If α , β are the roots of the equation x 2 - 2 x + 4 = 0 , then for any n ∈ N show that: α n + β n = 2 n + 1 cos n π 3

Given quadratic equation: x 2 - 2 x + 4 = 0 and to prove: α n + β n = 2 n + 1 cos n π 3 ⇒ x = 2 ± 4 - 16 2 = 2 ± 2 i 3 2 = 1 ± i 3 ∵ x = - b ± b 2 - 4 a c 2 a , when a x 2 + b x + c = 0 Given that, α , β are the roots of the given quadratic equation i.e., α = 1 + i 3 = 2 1 2 + i 3 2 = 2 cos π 3 + i sin π 3 and β = 1 - i 3 = 2 1 2 - i 3 2 = 2 cos π 3 - i sin π 3 Now, from LHS, we have, α n + β n = 2 cos π 3 + i sin π 3 n + 2 cos π 3 - i sin π 3 n = 2 n cos n π 3 + i sin n π 3 + cos n π 3 - i sin n π 3 ∵ ( cos θ + i sin θ ) n = cos n θ + i sin n θ , n ∈ Z = 2 n 2 cos n π 3 = 2 n + 1 cos n π 3 = RHS 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.

If cos α + cos β + cos γ = 0 = sin α + sin β + sin γ , prove that: cos 2 α + cos 2 β + cos 2 γ = 3 2 = sin 2 α + sin 2 β + sin 2 γ .

Given that cos α + cos β + cos γ = 0 = sin α + sin β + sin γ Let a = e i α = cos α + i sin α , b = e i β = cos β + i sin β , c = e i γ = cos γ + i sin γ e i α + e i β + e i γ = cos α + i sin α + cos β + i sin β + cos γ + i sin γ ⇒ e i α + e i β + e i γ = cos α + cos β + cos γ + i sin α + sin β + sin γ ⇒ e i α + e i β + e i γ = 0 ⇒ a + b + c = 0 Also, 1 a + 1 b + 1 c = e - i α + e - i β + e - i γ = cos α - i sin α + cos β - i sin β + cos γ - i sin γ = cos α + cos β + cos γ - i sin α + sin β + sin γ ⇒ 1 a + 1 b + 1 c = 0 We have, a + b + c = 0 ⇒ a + b + c 2 = 0 ⇒ a 2 + b 2 + c 2 + 2 a b + b c + c a = 0 ⇒ a 2 + b 2 + c 2 = - 2 a b + b c + c a ⇒ a 2 + b 2 + c 2 = - 2 a b c 1 c + 1 a + 1 b ⇒ a 2 + b 2 + c 2 = 0 ⇒ e i α 2 + e i β 2 + e i γ 2 = 0 ⇒ e i 2 α + e i 2 β + e i 2 γ = 0 ⇒ cos 2 α + i sin 2 α + cos 2 β + i sin 2 β + cos 2 γ + i sin 2 γ = 0 ⇒ cos 2 α + cos 2 β + cos 2 γ + i sin 2 α + sin 2 β + sin 2 γ = 0 + i 0 Hence, cos 2 α + cos 2 β + cos 2 γ = 0 , sin 2 α + sin 2 β + sin 2 γ = 0 Taken, cos 2 α + cos 2 β + cos 2 γ = 0 ⇒ 2 cos 2 α - 1 + 2 cos 2 β - 1 + 2 cos 2 γ - 1 = 0 ⇒ 2 cos 2 α + 2 cos 2 β + 2 cos 2 γ = 3 ⇒ cos 2 α + cos 2 β + cos 2 γ = 3 2 Similarly, cos 2 α + cos 2 β + cos 2 γ = 0 ⇒ 1 - 2 sin 2 α + 1 - 2 sin 2 β + 1 - 2 sin 2 γ = 0 ⇒ - 2 sin 2 α - 2 sin 2 β - 2 sin 2 γ = - 3 ⇒ 2 sin 2 α + 2 sin 2 β + 2 sin 2 γ = 3 ⇒ sin 2 α + sin 2 β + sin 2 γ = 3 2 Hence, cos 2 α + cos 2 β + cos 2 γ = 3 2 = sin 2 α + sin 2 β + sin 2 γ proved.

For n ∈ N , 1 + cos θ + i sin θ 1 + cos θ - i sin θ n =

- cos ( n θ ) + i sin ( n θ )

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 .

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.

