ) n = cos n A + i sin n A , n ∈ Z = - ( 1 - cos 2 n θ ) + i sin 2 n θ ( 1 + cos 2 n θ ) + i sin 2 n θ = i 2 2 sin 2 n θ + 2 i sin n θ · cos n θ 2 cos 2 n θ + 2 i sin n θ · cos n θ ∵ - 1 = i 2 , 1 + cos 2 A = 2 cos 2 A , 1 - cos 2 A = 2 sin 2 A , sin 2 A = 2 sin A cos A = 2 i sin n θ { cos n θ + i sin n θ } 2 cos n θ { cos n θ + i sin n θ } = i tan n θ = RHS Hence, proved.

Show that one value of 1 + sin π 8 + i cos π 8 1 + sin π 8 - i cos π 8 8 3 is - 1 .

Given, 1 + sin π 8 + i cos π 8 1 + sin π 8 - i cos π 8 8 3 = 1 + cos π 2 - π 8 + i sin π 2 - π 8 1 + cos π 2 - π 8 - i sin π 2 - π 8 8 3 ∵ sin π 2 - θ = cos θ & cos π 2 - θ = sin θ = 1 + cos 3 π 8 + i sin 3 π 8 1 + cos 3 π 8 - i sin 3 π 8 8 3 = 2 cos 2 3 π 16 + i 2 sin 3 π 16 cos 3 π 16 2 cos 2 3 π 16 - i 2 sin 3 π 16 cos 3 π 16 8 3 ∵ 1 + cos θ = 2 cos 2 θ 2 & sin θ = 2 sin θ 2 cos θ 2 = 2 cos 3 π 16 2 cos 3 π 16 cos 3 π 16 + i sin 3 π 16 cos 3 π 16 - i sin 3 π 16 8 3 = cos 3 π 16 + i sin 3 π 16 cos 3 π 16 - i sin 3 π 16 8 3 = cos 3 π 16 + i sin 3 π 16 cos 3 π 16 + i sin 3 π 16 cos 3 π 16 - i sin 3 π 16 cos 3 π 16 + i sin 3 π 16 8 3 = cos 3 π 16 + i sin 3 π 16 2 1 8 3 [For denominator, using a + b a - b = a 2 - b 2 & i 2 = - 1 ] = cos 3 π 16 + i sin 3 π 16 16 3 = cos 3 π 16 × 16 3 + i sin 3 π 16 × 16 3 ∵ c i s n θ = c i s n θ = cosπ + i sinπ = - 1 + i × 0 = - 1 ∴ 1 + sin π 8 + i cos π 8 1 + sin π 8 - i cos π 8 8 3 = - 1 .