De Moivre's Theorem MCQ Quiz - Objective Question with Answer for De Moivre's Theorem - Download Free PDF

Calculation

\(\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1 - 2i + i^2}{1 - i^2} = \frac{1 - 2i - 1}{1 - (-1)} = \frac{-2i}{2} = -i\)

\((-i)^k = -i\)

\(-i = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right)\)

Using De Moivre's Theorem:

\((-i)^k = \cos\left(\frac{3k\pi}{2}\right) + i\sin\left(\frac{3k\pi}{2}\right)\)

Comparing with -i, we have:

\(\cos\left(\frac{3k\pi}{2}\right) = 0\) and \(\sin\left(\frac{3k\pi}{2}\right) = -1\)

This implies:

\(\frac{3k\pi}{2} = (2n+1)\pi - \frac{\pi}{2}\)

⇒ \(\frac{3k}{2} = 2n + 1 - \frac{1}{2}\)

⇒ \(\frac{3k}{2} = \frac{4n+1}{2}\)

⇒ \(3k = 4n + 1\)

⇒ \(k = \frac{4n+1}{3}\)

For the least positive integer value of k, let n = 0:

\(k = \frac{1}{3}\)

Let n = 1:

\(k = \frac{5}{3}\)

Let n = 2:

\(k = \frac{9}{3} = 3\)

So, m = 3.

For the greatest negative integer value of k, we can analyze the values of k for negative n.

For n = -1:

\(k = \frac{-3}{3} = -1\)

So, n = -1.

\(m - n = 3 - (-1) = 3 + 1 = 4\)

Formula Used:

1. Euler's formula: \(e^{i\theta} = \cos \theta + i \sin \theta\)

2. De Moivre's theorem: \((\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta\)

3. \(\sin \theta = \cos(\frac{\pi}{2} - \theta)\) and \(\cos \theta = \sin(\frac{\pi}{2} - \theta)\)

Calculation:

 \(\frac{(\cos a + i \sin a)^6}{(\sin b + i \cos b)^8} = \frac{(e^{ia})^6}{(\cos(\frac{\pi}{2}-b) + i \sin(\frac{\pi}{2}-b))^8}\)

⇒ \(= \frac{e^{i6a}}{(e^{i(\frac{\pi}{2}-b)})^8}\)

⇒ \(= \frac{e^{i6a}}{e^{i(4\pi - 8b)}}\)

⇒ \(= e^{i(6a - 4\pi + 8b)}\)

⇒ \(= e^{i(6a + 8b)}\) (Since \(e^{i(-4\pi)} = \cos(-4\pi) + i\sin(-4\pi) = 1\))

⇒ \(= \cos(6a + 8b) + i \sin(6a + 8b)\)

⇒ Real part = \(\cos(6a + 8b)\)

∴ The real part of the given expression is \(\cos(6a + 8b)\).

Hence option 4 is correct

Calculation

\(\frac{1-i}{1+i} = \frac{(1-i)(1-i)}{(1+i)(1-i)} = \frac{1 - 2i + i^2}{1 - i^2} = \frac{1 - 2i - 1}{1 - (-1)} = \frac{-2i}{2} = -i\)

\((-i)^k = -i\)

\(-i = \cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right)\)

Using De Moivre's Theorem:

\((-i)^k = \cos\left(\frac{3k\pi}{2}\right) + i\sin\left(\frac{3k\pi}{2}\right)\)

Comparing with -i, we have:

\(\cos\left(\frac{3k\pi}{2}\right) = 0\) and \(\sin\left(\frac{3k\pi}{2}\right) = -1\)

This implies:

\(\frac{3k\pi}{2} = (2n+1)\pi - \frac{\pi}{2}\)

⇒ \(\frac{3k}{2} = 2n + 1 - \frac{1}{2}\)

⇒ \(\frac{3k}{2} = \frac{4n+1}{2}\)

⇒ \(3k = 4n + 1\)

⇒ \(k = \frac{4n+1}{3}\)

For the least positive integer value of k, let n = 0:

\(k = \frac{1}{3}\)

Let n = 1:

\(k = \frac{5}{3}\)

Let n = 2:

\(k = \frac{9}{3} = 3\)

So, m = 3.

For the greatest negative integer value of k, we can analyze the values of k for negative n.

For n = -1:

\(k = \frac{-3}{3} = -1\)

So, n = -1.

