QBManager

Complex Numbers and Quadratic Equation

118002 Which of the following is a fourth root of $\frac{1}{2} + \frac{i \sqrt{3}}{2}$ ?

1

\cos \frac{π}{12}

2

\cos \frac{π}{2}

3

\cos \frac{π}{6}

4

\cos \frac{π}{3}

Explanation:

Exp: (A): We have,
$(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = \cos \frac{π}{3} + i \sin \frac{π}{3}$
$[∵ \cos θ + i \sin θ = e^{i θ} \to DeMoivre theorem]$
$= e^{i π / 3}$
Fourth root $= {(e^{i π / 3})}^{1 / 4}$
$= e^{i π / 12}$
$= \cos \frac{π}{12} + i \sin \frac{π}{12}$
$= \cos \frac{π}{12}$

APEAPCET-20.08.2021

Complex Numbers and Quadratic Equation

117989 The value of $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1) π}{15} + i \sin \frac{(2 k + 1) π}{15}}$ is

1 0

2 -1

3 1

4 i
[SRM JEE-2007]

Explanation:

C We have,
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1) π}{15} + i \sin \frac{(2 k + 1) π}{15}}$
$= 1 + \sum_{k = 0}^{14} e^{i (2 k + 1) π / 15}$
$= 1 + e^{i π / 15} + e^{i 3 π / 15} + \dots \dots + e^{i 29 π / 15}$
$= 1 + e^{i π / 15} {\frac{1 - {(e^{i 2 π / 15})}^{15}}{1 - e^{i 2 π / 15}}}$
$= 1 + e^{i π / 15} {\frac{1 - e^{i 2 π}}{1 - e^{i 2 π / 15}}} (∵ e^{i 2 π} = 1)$
$= 1 + 0 = 1$

Complex Numbers and Quadratic Equation

117990 If ${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = 3^{25} (x + i y)$ , where $x$ and $y$ are real, then the ordered pair $(x, y)$ is given by

1

(0, 3)

2

(1 / 2, \sqrt{3} / 2)

3

(- 3, 0)

4

(0, - 3)

Explanation:

(B) Given, ${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = 3^{25} (x + i y)$
${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = (\sqrt{3})^{50} (x + i y)$
${(\frac{3}{2 \sqrt{3}} + \frac{i \sqrt{3}}{2 \sqrt{3}})}^{50} = x + i y$
${(\frac{\sqrt{3}}{2} + i \frac{1}{2})}^{50} = x + i y$
${(\cos \frac{π}{6} + i \sin \frac{π}{6})}^{50} = x + i y$
$\cos \frac{50 π}{6} + i \sin \frac{50 π}{6} = x + i y$
$\cos \frac{25 π}{3} + i \sin \frac{25 π}{3} = x + i y$
$\cos (8 π + \frac{π}{3}) + i \sin (8 π + \frac{π}{3}) = x + i y$
$\cos \frac{π}{3} + i \sin (\frac{π}{3}) = x + i y$
$\frac{1}{2} + \frac{i \sqrt{3}}{2} = x + i y$
$x = \frac{1}{2}, y = \frac{\sqrt{3}}{2}$

SRM JEEE-2008

Complex Numbers and Quadratic Equation

117991 The least positive integer n. for which $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is positive is

1 4

2 3

3 2

4 1

Explanation:

D $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is positive for what value of $n$
${(\frac{1 + i}{1 - i})}^{n} \times (1 - i)^{2} = {[\frac{(1 + i)^{2}}{(1^{2} - i^{2})}]}^{n} \times (1 - i)^{2}$
$= \frac{(1 - 1 + 2 i)^{n}}{2^{n}} \times (- 2 i)$
$= - \frac{(2 i)^{n + 1}}{2^{n}} = \frac{2^{n + 1}}{2^{n}} [- 1 (i)^{n + 1}]$
$= - 2 (i)^{n + 1} = 2$
For least positive integer, $n + 1 = 2$
$n = 1$

