De Moivre's Theorem MCQ Quiz in বাংলা - Objective Question with Answer for De Moivre's Theorem - বিনামূল্যে ডাউনলোড করুন [PDF]

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