MEDIUM

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${(\frac{1 + \cos \frac{π}{8} - i \sin \frac{π}{8}}{1 + \cos \frac{π}{8} + i \sin \frac{π}{8}})}^{12} =$

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Important Questions on Complex Numbers

MEDIUM

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

If $α, β$ are the roots of the equation $x^{2} - 2 x + 4 = 0$ , then for any $n \in N$ show that:

$α^{n} + β^{n} = 2^{n + 1} \cos (\frac{n π}{3})$

MEDIUM

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

{(\frac{\sqrt{6} - \sqrt{2}}{4} + \frac{\sqrt{6} + \sqrt{2}}{4} i)}^{2020} =

MEDIUM

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

n

is a positive integer, prove that

{(\frac{1 + \sin θ - i \cos θ}{1 + \sin θ + i \cos θ})}^{n} = \cos n (\frac{π}{2} - θ) - i \sin n (\frac{π}{2} - θ)

MEDIUM

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

n

is an integer and

Z = cis θ, (θ \neq (2 n + 1) \frac{π}{2})

then show that

\frac{Z^{2 n} - 1}{Z^{2 n} + 1} = i \tan n θ

MEDIUM

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

(\sqrt{3} + i)^{100} = 2^{99} (p + i q),

then

p

and

q

are roots of the equation :

HARD

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

If $\cos α + \cos β + \cos γ = 0 = \sin α + \sin β + \sin γ$ , prove that:

$\cos^{2} α + \cos^{2} β + \cos^{2} γ = \frac{3}{2} = \sin^{2} α + \sin^{2} β + \sin^{2} γ$ .

HARD

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

For

n \in N, {(\frac{1 + \cos θ + i \sin θ}{1 + \cos θ - i \sin θ})}^{n} =

HARD

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

z

is a non-real complex number, then the minimum value of

\frac{I m z^{5}}{{(I m z)}^{5}}

is (Where

I m z =

Imaginary part of

z

)

EASY

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

Real part of

(\cos 4 + i \sin 4 + 1)^{2020}

is_______

EASY

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

(\sin θ - i \cos θ)^{3}

is equal to

MEDIUM

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

Show that one value of

{[\frac{1 + \sin \frac{π}{8} + i \cos \frac{π}{8}}{1 + \sin \frac{π}{8} - i \cos \frac{π}{8}}]}^{\frac{8}{3}}

- 1

MEDIUM

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

The product of all values of

(\cos α + i \sin α)^{3 / 5}

MEDIUM

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

One of the values of

{(\frac{1 + i}{\sqrt{2}})}^{\frac{2}{3}}

HARD

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

n

is a positive integer, show that :

{(P + i Q)}^{\frac{1}{n}} + {(P - i Q)}^{\frac{1}{n}} = 2 {(P^{2} + Q^{2})}^{\frac{1}{2 n}} \cdot \cos [\frac{1}{n} \tan^{- 1} \frac{Q}{P}]

HARD

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

Let

α

and

β

be the roots of the equation

x^{2} + 2 x + 2 = 0,

then

α^{15} + β^{15}

is equal to

HARD

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

1 + x^{2} = \sqrt{3} x

, then

\sum_{n = 1}^{24} {(x^{n} - \frac{1}{x^{n}})}^{2}

is equal to

EASY

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

The value of

\frac{2 (\cos 75^{o} + i s i n 75^{o})}{0.2 (\cos 30^{o} + i s i n 30^{o})}

MEDIUM

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

What is the value of

(1 - i \sqrt{3})^{9} =

EASY

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

Which of the following can be the equal expression for

\cos (3 T) + isin (3 T)

using the De Moivre's theorem and binomial theorem concepts?

MEDIUM

Mathematics>Algebra>Complex Numbers>De Moivre's Theorem

If $e^{i x}$ is a solution of the equation $z^{n} + p_{1} z^{n - 1} + p_{2} z^{n - 2} + \dots \dots + p_{n} = 0$ , where $p_{i}$ are real $(i = 1, 2, 3, \dots \dots \dots n)$ ,then

$p_{n} \sin n x + p_{n - 1} \sin (n - 1) x + \dots + p_{1} \sin x + 1 =$

1+cosπ8-isinπ81+cosπ8+isinπ812=

Important Questions on Complex Numbers

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${(\frac{1 + \cos \frac{π}{8} - i \sin \frac{π}{8}}{1 + \cos \frac{π}{8} + i \sin \frac{π}{8}})}^{12} =$