Karnataka CET-2010

Complex Numbers and Quadratic Equation

117992 If $x + i y = (- 1 + i \sqrt{3})^{2010}$ , then $x =$

1

2^{2010}

2

- 2^{2010}

3 -1

4 1

Explanation:

(A) $Given,$
$x + i y = (- 1 + i \sqrt{3})^{2010}$
$= {[\frac{2 (- 1 + i \sqrt{3})}{2}]}^{2010}$
$= 2^{2010} {[\frac{- 1 + i \sqrt{3}}{2}]}^{2010}$
$x + iy = 2^{2010} ω^{2010} [∵ ω = \frac{- 1 + i \sqrt{3}}{2}]$
$= 2^{2010} \times {(ω^{3})}^{670}$
$= 2^{2010} \times (1)^{670}$
$x + iy = 2^{2010} + i \times 0$
$So, = 2^{2010}, y = 0$
So, $x = 2^{2010}, y = 0$

Karnataka CET-2010

Complex Numbers and Quadratic Equation

118002 Which of the following is a fourth root of $\frac{1}{2} + \frac{i \sqrt{3}}{2}$ ?

1

\cos \frac{π}{12}

2

\cos \frac{π}{2}

3

\cos \frac{π}{6}

4

\cos \frac{π}{3}

Explanation:

Exp: (A): We have,
$(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = \cos \frac{π}{3} + i \sin \frac{π}{3}$
$[∵ \cos θ + i \sin θ = e^{i θ} \to DeMoivre theorem]$
$= e^{i π / 3}$
Fourth root $= {(e^{i π / 3})}^{1 / 4}$
$= e^{i π / 12}$
$= \cos \frac{π}{12} + i \sin \frac{π}{12}$
$= \cos \frac{π}{12}$

APEAPCET-20.08.2021

Complex Numbers and Quadratic Equation

117989 The value of $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1) π}{15} + i \sin \frac{(2 k + 1) π}{15}}$ is

1 0

2 -1

3 1

4 i
[SRM JEE-2007]

Explanation:

C We have,
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1) π}{15} + i \sin \frac{(2 k + 1) π}{15}}$
$= 1 + \sum_{k = 0}^{14} e^{i (2 k + 1) π / 15}$
$= 1 + e^{i π / 15} + e^{i 3 π / 15} + \dots \dots + e^{i 29 π / 15}$
$= 1 + e^{i π / 15} {\frac{1 - {(e^{i 2 π / 15})}^{15}}{1 - e^{i 2 π / 15}}}$
$= 1 + e^{i π / 15} {\frac{1 - e^{i 2 π}}{1 - e^{i 2 π / 15}}} (∵ e^{i 2 π} = 1)$
$= 1 + 0 = 1$

Complex Numbers and Quadratic Equation

117990 If ${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = 3^{25} (x + i y)$ , where $x$ and $y$ are real, then the ordered pair $(x, y)$ is given by

1

(0, 3)

2

(1 / 2, \sqrt{3} / 2)

3

(- 3, 0)

4

(0, - 3)

Explanation:

(B) Given, ${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = 3^{25} (x + i y)$
${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = (\sqrt{3})^{50} (x + i y)$
${(\frac{3}{2 \sqrt{3}} + \frac{i \sqrt{3}}{2 \sqrt{3}})}^{50} = x + i y$
${(\frac{\sqrt{3}}{2} + i \frac{1}{2})}^{50} = x + i y$
${(\cos \frac{π}{6} + i \sin \frac{π}{6})}^{50} = x + i y$
$\cos \frac{50 π}{6} + i \sin \frac{50 π}{6} = x + i y$
$\cos \frac{25 π}{3} + i \sin \frac{25 π}{3} = x + i y$
$\cos (8 π + \frac{π}{3}) + i \sin (8 π + \frac{π}{3}) = x + i y$
$\cos \frac{π}{3} + i \sin (\frac{π}{3}) = x + i y$
$\frac{1}{2} + \frac{i \sqrt{3}}{2} = x + i y$
$x = \frac{1}{2}, y = \frac{\sqrt{3}}{2}$

