QBManager

Complex Numbers and Quadratic Equation

117967 If $x_{r} = \cos \frac{π}{2^{r}} + i \sin \frac{π}{2^{r}}$ , then $x_{1} \cdot x_{2} \cdot x_{3} \dots$ to $\infty =$

1

1 / 2

2 -1

3 0

4 1

Explanation:

B $x_{r} = \cos \frac{π}{2^{r}} + i \sin \frac{π}{2^{r}}$
$x_{r} = e^{i π / 2^{I}}$
Now,
$x_{1} x_{2} x_{3} \dots \dots \infty = e^{i π / 2} e^{i π / 2^{2}} e^{i π / 2^{3}} + \dots \dots \infty$
$= e^{i π [\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots \dots]} = e^{i π [\frac{1 / 2}{1 - 1 / 2}]} = e^{i π}$
$= \cos π + i \sin π = - 1$

COMEDK-2011

Complex Numbers and Quadratic Equation

117968 The continued product of the four values of ${(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{3 / 4}$ is

1 1

2 -1

3 2

4 -2

Explanation:

A ${[{(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{3}]}^{1 / 4} = (\cos π + i \sin π)^{1 / 4}$
$= [\cos (2 k π + π) + i \sin (2 k π + π)]^{1 / 4}, k = 0, 1, 2, 3$
$= \cos (2 k + 1) \frac{π}{4} + i \sin (2 k + 1) \frac{π}{4}, k = 0, 1, 2, 3$
The continued product of the four values is
$\cos (\frac{π}{4} + \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4}) + i \sin (\frac{π}{4} + \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4})$
$= \cos 4 π + i \sin 4 π = 1 + i .0 = 1$

COMEDK-2011

Complex Numbers and Quadratic Equation

117970 What is the principal value of amplitude of $1 -$ i?

1

- π / 4

2

π / 3

3

π / 4

4

- π / 3

Explanation:

A Let $z = x + iy = 1 - i$
We know that, $amp (z) = \tan^{- 1} (\frac{y}{x}) = θ$
then, $amp (z) = \tan^{- 1} (| - 1 |)$
$amp (z) = (\frac{π}{4}) = θ$
$∵$ Given complex number lie in IV quadrant
$∴ amp (z) = - θ$
$= - \frac{π}{4}$

COMEDK-2018

Complex Numbers and Quadratic Equation

117971 The amplitude of $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5})$ is

1

\frac{2 π}{5}

2

\frac{π}{5}

3

\frac{π}{15}

4

\frac{π}{10}

Explanation:

D We have, $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5})$
Here,
$r \cos θ = \sin (\frac{π}{5})$
And,
$r \sin θ = 1 - \cos \frac{π}{5}$
$∴ \tan θ = \frac{1 - \cos (π / 5)}{\sin (π / 5)}$
$= \frac{2 \sin^{2} (π / 10)}{2 \sin (π / 10) \cos (π / 10)}$
$\tan θ = \tan (π / 10)$
$θ = π / 10$

COMEDK-2020

Complex Numbers and Quadratic Equation

117967 If $x_{r} = \cos \frac{π}{2^{r}} + i \sin \frac{π}{2^{r}}$ , then $x_{1} \cdot x_{2} \cdot x_{3} \dots$ to $\infty =$

1

1 / 2

2 -1

3 0

4 1

Explanation:

B $x_{r} = \cos \frac{π}{2^{r}} + i \sin \frac{π}{2^{r}}$
$x_{r} = e^{i π / 2^{I}}$
Now,
$x_{1} x_{2} x_{3} \dots \dots \infty = e^{i π / 2} e^{i π / 2^{2}} e^{i π / 2^{3}} + \dots \dots \infty$
$= e^{i π [\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots \dots]} = e^{i π [\frac{1 / 2}{1 - 1 / 2}]} = e^{i π}$
$= \cos π + i \sin π = - 1$

COMEDK-2011

Complex Numbers and Quadratic Equation

117968 The continued product of the four values of ${(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{3 / 4}$ is

1 1

2 -1

3 2

4 -2

Explanation:

A ${[{(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{3}]}^{1 / 4} = (\cos π + i \sin π)^{1 / 4}$
$= [\cos (2 k π + π) + i \sin (2 k π + π)]^{1 / 4}, k = 0, 1, 2, 3$
$= \cos (2 k + 1) \frac{π}{4} + i \sin (2 k + 1) \frac{π}{4}, k = 0, 1, 2, 3$
The continued product of the four values is
$\cos (\frac{π}{4} + \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4}) + i \sin (\frac{π}{4} + \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4})$
$= \cos 4 π + i \sin 4 π = 1 + i .0 = 1$

COMEDK-2011

Complex Numbers and Quadratic Equation

117970 What is the principal value of amplitude of $1 -$ i?

1

- π / 4

2

π / 3

3

π / 4

4

- π / 3

Explanation:

A Let $z = x + iy = 1 - i$
We know that, $amp (z) = \tan^{- 1} (\frac{y}{x}) = θ$
then, $amp (z) = \tan^{- 1} (| - 1 |)$
$amp (z) = (\frac{π}{4}) = θ$
$∵$ Given complex number lie in IV quadrant
$∴ amp (z) = - θ$
$= - \frac{π}{4}$

COMEDK-2018

Complex Numbers and Quadratic Equation

117971 The amplitude of $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5})$ is

1

\frac{2 π}{5}

2

\frac{π}{5}

3

\frac{π}{15}

4

\frac{π}{10}

Explanation:

D We have, $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5})$
Here,
$r \cos θ = \sin (\frac{π}{5})$
And,
$r \sin θ = 1 - \cos \frac{π}{5}$
$∴ \tan θ = \frac{1 - \cos (π / 5)}{\sin (π / 5)}$
$= \frac{2 \sin^{2} (π / 10)}{2 \sin (π / 10) \cos (π / 10)}$
$\tan θ = \tan (π / 10)$
$θ = π / 10$

COMEDK-2020

Complex Numbers and Quadratic Equation

117967 If $x_{r} = \cos \frac{π}{2^{r}} + i \sin \frac{π}{2^{r}}$ , then $x_{1} \cdot x_{2} \cdot x_{3} \dots$ to $\infty =$

1

1 / 2

2 -1

3 0

4 1

Explanation:

B $x_{r} = \cos \frac{π}{2^{r}} + i \sin \frac{π}{2^{r}}$
$x_{r} = e^{i π / 2^{I}}$
Now,
$x_{1} x_{2} x_{3} \dots \dots \infty = e^{i π / 2} e^{i π / 2^{2}} e^{i π / 2^{3}} + \dots \dots \infty$
$= e^{i π [\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots \dots]} = e^{i π [\frac{1 / 2}{1 - 1 / 2}]} = e^{i π}$
$= \cos π + i \sin π = - 1$

COMEDK-2011

Complex Numbers and Quadratic Equation

117968 The continued product of the four values of ${(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{3 / 4}$ is

1 1

2 -1

3 2

4 -2

Explanation:

A ${[{(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{3}]}^{1 / 4} = (\cos π + i \sin π)^{1 / 4}$
$= [\cos (2 k π + π) + i \sin (2 k π + π)]^{1 / 4}, k = 0, 1, 2, 3$
$= \cos (2 k + 1) \frac{π}{4} + i \sin (2 k + 1) \frac{π}{4}, k = 0, 1, 2, 3$
The continued product of the four values is
$\cos (\frac{π}{4} + \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4}) + i \sin (\frac{π}{4} + \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4})$
$= \cos 4 π + i \sin 4 π = 1 + i .0 = 1$

COMEDK-2011

Complex Numbers and Quadratic Equation

117970 What is the principal value of amplitude of $1 -$ i?

1

- π / 4

2

π / 3

3

π / 4

4

- π / 3

Explanation:

A Let $z = x + iy = 1 - i$
We know that, $amp (z) = \tan^{- 1} (\frac{y}{x}) = θ$
then, $amp (z) = \tan^{- 1} (| - 1 |)$
$amp (z) = (\frac{π}{4}) = θ$
$∵$ Given complex number lie in IV quadrant
$∴ amp (z) = - θ$
$= - \frac{π}{4}$

