De Moivre's Theorem MCQ Quiz in తెలుగు - Objective Question with Answer for De Moivre's Theorem - ముఫ్త్ [PDF] డౌన్‌లోడ్ కరెన్

\(= \, |1 \, - \, e^{2(r-1)\pi i/n}| \, \, \, \, \, \, [\because |z_1| \, = \, 1]\)

Hence \(|A_1A_2| \cdot |A_1A_3| \ldots |A_1A_n|\)

\(= \, |1 \, - \, e^{2 \pi i/n}||1 \, - \, e^{4 \pi i/n}|\ldots|1 \, - \, e^{2(n-1)\pi i/n}|.....(1)\)

Since \(e^{2 \pi i/n}, e^{4 \pi i/n}\ldots e^{2(n-1)\pi i/n}\) are the \(n^{th}\) roots of unity, we have the identity

\(z^n \, - \, 1 \, \equiv \, (z \, - \, 1 )(z \, - \, e^{2 \pi i/n})(z \, - \, e^{4 \pi i/n})\ldots(z \, - \, e^{2 (n-1)\pi i/n})\)

or \(\dfrac{z^n \, - \, 1 }{z \, - \, 1} \, \equiv \, (z \, - \, e^{2 \pi i/n})(z \, - \, e^{4 \pi i/n})\ldots(z \, - \, e^{2 (n-1)\pi i/n})\)

or \(1 \,+ \, z \, + \, z^2 \, + \, \ldots \,+ \, z^{n \, - \, 1} \equiv \, (z \, - \, e^{2 \pi i/n})\ldots(z \, - \, e^{2 (n-1)\pi i/n})\)

Putting \(z =1\) in the above identity, we get

\(n \, = \, (1 \, - \, e^{2 \pi i/n})(1 \, - \, e^{4 \pi i/n})\ldots(1 \, - \, e^{2 (n-1)\pi i/n})\)

Hence \(n \, = \, |n| \, = \, |1 \, - \, e^{2 \pi i/n}| \, |1 \, - \, e^{4 \pi i/n}|\ldots|1 \, - \, e^{2 (n-1)\pi i/n}|....(2)\)

From (1) and (2), we get

\(\therefore |A_1A_2| \cdot |A_1A_3| \ldots |A_1A_n| \, = \, n\)

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:

Euler's theorem:

e^iθ = cosθ + isinθ 

De Moivre's Theorem:

(cosθ + isinθ)ⁿ = cos(nθ) + isin(nθ) 

Calculation:

Given, \((1+z)^n=1+{}^nC_1z+{}^nC_2z^2+\cdots+{}^nC_nz^n=\sum_{r=0}^{n}{}^{n}C_rz^r\)

Putting z = e^ir, we get:

\((1+e^{ir})^n=1+{}^nC_1e^{ir}+{}^nC_2e^{2ir}+\cdots+{}^nC_ne^{nir}=\sum_{r=0}^{n}{}^{n}C_r(e^{ix})^r\)...(i)

Now \(\sum_{r=0}^{100}{}^{100}C_rz^r(\sin rx)\)

= \(\rm Img[\sum_{r=0}^{100}{}^{100}C_rz^r(e^{ix})^r]\), where Img(z) represents the imaginary part of z

= \(\rm Img[(1+e^{ix})^{100}]\)

= \(\rm Img[(1+\cos x +i\sin x)^{100}]\)

= \(\rm Img\left[\left(2\cos^2\left(\frac{x}{2} \right )+2i\sin\left(\frac{x}{2} \right )\cos\left(\frac{x}{2} \right )\right)^{100}\right]\)

= \(\rm Img\left[\left(2cos\left(\frac{x}{2} \right )\right)^{100}\left(\cos\left(\frac{x}{2} \right )+i\sin\left(\frac{x}{2} \right)\right)^{100}\right]\)

= \(\rm \left(2\cos\left(\frac{x}{2} \right )\right)^{100}Img\left[\cos\left(\frac{100x}{2} \right )+i\sin\left(\frac{100x}{2} \right)\right]\)

= \(\rm \left(2\cos\left(\frac{x}{2} \right )\right)^{100}Img\left[\cos50x+i\sin50x\right]\)

= \(\rm \left(2\cos\left(\frac{x}{2} \right )\right)^{100}\sin50x\) = \(\left(2 \cos \left(\frac{x}{2}\right)\right)^{100} \sin k x\) (given)

Comparing both sides we get, k = 50

∴ The value of k is 50.

Concept:

• If \(\rm \frac{a}{b}=\frac{c}{d}\) then by componendo and dividendo rule  \(\rm \frac{a+b}{a-b}=\frac{c+d}{c-d}\)

Calculation:

Given z = e^iθ and  \(\dfrac{3 \cos 3 θ + 2\cos2θ + 5\cos5θ}{3 \sin3θ +2\sin2θ + 5\sin5θ}\) = \(\dfrac{{i\sum\limits_{r = 0}^{10} {a_rz^r} }}{{\sum\limits_{r = 0}^{10} {b_rz^r} }}\)

⇒   \(\dfrac{{\sum\limits_{r = 0}^{10} {a_rz^r} }}{{\sum\limits_{r = 0}^{10} {b_rz^r} }}\) = \(\dfrac{3 \cos 3 θ + 2\cos2θ + 5\cos5θ}{i(3 \sin3θ +2\sin2θ + 5\sin5θ)}\)