SRM JEEE-2008

Complex Numbers and Quadratic Equation

117991 The least positive integer n. for which $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is positive is

1 4

2 3

3 2

4 1

Explanation:

D $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is positive for what value of $n$
${(\frac{1 + i}{1 - i})}^{n} \times (1 - i)^{2} = {[\frac{(1 + i)^{2}}{(1^{2} - i^{2})}]}^{n} \times (1 - i)^{2}$
$= \frac{(1 - 1 + 2 i)^{n}}{2^{n}} \times (- 2 i)$
$= - \frac{(2 i)^{n + 1}}{2^{n}} = \frac{2^{n + 1}}{2^{n}} [- 1 (i)^{n + 1}]$
$= - 2 (i)^{n + 1} = 2$
For least positive integer, $n + 1 = 2$
$n = 1$

Karnataka CET-2010

Complex Numbers and Quadratic Equation

117992 If $x + i y = (- 1 + i \sqrt{3})^{2010}$ , then $x =$

1

2^{2010}

2

- 2^{2010}

3 -1

4 1

Explanation:

(A) $Given,$
$x + i y = (- 1 + i \sqrt{3})^{2010}$
$= {[\frac{2 (- 1 + i \sqrt{3})}{2}]}^{2010}$
$= 2^{2010} {[\frac{- 1 + i \sqrt{3}}{2}]}^{2010}$
$x + iy = 2^{2010} ω^{2010} [∵ ω = \frac{- 1 + i \sqrt{3}}{2}]$
$= 2^{2010} \times {(ω^{3})}^{670}$
$= 2^{2010} \times (1)^{670}$
$x + iy = 2^{2010} + i \times 0$
$So, = 2^{2010}, y = 0$
So, $x = 2^{2010}, y = 0$

Karnataka CET-2010

Complex Numbers and Quadratic Equation

118002 Which of the following is a fourth root of $\frac{1}{2} + \frac{i \sqrt{3}}{2}$ ?

1

\cos \frac{π}{12}

2

\cos \frac{π}{2}

3

\cos \frac{π}{6}

4

\cos \frac{π}{3}

Explanation:

Exp: (A): We have,
$(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = \cos \frac{π}{3} + i \sin \frac{π}{3}$
$[∵ \cos θ + i \sin θ = e^{i θ} \to DeMoivre theorem]$
$= e^{i π / 3}$
Fourth root $= {(e^{i π / 3})}^{1 / 4}$
$= e^{i π / 12}$
$= \cos \frac{π}{12} + i \sin \frac{π}{12}$
$= \cos \frac{π}{12}$

APEAPCET-20.08.2021

Complex Numbers and Quadratic Equation

117989 The value of $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1) π}{15} + i \sin \frac{(2 k + 1) π}{15}}$ is

1 0

2 -1

3 1

4 i
[SRM JEE-2007]

Explanation:

C We have,
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1) π}{15} + i \sin \frac{(2 k + 1) π}{15}}$
$= 1 + \sum_{k = 0}^{14} e^{i (2 k + 1) π / 15}$
$= 1 + e^{i π / 15} + e^{i 3 π / 15} + \dots \dots + e^{i 29 π / 15}$
$= 1 + e^{i π / 15} {\frac{1 - {(e^{i 2 π / 15})}^{15}}{1 - e^{i 2 π / 15}}}$
$= 1 + e^{i π / 15} {\frac{1 - e^{i 2 π}}{1 - e^{i 2 π / 15}}} (∵ e^{i 2 π} = 1)$
$= 1 + 0 = 1$

Complex Numbers and Quadratic Equation

117990 If ${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = 3^{25} (x + i y)$ , where $x$ and $y$ are real, then the ordered pair $(x, y)$ is given by