COMEDK-2018

Complex Numbers and Quadratic Equation

117971 The amplitude of $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5})$ is

1

\frac{2 π}{5}

2

\frac{π}{5}

3

\frac{π}{15}

4

\frac{π}{10}

Explanation:

D We have, $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5})$
Here,
$r \cos θ = \sin (\frac{π}{5})$
And,
$r \sin θ = 1 - \cos \frac{π}{5}$
$∴ \tan θ = \frac{1 - \cos (π / 5)}{\sin (π / 5)}$
$= \frac{2 \sin^{2} (π / 10)}{2 \sin (π / 10) \cos (π / 10)}$
$\tan θ = \tan (π / 10)$
$θ = π / 10$

COMEDK-2020

Complex Numbers and Quadratic Equation

117967 If $x_{r} = \cos \frac{π}{2^{r}} + i \sin \frac{π}{2^{r}}$ , then $x_{1} \cdot x_{2} \cdot x_{3} \dots$ to $\infty =$

1

1 / 2

2 -1

3 0

4 1

Explanation:

B $x_{r} = \cos \frac{π}{2^{r}} + i \sin \frac{π}{2^{r}}$
$x_{r} = e^{i π / 2^{I}}$
Now,
$x_{1} x_{2} x_{3} \dots \dots \infty = e^{i π / 2} e^{i π / 2^{2}} e^{i π / 2^{3}} + \dots \dots \infty$
$= e^{i π [\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + \dots \dots]} = e^{i π [\frac{1 / 2}{1 - 1 / 2}]} = e^{i π}$
$= \cos π + i \sin π = - 1$

COMEDK-2011

Complex Numbers and Quadratic Equation

117968 The continued product of the four values of ${(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{3 / 4}$ is

1 1

2 -1

3 2

4 -2

Explanation:

A ${[{(\cos \frac{π}{3} + i \sin \frac{π}{3})}^{3}]}^{1 / 4} = (\cos π + i \sin π)^{1 / 4}$
$= [\cos (2 k π + π) + i \sin (2 k π + π)]^{1 / 4}, k = 0, 1, 2, 3$
$= \cos (2 k + 1) \frac{π}{4} + i \sin (2 k + 1) \frac{π}{4}, k = 0, 1, 2, 3$
The continued product of the four values is
$\cos (\frac{π}{4} + \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4}) + i \sin (\frac{π}{4} + \frac{3 π}{4} + \frac{5 π}{4} + \frac{7 π}{4})$
$= \cos 4 π + i \sin 4 π = 1 + i .0 = 1$

COMEDK-2011

Complex Numbers and Quadratic Equation

117970 What is the principal value of amplitude of $1 -$ i?

1

- π / 4

2

π / 3

3

π / 4

4

- π / 3

Explanation:

A Let $z = x + iy = 1 - i$
We know that, $amp (z) = \tan^{- 1} (\frac{y}{x}) = θ$
then, $amp (z) = \tan^{- 1} (| - 1 |)$
$amp (z) = (\frac{π}{4}) = θ$
$∵$ Given complex number lie in IV quadrant
$∴ amp (z) = - θ$
$= - \frac{π}{4}$

COMEDK-2018

Complex Numbers and Quadratic Equation

117971 The amplitude of $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5})$ is

1

\frac{2 π}{5}

2

\frac{π}{5}

3

\frac{π}{15}

4

\frac{π}{10}

Explanation:

D We have, $\sin \frac{π}{5} + i (1 - \cos \frac{π}{5})$
Here,
$r \cos θ = \sin (\frac{π}{5})$
And,
$r \sin θ = 1 - \cos \frac{π}{5}$
$∴ \tan θ = \frac{1 - \cos (π / 5)}{\sin (π / 5)}$
$= \frac{2 \sin^{2} (π / 10)}{2 \sin (π / 10) \cos (π / 10)}$
$\tan θ = \tan (π / 10)$
$θ = π / 10$

COMEDK-2020

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