\(m - n = 3 - (-1) = 3 + 1 = 4\)

Hence option 1 is correct

Concept Used:

Factoring the equation and expressing roots in cis form. Using De Moivre's Theorem.

cis(θ) = cos(θ) + i sin(θ)

Calculation

\(x^{14} + x^9 - x^5 - 1 = 0\)

⇒ \(x^9(x^5 + 1) - 1(x^5 + 1) = 0\)

⇒ \((x^9 - 1)(x^5 + 1) = 0\)

Thus, \(x^9 = 1\) or \(x^5 = -1\)

Let's analyze the options:

Option 1: \(\frac{1 + \sqrt{3}i}{2} = \text{cis}(\frac{\pi}{3})\)

\(x^5 = \text{cis}(\frac{5\pi}{3}) = \cos(\frac{5\pi}{3}) + i\sin(\frac{5\pi}{3}) = \frac{1}{2} - i\frac{\sqrt{3}}{2} \ne -1\)

Option 3: \(\frac{1 - \sqrt{3}i}{2} = \text{cis}(-\frac{\pi}{3})\)

\(x^5 = \text{cis}(-\frac{5\pi}{3}) = \cos(-\frac{5\pi}{3}) + i\sin(-\frac{5\pi}{3}) = \frac{1}{2} + i\frac{\sqrt{3}}{2} \ne -1\)

Option 4: \(\frac{\sqrt{5}+1}{4} + i\frac{\sqrt{10-2\sqrt{5}}}{4} = \text{cis}(\frac{\pi}{5})\)

\(x^5 = \text{cis}(\pi) = \cos(\pi) + i\sin(\pi) = -1\)

Option 2: \(\frac{\sqrt{5}-1}{4} + i\frac{\sqrt{10+2\sqrt{5}}}{4} = \text{cis}(\frac{2\pi}{5})\)

\(x^5 = \text{cis}(2\pi) = \cos(2\pi) + i\sin(2\pi) = 1 \ne -1\)

Since option 4 satisfies \(x^5 = -1\), it is a root of the equation.

∴ The root is \(\frac{\sqrt{5}+1}{4} + i\frac{\sqrt{10-2\sqrt{5}}}{4}\)

Hence option 4 is correct

Concept Used:

Complex numbers in polar form: z = |z|(cos(θ) + i sin(θ))

De Moivre's Theorem: (cos(θ) + i sin(θ))^n = cos(nθ) + i sin(nθ)

Calculation

Given:

|x| = |y| = 1

Arg(x) = 2α

Arg(y) = 3β

α + β = π/36

\(x = \cos(2\alpha) + i \sin(2\alpha)\)

\(y = \cos(3\beta) + i \sin(3\beta)\)

\(x^6y^4 = (\cos(2\alpha) + i \sin(2\alpha))^6 (\cos(3\beta) + i \sin(3\beta))^4\)

⇒ \(x^6y^4 = (\cos(12\alpha) + i \sin(12\alpha)) (\cos(12\beta) + i \sin(12\beta))\)

⇒ \(x^6y^4 = \cos(12\alpha + 12\beta) + i \sin(12\alpha + 12\beta)\)

⇒ \(x^6y^4 = \cos(12(\alpha + \beta)) + i \sin(12(\alpha + \beta))\)

⇒ \(x^6y^4 = \cos(12 \times \frac{\pi}{36}) + i \sin(12 \times \frac{\pi}{36})\)

⇒ \(x^6y^4 = \cos(\frac{\pi}{3}) + i \sin(\frac{\pi}{3})\)

⇒ \(x^6y^4 = \frac{1}{2} + i\frac{\sqrt{3}}{2}\)

\(\frac{1}{x^6y^4} = \frac{1}{\frac{1}{2} + i\frac{\sqrt{3}}{2}}\)

⇒ \(\frac{1}{x^6y^4} = \frac{\frac{1}{2} - i\frac{\sqrt{3}}{2}}{(\frac{1}{2} + i\frac{\sqrt{3}}{2})(\frac{1}{2} - i\frac{\sqrt{3}}{2})}\)

⇒ \(\frac{1}{x^6y^4} = \frac{\frac{1}{2} - i\frac{\sqrt{3}}{2}}{\frac{1}{4} + \frac{3}{4}}\)

⇒ \(\frac{1}{x^6y^4} = \frac{1}{2} - i\frac{\sqrt{3}}{2}\)

\(x^6y^4 + \frac{1}{x^6y^4} = (\frac{1}{2} + i\frac{\sqrt{3}}{2}) + (\frac{1}{2} - i\frac{\sqrt{3}}{2})\)

⇒ \(x^6y^4 + \frac{1}{x^6y^4} = 1\) 

Hence option 3 is correct

Concept:

Euler's formula:

A complex number z = cos θ + i sin θ can also be written as eiθ.