1

(0, 3)

2

(1 / 2, \sqrt{3} / 2)

3

(- 3, 0)

4

(0, - 3)

Explanation:

(B) Given, ${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = 3^{25} (x + i y)$
${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = (\sqrt{3})^{50} (x + i y)$
${(\frac{3}{2 \sqrt{3}} + \frac{i \sqrt{3}}{2 \sqrt{3}})}^{50} = x + i y$
${(\frac{\sqrt{3}}{2} + i \frac{1}{2})}^{50} = x + i y$
${(\cos \frac{π}{6} + i \sin \frac{π}{6})}^{50} = x + i y$
$\cos \frac{50 π}{6} + i \sin \frac{50 π}{6} = x + i y$
$\cos \frac{25 π}{3} + i \sin \frac{25 π}{3} = x + i y$
$\cos (8 π + \frac{π}{3}) + i \sin (8 π + \frac{π}{3}) = x + i y$
$\cos \frac{π}{3} + i \sin (\frac{π}{3}) = x + i y$
$\frac{1}{2} + \frac{i \sqrt{3}}{2} = x + i y$
$x = \frac{1}{2}, y = \frac{\sqrt{3}}{2}$

SRM JEEE-2008

Complex Numbers and Quadratic Equation

117991 The least positive integer n. for which $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is positive is

1 4

2 3

3 2

4 1

Explanation:

D $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is positive for what value of $n$
${(\frac{1 + i}{1 - i})}^{n} \times (1 - i)^{2} = {[\frac{(1 + i)^{2}}{(1^{2} - i^{2})}]}^{n} \times (1 - i)^{2}$
$= \frac{(1 - 1 + 2 i)^{n}}{2^{n}} \times (- 2 i)$
$= - \frac{(2 i)^{n + 1}}{2^{n}} = \frac{2^{n + 1}}{2^{n}} [- 1 (i)^{n + 1}]$
$= - 2 (i)^{n + 1} = 2$
For least positive integer, $n + 1 = 2$
$n = 1$

Karnataka CET-2010

Complex Numbers and Quadratic Equation

117992 If $x + i y = (- 1 + i \sqrt{3})^{2010}$ , then $x =$

1

2^{2010}

2

- 2^{2010}

3 -1

4 1

Explanation:

(A) $Given,$
$x + i y = (- 1 + i \sqrt{3})^{2010}$
$= {[\frac{2 (- 1 + i \sqrt{3})}{2}]}^{2010}$
$= 2^{2010} {[\frac{- 1 + i \sqrt{3}}{2}]}^{2010}$
$x + iy = 2^{2010} ω^{2010} [∵ ω = \frac{- 1 + i \sqrt{3}}{2}]$
$= 2^{2010} \times {(ω^{3})}^{670}$
$= 2^{2010} \times (1)^{670}$
$x + iy = 2^{2010} + i \times 0$
$So, = 2^{2010}, y = 0$
So, $x = 2^{2010}, y = 0$

Karnataka CET-2010

Complex Numbers and Quadratic Equation

118002 Which of the following is a fourth root of $\frac{1}{2} + \frac{i \sqrt{3}}{2}$ ?

1

\cos \frac{π}{12}

2

\cos \frac{π}{2}

3

\cos \frac{π}{6}

4

\cos \frac{π}{3}

Explanation:

Exp: (A): We have,
$(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = \cos \frac{π}{3} + i \sin \frac{π}{3}$
$[∵ \cos θ + i \sin θ = e^{i θ} \to DeMoivre theorem]$
$= e^{i π / 3}$
Fourth root $= {(e^{i π / 3})}^{1 / 4}$
$= e^{i π / 12}$
$= \cos \frac{π}{12} + i \sin \frac{π}{12}$
$= \cos \frac{π}{12}$

