If z 2 + z + 1 = 0 , where z is a complex number, prove that z + 1 z 2 + z 2 + 1 z 2 2 + z 3 + 1 z 3 2 + z 4 + 1 z 4 2 + z 5 + 1 z 5 2 + z 6 + 1 z 6 2 = 12 .

Given equation: z 2 + z + 1 = 0 , where z is a complex number, so clearly ω , ω 2 are the two roots which satisfies the given equation. Let z = ω To prove: z + 1 z 2 + z 2 + 1 z 2 2 + z 3 + 1 z 3 2 + z 4 + 1 z 4 2 + z 5 + 1 z 5 2 + z 6 + 1 z 6 2 = 12 Now, from LHS, we have z + 1 z 2 + z 2 + 1 z 2 2 + z 3 + 1 z 3 2 + z 4 + 1 z 4 2 + z 5 + 1 z 5 2 + z 6 + 1 z 6 2 = ω + 1 ω 2 + ω 2 + 1 ω 2 2 + ω 3 + 1 ω 3 2 + ω 4 + 1 ω 4 2 + ω 5 + 1 ω 5 2 + ω 6 + 1 ω 6 2 = ω + 1 ω 2 + ω 2 + 1 ω 2 2 + ω 3 + 1 ω 3 2 + ω 3 · ω + 1 ω 3 · ω 2 + ω 3 · ω 2 + 1 ω 3 · ω 2 2 + ω 6 + 1 ω 6 2 = ω + 1 ω 2 + ω 2 + 1 ω 2 2 + 1 + 1 2 + ω + 1 ω 2 + ω 2 + 1 ω 2 2 + 1 + 1 2 ∵ ω 3 = 1 , ω 6 = 1 = ω + ω 3 ω 2 + ω 2 + ω 3 ω 2 2 + 8 + ω + ω 3 ω 2 + ω 2 + ω 3 ω 2 2 = ω + ω 2 2 + ω 2 + ω 2 + 8 + ω + ω 2 2 + ω 2 + ω 2 = - 1 2 + - 1 2 + 8 + - 1 2 + - 1 2 ∵ 1 + ω + ω 2 = 0 = 12 = RHS Hence, proved.

If 1 , α , α 2 , α 3 , … , α n - 1 be the n th roots of unity, then prove that 1 p + α p + α 2 p + α 3 p + … + α n - 1 p = 0 ; if p ≠ k n n ; if p = k n , where p , k ∈ N .

We know that, if x is n th root of unity, then it satisfies the relation x n - 1 = 0 It is given that, 1 , α , α 2 , α 3 , … , α n - 1 are the n th roots of unity, therefore, α n = 1 and use α = e i 2 π n Case (i): when p ≠ k n ⇒ 1 p + α p + α 2 p + α 3 p + … + α n - 1 p = e i 2 π n n p - 1 e i 2 π n p - 1 (Since α = e i 2 π n and S n = a r n - 1 r - 1 , r > 1 ) ⇒ 1 p + α p + α 2 p + α 3 p + … + α n - 1 p = e i 2 π p - 1 e i 2 π n p - 1 ⇒ 1 p + α p + α 2 p + α 3 p + … + α n - 1 p = 1 - 1 e i 2 π n p - 1 = 0 [Since, p , k ∈ N and e i 2 π p = cos 2 π p + i sin 2 π p = 1 + 0 = 1 , Since denominator cannot be zero because p ≠ k n ] Case (ii): when p = k n Therefore, 1 p + α p + α 2 p + α 3 p + … + α n - 1 p = 1 k n + α k n + α 2 k n + α 3 k n + … + α n - 1 k n = 1 + α n k + α n 2 k + α n 3 k + … + α n n - 1 k = 1 + 1 + 1 + 1 + … + 1 [since α n = 1 ] = n Hence, proved.

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 Therefore, ( 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 = RHS Hence, proved.

Show that 1 + sin π 8 + i cos π 8 1 + sin π 8 - i cos π 8 8 3 = - 1 .

To prove: 1 + sin π 8 + i cos π 8 1 + sin π 8 - i cos π 8 8 3 = - 1 We have LHS = 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 = 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 + 2 i sin 3 π 16 cos 3 π 16 2 cos 2 3 π 16 - 2 i sin 3 π 16 cos 3 π 16 8 3 ∵ 1 + cos x = 2 cos 2 x 2 , sin x = 2 sin x 2 cos x 2 = 2 cos 3 π 16 cos 3 π 16 + i sin 3 π 16 2 cos 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 {Multiplying and dividing with the conjugate of the denominator} = cos 3 π 16 + i sin 3 π 16 2 cos 2 3 π 16 + sin 2 3 π 16 8 3 = cos 3 π 8 + i sin 3 π 8 8 3 ∵ sin 2 x + cos 2 x = 1 , x ∈ R = cos π + i sin π = - 1 {Using De Moivre's Theorem when N is a Fractional Number} = RHS Hence, proved.