We know that if \(\rm \frac{a}{b}=\frac{c}{d}\) then \(\rm \frac{a+b}{a-b}=\frac{c+d}{c-d}\)

⇒ \(\rm\frac{{\sum\limits_{r = 0}^{10} {a_rz^r} }+{\sum\limits_{r = 0}^{10} {b_rz^r} }}{{\sum\limits_{r = 0}^{10} {a_rz^r} }-{\sum\limits_{r = 0}^{10} {b_rz^r} }}\) = \(\rm\frac{(3 \cos 3 θ + 2\cos2θ + 5\cos5θ)+i(3 \sin3θ +2\sin2θ + 5\sin5θ)}{(3 \cos 3 θ + 2\cos2θ + 5\cos5θ)-i(3 \sin3θ +2\sin2θ + 5\sin5θ)}\)

⇒ \(\rm\frac{{\sum\limits_{r = 0}^{10} {a_rz^r} }+{\sum\limits_{r = 0}^{10} {b_rz^r} }}{{\sum\limits_{r = 0}^{10} {a_rz^r} }-{\sum\limits_{r = 0}^{10} {b_rz^r} }}\) = \(\rm\frac{3e^{iθ}+2e^{iθ}+5e^{iθ}}{3e^{-iθ}+2e^{-iθ}+5e^{-iθ}}\)  [e^iθ = cos θ + i sin θ ]

⇒ \(\rm\frac{{\sum\limits_{r = 0}^{10} {z^r(a_r + b_r)} }}{{\sum\limits_{r = 0}^{10} {z^r(a_r -b_r)} }}\) = \(\rm\frac{3e^{iθ}+2e^{iθ}+5e^{iθ}}{3e^{-iθ}+2e^{-iθ}+5e^{-iθ}}\) 

We know z^r = e^irθ and in the R.H.S the value of r are 2, 3 and 5

∴ a₀ + b₀ = a₁ + b₁ = a₄ + b₄ = a₆ + b₆ = a₇ + b₇ = a₈ + b₈ = a₉ + b₉ = a₁₀ + b₁₀ = 0

∴ a₂ + b₂ = 2 and a₃ + b₃ = 3 and a₅ + b₅ = 5 .

⇒ \(\frac{\Bigg({\sum\limits_{r = 0}^{10} {a_r + } \sum\limits_{r = 0}^{10} {b_r} }\Bigg)}{{10}}\)= \(\frac{2+3+5}{10}\) = 1

Required value of the expression is 1

Calculation:

Given,

\( (1 + i)^n + (1 - i)^n \)

We know from De Moivre's theorem that:

\( (1 + i)^n = (\sqrt{2})^n \left( \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} \right) \)

\( (1 - i)^n = (\sqrt{2})^n \left( \cos \frac{-n\pi}{4} + i \sin \frac{-n\pi}{4} \right) \)

Now, adding both the equations:

\( (1 + i)^n + (1 - i)^n = (\sqrt{2})^n \left( \cos \frac{n\pi}{4} + i \sin \frac{n\pi}{4} \right) + (\sqrt{2})^n \left( \cos \frac{-n\pi}{4} + i \sin \frac{-n\pi}{4} \right) \)

Simplifying, the imaginary parts cancel each other, and we are left with:

\( (1 + i)^n + (1 - i)^n = (\sqrt{2})^{n+2} \cos \frac{n\pi}{4} \)

Hence, the correct answer is Option 3.

Calculation:

Given,

\( x + \frac{1}{x} = 2 \cos \theta \)

We know that:

\( x^2 + 2x \cos \theta + 1 = 0 \)

Rearranging the terms, we get:

\( x = \cos \theta + i \sin \theta \)

Then, the value of \(x^n + \frac{1}{x^n} \) is:

\( x^n = (\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta \)

\( \frac{1}{x^n} = \cos n\theta - i \sin n\theta \)

Thus,

\( x^n + \frac{1}{x^n} = 2 \cos n \theta \)

Hence, the correct answer is Option 1

Calculation:

Given,

\( \frac{1 + \cos \phi + i \sin \phi}{1 + \cos \phi - i \sin \phi} \)

We need to simplify this expression:

\( \frac{1 + \cos \phi + i \sin \phi}{1 + \cos \phi - i \sin \phi} \times \frac{1 + \cos \phi + i \sin \phi}{1 + \cos \phi + i \sin \phi} = \frac{(1 + \cos \phi)^2 + \sin^2 \phi + 2i(1 + \cos \phi) \sin \phi}{(1 + \cos \phi)^2 + \sin^2 \phi} \)

By simplifying the denominator:

\( (1 + \cos \phi)^2 + \sin^2 \phi = 2(1 + \cos \phi) = 2A \)

Now, the numerator becomes:

\( 2A \cos \phi + 2i A \sin \phi \)

Thus the expression simplifies to:

\( \frac{A \cos \phi + i A \sin \phi}{A} = e^{i \phi} \)

Now, we raise this expression to the power \(n\):

\( e^{i \phi} \)

By using De Moivre's Theorem:

\( (e^{i \phi})^n = e^{in \phi} = \cos(n \phi) + i \sin(n \phi) \)

Thus, the real part\(u = \cos(n \phi) \)

Hence, the correct answer is Option (b).