APEAPCET-20.08.2021

Complex Numbers and Quadratic Equation

117989 The value of $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1) π}{15} + i \sin \frac{(2 k + 1) π}{15}}$ is

1 0

2 -1

3 1

4 i
[SRM JEE-2007]

Explanation:

C We have,
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1) π}{15} + i \sin \frac{(2 k + 1) π}{15}}$
$= 1 + \sum_{k = 0}^{14} e^{i (2 k + 1) π / 15}$
$= 1 + e^{i π / 15} + e^{i 3 π / 15} + \dots \dots + e^{i 29 π / 15}$
$= 1 + e^{i π / 15} {\frac{1 - {(e^{i 2 π / 15})}^{15}}{1 - e^{i 2 π / 15}}}$
$= 1 + e^{i π / 15} {\frac{1 - e^{i 2 π}}{1 - e^{i 2 π / 15}}} (∵ e^{i 2 π} = 1)$
$= 1 + 0 = 1$

Complex Numbers and Quadratic Equation

117990 If ${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = 3^{25} (x + i y)$ , where $x$ and $y$ are real, then the ordered pair $(x, y)$ is given by

1

(0, 3)

2

(1 / 2, \sqrt{3} / 2)

3

(- 3, 0)

4

(0, - 3)

Explanation:

(B) Given, ${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = 3^{25} (x + i y)$
${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = (\sqrt{3})^{50} (x + i y)$
${(\frac{3}{2 \sqrt{3}} + \frac{i \sqrt{3}}{2 \sqrt{3}})}^{50} = x + i y$
${(\frac{\sqrt{3}}{2} + i \frac{1}{2})}^{50} = x + i y$
${(\cos \frac{π}{6} + i \sin \frac{π}{6})}^{50} = x + i y$
$\cos \frac{50 π}{6} + i \sin \frac{50 π}{6} = x + i y$
$\cos \frac{25 π}{3} + i \sin \frac{25 π}{3} = x + i y$
$\cos (8 π + \frac{π}{3}) + i \sin (8 π + \frac{π}{3}) = x + i y$
$\cos \frac{π}{3} + i \sin (\frac{π}{3}) = x + i y$
$\frac{1}{2} + \frac{i \sqrt{3}}{2} = x + i y$
$x = \frac{1}{2}, y = \frac{\sqrt{3}}{2}$

SRM JEEE-2008

Complex Numbers and Quadratic Equation

117991 The least positive integer n. for which $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is positive is

1 4

2 3

3 2

4 1

Explanation:

D $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is positive for what value of $n$
${(\frac{1 + i}{1 - i})}^{n} \times (1 - i)^{2} = {[\frac{(1 + i)^{2}}{(1^{2} - i^{2})}]}^{n} \times (1 - i)^{2}$
$= \frac{(1 - 1 + 2 i)^{n}}{2^{n}} \times (- 2 i)$
$= - \frac{(2 i)^{n + 1}}{2^{n}} = \frac{2^{n + 1}}{2^{n}} [- 1 (i)^{n + 1}]$
$= - 2 (i)^{n + 1} = 2$
For least positive integer, $n + 1 = 2$
$n = 1$

Karnataka CET-2010

Complex Numbers and Quadratic Equation

117992 If $x + i y = (- 1 + i \sqrt{3})^{2010}$ , then $x =$

1

2^{2010}

2

- 2^{2010}

3 -1

4 1

Explanation:

(A) $Given,$
$x + i y = (- 1 + i \sqrt{3})^{2010}$
$= {[\frac{2 (- 1 + i \sqrt{3})}{2}]}^{2010}$
$= 2^{2010} {[\frac{- 1 + i \sqrt{3}}{2}]}^{2010}$
$x + iy = 2^{2010} ω^{2010} [∵ ω = \frac{- 1 + i \sqrt{3}}{2}]$
$= 2^{2010} \times {(ω^{3})}^{670}$
$= 2^{2010} \times (1)^{670}$
$x + iy = 2^{2010} + i \times 0$
$So, = 2^{2010}, y = 0$
So, $x = 2^{2010}, y = 0$