Calculation:

From the Euler's formula, we know that:

\(\rm \frac{\cos\theta+i\sin\theta}{\cos\theta-i\sin\theta}=\frac{e^{i\theta}}{e^{-i\theta}}\) = e^2iθ = cos 2θ + i sin 2θ.

Concept:

1. Euler's Formula on Complex Numbers:

• e^ix = cos x + i sin x
• e^-ix = cos x - i sin x

2. Trigonometry formulas:

• 1 – cos θ = 2 sin² (θ/2)
• sin θ = 2 sin (θ/2) cos (θ/2

Calculation:

We have to find the value of \({\left[ {\frac{{\sin \frac{{\rm{\pi }}}{6} + {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}{{\sin \frac{{\rm{\pi }}}{6} - {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}} \right]^3}\)

\( \Rightarrow {\left[ {\frac{{\sin \frac{{\rm{\pi }}}{6} + {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}{{\sin \frac{{\rm{\pi }}}{6} - {\rm{i}}\left( {1 - \cos \frac{{\rm{\pi }}}{6}} \right)}}} \right]^3} = {\rm{\;}}{\left[ {\frac{{2 \times \sin \frac{{\rm{\pi }}}{{12}} \times \cos \frac{{\rm{\pi }}}{{12}} + {\rm{i}}\left( {{\rm{\;}}2{{\sin }^2}\frac{{\rm{\pi }}}{{12}}} \right)}}{{2 \times \sin \frac{{\rm{\pi }}}{{12}} \times \cos \frac{{\rm{\pi }}}{{12}} - {\rm{i}}\left( {{\rm{\;}}2{{\sin }^2}\frac{{\rm{\pi }}}{{12}}} \right)}}} \right]^3}{\rm{\;}}\)

\( = {\rm{\;}}{\left[ {\frac{{2 \times \sin \frac{{\rm{\pi }}}{{12}} \times \left( {\cos \frac{{\rm{\pi }}}{{12}} + {\rm{i}}\sin \frac{{\rm{\pi }}}{{12}}} \right)}}{{2 \times \sin \frac{{\rm{\pi }}}{{12}} \times \left( {\cos \frac{{\rm{\pi }}}{{12}} - {\rm{i}}\sin \frac{{\rm{\pi }}}{{12}}} \right)}}} \right]^3}\)

\(= {\rm{\;}}{\left[ {\frac{{\left( {\cos \frac{{\rm{\pi }}}{{12}} + {\rm{i}}\sin \frac{{\rm{\pi }}}{{12}}} \right)}}{{\left( {\cos \frac{{\rm{\pi }}}{{12}} - {\rm{i}}\sin \frac{{\rm{\pi }}}{{12}}} \right)}}} \right]^3}\)                          (∵e^ix = cos x + i sin x and e^-ix = cos x - i sin x)

\(= {\rm{\;}}{\left[ {\frac{{{{\rm{e}}^{\frac{{{\rm{i\pi }}}}{{12}}}}}}{{{{\rm{e}}^{\frac{{ - {\rm{i\pi }}}}{{12}}}}}}} \right]^3} = {\left[ {{{\rm{e}}^{\frac{{{\rm{i\pi }}}}{{12}}}}{\rm{\;}} \times {\rm{\;}}{{\rm{e}}^{\frac{{{\rm{i\pi }}}}{{12}}}}} \right]^3} = {\left[ {{{\rm{e}}^{\frac{{{\rm{i\pi }}}}{6}}}{\rm{\;}}} \right]^3} = {\rm{\;}}{{\rm{e}}^{\frac{{{\rm{i\pi }}}}{2}}}\)             

= cos (π/2) + i sin (π/2) = 0 + i  = i

Concept:

\({e^{i\; ⋅ \;\theta }} = \cos \theta + i\sin \theta \)

Calculation:

Given: 

x = (cos π/14 + i sin π/14), y = (cos 9π/14 + i sin 9π/14)

As we know that, 

\({e^{i\; ⋅ \;\theta }} = \cos \theta + i\sin \theta \) 

We can re-write x and y as:

⇒ x = e^{i ⋅ π/14} and y = e^{i ⋅ 9π/14}

⇒ x⁵ = e^{i ⋅ 5π/14} and y¹⁵ = e^{i ⋅ 135π/14}

⇒  x5 ⋅ y¹⁵ = e^{i ⋅ 10π} 

Concept:

\({e^{i\theta }} = \cos \theta + i\sin \theta \)

Calculation:

Given: x =  (cos π/9 + i sin π/9 )18 and y = (cos π/16 + i sin π/16 )⁸

As we know that, \({e^{i\theta }} = \cos \theta + i\sin \theta \)

⇒ y = (cos π/16 + i sin π/16 )8 = [ei ⋅ π/16]8 = ei ⋅ π/2

⇒ ei ⋅ π/2 = cos π/2 + i sin π/2 = i

⇒ y = i 

So, y^-2 = i^-2 = - 1  ----(1)

Similarly, 

As we know that, \({e^{i\theta }} = \cos \theta + i\sin \theta \)

⇒ x = (cos π/9 + i sin π/9 )18 = [ei ⋅ π/9]18 = ei ⋅ 2π

⇒ ei ⋅ 2π = cos 2π + i sin 2π = 1

⇒ x = 1     -------(2)

So, from (1) and (2), we get

⇒ x ⋅ y^-2 = 1 × - 1 = - 1

Concept:

\({e^{i\; ⋅ \;\theta }} = \cos \theta + i\sin \theta \)

Calculation:

As we know that, 

\({e^{i\; ⋅ \;\theta }} = \cos \theta + i\sin \theta \)

⇒ (cos π/9 + i sin π/9 )18 

= [e^i ⋅ π/9]¹⁸ = e^{i ⋅ 2π}

⇒ e^{i ⋅ 2π} 

= cos 2π + i sin 2π = 1

Concept:

\({e^{i\; ⋅ \;\theta }} = \cos \theta + i\sin \theta \)

Calculation:

As we know that, \({e^{i\; ⋅ \;\theta }} = \cos \theta + i\sin \theta \)

⇒ (cos π/16 + i sin π/16 )8 = [e^{i ⋅ π/16}]⁸ = e^{i ⋅ π/2}

⇒ e^{i ⋅ π/2} = cos π/2 + i sin π/2 = i

Concept:

Relation between modulus and argument of a complex number:

For any complex number \(z\) if the modulus is given by \(|z|\) and argument is \(\theta\) then the following relation always holds:

\(z = |z|e^{i\theta}\)

\(e^{i\theta} = \cos \theta+i\sin\theta\)

Calculation:

Let the given complex number be \(z\) then we have \(|z| = 4\) and arg \(z = \dfrac{5\pi}{6}\). 

Therefore, \(\theta = \dfrac{5\pi}{6}\).

Now using the above formulae,

\(\begin{align*} z &= |z|e^{i\theta}\\ &= 4\left(e^{i\frac{5\pi}{6}}\right)\\ &= 4\left(\cos\dfrac{5\pi}{6}+i\sin\dfrac{5\pi}{6}\right)\\ &= 4\left(-\dfrac{\sqrt3}{2} + i\dfrac{1}{2}\right)\\ &= -2\sqrt3+2i \end{align*}\)

Therefore, the required complex number is \(z = -2\sqrt3+2i\).

Concept:

Let z = x + iy be any complex number then its polar form is \(\rm z = r(cos\;\theta + isin\;\theta) \), where r = \(\rm \sqrt{x^2+y^2}\) and \(\rm \theta = tan^{-1}{(\dfrac y x)}\)

De Moivre’s Theorem 

Given any complex number cos θ + i sin θ and any integer n,

(cos θ + i sin θ )n = cos nθ + i sin nθ 

Calculations:

To Find \(\rm (1+i\sqrt{3})^n + (1-i\sqrt{3})^n\)

First write the complex numbers \((\rm 1+i\sqrt{3})\) and \((\rm 1-i\sqrt{3})\) in polar form and apply De - Moivers theorem.

