Questions - Answers
3. Show that (−1 + i )3n + (−1 − i )3n = (−1)3n 23n+1 cos nπ .
Sol. L.H.S = (−1 + i )3n + (−1 − i )3n
= [−1 (1 − i )]3n + [−1 (1 + i )]3n
= (−1)3n [(1 − i )3n + (1 + i )3n]
Let us find the Mod - amplitude form of 1 + i .
Let 1 + i = x + iy.
x = 1 y =
4. If 'n' is a positive integer, show that
Sol. Let p + iq = r (cosθ + i sinθ).
Then, r cosθ = p ....... (1) r sinθ = q ....... (2)
(1)2 + (2)2
⇒ r 2 cos2θ + r 2 sin2θ = p2 + q2
⇒ r 2 = p2 + q2
9. If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ then Prove that
i. cos 3α + cos 3β + cos 3 γ = 3 cos(α + β + γ)
ii. sin 3α + sin 3β + sin 3γ = 3 sin(α + β + γ) .
Sol: Given cos α + cos β + cos γ = 0 = sin α + sin β + sin γ .................(1)
Let a = cos α + i sin α = cis α
b = cos β + i sin β = cis β
c = cos γ + i sin γ = cis γ
Now, a + b + c = (cos α + i sin α) + (cos β + i sin β) + (cos γ + i sin γ)
⇒ a + b + c = (cos α + cos β + cos γ) + i (sin α + sin β + sin γ)
⇒ a + b + c = 0 + i(0) [ From (1) ]
⇒ a + b + c = 0
⇒ a3 + b3 + c3 = 3abc
⇒ (cos α + i sin α)3 + (cos β + i sin β)3 + (cos γ + i sin γ)3 = 3 cis α cis β cis γ
⇒ (cos 3α + i sin 3α) + (cos 3β + i sin 3β) + (cos 3γ + i sin 3γ) = 3 cis (α + β + γ )
⇒ (cos 3α + cos 3β + cos 3γ) + i(sin 3α + sin 3β + sin 3γ) = 3[cos (α + β + γ) + i sin (α + β + γ)]
Equating real and imaginary parts on both sides
i. cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ)
ii. sin 3α + sin 3β + sin 3γ = 3 sin(α + β + γ).
10. If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ Then prove that
i. cos2 α + cos2 β + cos2 γ = 3/2
ii. sin2 α + sin2 β + sin2 γ = 3/2
Sol : Given : cos α + cos β + cos γ = 0 = sin α + sin β + sin γ ........ (1)
We know that a + b + c = 0
(a + b + c)2 = 0
⇒ a2 + b2 + c2 + 2ab + 2bc + 2ca = 0
⇒ a2 + b2 + c2 = - 2ab - 2bc - 2ca
⇒ a2 + b2 + c2 = -2abc
⇒ (cos α + i sin α)2 + (cos β + i sin β)2 + (cos γ + i sin γ)2
= -2abc [(cos γ - i sin γ) + (cos α - i sin α) + (cos β - i sin β)]
⇒ (cos 2α + i sin 2α) + (cos 2β + i sin 2β) + (cos 2γ + i sin 2γ)
= -2abc [(cos α + cos β + cos γ) - i(sin α + sin β + sin γ)]
⇒ (cos 2α + cos 2β + cos 2γ) + i(sin 2α + sin 2β + sin 2γ)
= -2abc [0 - i (0)] [From (1)]
⇒ (cos 2α + cos 2β + cos 2γ) + i(sin 2α + sin 2β + sin 2γ) = 0 = 0 + i (0)
Equating real parts on both sides
cos 2α + cos 2β + cos 2γ = 0
⇒ 2cos2 α - 1 + 2cos2 β - 1 + 2cos2 γ - 1 = 0
⇒ 2(cos2 α + cos2 β + cos2 γ) = 3
∴ (i) cos2 α + cos2 β + cos2 γ = 3/2
⇒ 1 - sin2 α + 1 - sin2 β + 1 - sin2 γ = 3/2
⇒ sin2 α + sin2 β + sin2 γ = 3 - 3/2
∴ (ii) sin2 α + sin2 β + sin2 γ = 3