Let z = x + iy be any complex number then its polar form is  \(\rm z = r(cos\;\theta + isin\;\theta) \), where r = \(\rm \sqrt{x^2+y^2}\) and \(\rm \theta = tan^{-1}{(\dfrac y x)}\)

The polar form of \((\rm 1+i\sqrt{3})\) is \(\rm 2(cos\;\dfrac{\pi}{3} + isin\;\dfrac{\pi}{3}) \)

The polar form of \((\rm 1-i\sqrt{3})\) is \(\rm 2(cos\;\dfrac{\pi}{3} - isin\;\dfrac{\pi}{3}) \)

Consider, \((1+i\sqrt{3})^n + (1-i\sqrt{3})^n\)

 = \(\Big[\rm 2(cos\;\dfrac{\pi}{3} + isin\;\dfrac{\pi}{3})\Big]^n\) + \(\Big[\rm 2(cos\;\dfrac{\pi}{3} - isin\;\dfrac{\pi}{3})\Big]^n\)

Apply De Moivre’s Theorem 

Given any complex number cos θ + i sin θ and any integer n,

(cos θ + i sin θ )n = cos nθ + i sin nθ .

= \(\rm 2^n(cos\;\dfrac{n\pi}{3} + isin\;\dfrac{n\pi}{3})\) + \(\rm 2^n(cos\;\dfrac{n\pi}{3} - isin\;\dfrac{n\pi}{3})\)

= \(\rm2^{n+1} \cos \dfrac{n\pi}{3}\)

Hence, \((1+i\sqrt{3})^n + (1-i\sqrt{3})^n\)= \(\rm2^{n+1} \cos \dfrac{n\pi}{3}\)

Concept:

\({e^{i\; ⋅ \;\theta }} = \cos \theta + i\sin \theta \)

Calculation:

The given expression:\({\left( {\frac{{√ 2 + i\;√ 2 }}{2}} \right)^{64}}\) can re-written as

 =  \((\frac{\sqrt 2}{2}\ +\ \frac{\sqrt 2}{2}i)^{64}\)

 = \((\frac{1}{\sqrt 2}\ +\ \frac{1}{\sqrt 2}i)^{64}\)

As we know that sin π/4 = 1/√2 = cos π/4

So, we can write the given expression \({\left( {\frac{{√ 2 + i\;√ 2 }}{2}} \right)^{64}}\)as

 = (cos π/4 + i sin π/4)⁶⁴

As we know that, \({e^{i\; ⋅ \;\theta }} = \cos \theta + i\sin \theta \)

⇒ (cos π/4 + i sin π/4)⁶⁴ = (e^i ⋅ π/4 )⁶⁴

⇒ (cos π/4 + i sin π/4)64 = e^{i ⋅ 16π}

Concept:

De Moivre's formula:

If z = re^iθ = r(cos θ + i sin θ) , then zⁿ = rⁿe^inθ(cos nθ + i sin nθ)

Calculation:

Given, z = \((\frac{√3}{2}+\frac{i}{2})^5+(\frac{√3}{2}-\frac{i}{2})^5\)

⇒ z = \(\left(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6}\right)^5+\left(\cos\frac{-\pi}{6}+i\sin\frac{-\pi}{6}\right)^5\)

⇒ z = \(\left(e^{i\frac{\pi}{6}}\right)^5+\left(e^{i\frac{-\pi}{6}}\right)^5\)

⇒ z = \(\left(e^{i\frac{5\pi}{6}}\right)+\left(e^{i\frac{-5\pi}{6}}\right)\)

⇒ z = \(\left(\cos \frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)+\left(\cos\frac{-5\pi}{6}+i\sin\frac{-5\pi}{6}\right)\)

⇒ z = \(\left(\cos \frac{5\pi}{6}+i\sin\frac{5\pi}{6}\right)+\left(\cos\frac{5\pi}{6}-i\sin\frac{5\pi}{6}\right)\)

⇒ z = \(\left[\cos (\pi-\frac{5\pi}{6})+i\sin(\pi-\frac{5\pi}{6})\right]+\left[\cos (\pi-\frac{5\pi}{6})-i\sin(\pi-\frac{5\pi}{6})\right]\)

⇒ z = \(\left(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6}\right)+\left(\cos\frac{\pi}{6}-i\sin\frac{\pi}{6}\right)\)

⇒ z = 2 cos \(\frac{\pi}{6}\) = √3

∴ Im (z) = 0