Karnataka CET-2010

Complex Numbers and Quadratic Equation

118002 Which of the following is a fourth root of $\frac{1}{2} + \frac{i \sqrt{3}}{2}$ ?

1

\cos \frac{π}{12}

2

\cos \frac{π}{2}

3

\cos \frac{π}{6}

4

\cos \frac{π}{3}

Explanation:

Exp: (A): We have,
$(\frac{1}{2} + i \frac{\sqrt{3}}{2}) = \cos \frac{π}{3} + i \sin \frac{π}{3}$
$[∵ \cos θ + i \sin θ = e^{i θ} \to DeMoivre theorem]$
$= e^{i π / 3}$
Fourth root $= {(e^{i π / 3})}^{1 / 4}$
$= e^{i π / 12}$
$= \cos \frac{π}{12} + i \sin \frac{π}{12}$
$= \cos \frac{π}{12}$

APEAPCET-20.08.2021

Complex Numbers and Quadratic Equation

117989 The value of $1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1) π}{15} + i \sin \frac{(2 k + 1) π}{15}}$ is

1 0

2 -1

3 1

4 i
[SRM JEE-2007]

Explanation:

C We have,
$1 + \sum_{k = 0}^{14} {\cos \frac{(2 k + 1) π}{15} + i \sin \frac{(2 k + 1) π}{15}}$
$= 1 + \sum_{k = 0}^{14} e^{i (2 k + 1) π / 15}$
$= 1 + e^{i π / 15} + e^{i 3 π / 15} + \dots \dots + e^{i 29 π / 15}$
$= 1 + e^{i π / 15} {\frac{1 - {(e^{i 2 π / 15})}^{15}}{1 - e^{i 2 π / 15}}}$
$= 1 + e^{i π / 15} {\frac{1 - e^{i 2 π}}{1 - e^{i 2 π / 15}}} (∵ e^{i 2 π} = 1)$
$= 1 + 0 = 1$

Complex Numbers and Quadratic Equation

117990 If ${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = 3^{25} (x + i y)$ , where $x$ and $y$ are real, then the ordered pair $(x, y)$ is given by

1

(0, 3)

2

(1 / 2, \sqrt{3} / 2)

3

(- 3, 0)

4

(0, - 3)

Explanation:

(B) Given, ${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = 3^{25} (x + i y)$
${(\frac{3}{2} + \frac{i \sqrt{3}}{2})}^{50} = (\sqrt{3})^{50} (x + i y)$
${(\frac{3}{2 \sqrt{3}} + \frac{i \sqrt{3}}{2 \sqrt{3}})}^{50} = x + i y$
${(\frac{\sqrt{3}}{2} + i \frac{1}{2})}^{50} = x + i y$
${(\cos \frac{π}{6} + i \sin \frac{π}{6})}^{50} = x + i y$
$\cos \frac{50 π}{6} + i \sin \frac{50 π}{6} = x + i y$
$\cos \frac{25 π}{3} + i \sin \frac{25 π}{3} = x + i y$
$\cos (8 π + \frac{π}{3}) + i \sin (8 π + \frac{π}{3}) = x + i y$
$\cos \frac{π}{3} + i \sin (\frac{π}{3}) = x + i y$
$\frac{1}{2} + \frac{i \sqrt{3}}{2} = x + i y$
$x = \frac{1}{2}, y = \frac{\sqrt{3}}{2}$

SRM JEEE-2008

Complex Numbers and Quadratic Equation

117991 The least positive integer n. for which $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is positive is

1 4

2 3

3 2

4 1

Explanation:

D $\frac{(1 + i)^{n}}{(1 - i)^{n - 2}}$ is positive for what value of $n$
${(\frac{1 + i}{1 - i})}^{n} \times (1 - i)^{2} = {[\frac{(1 + i)^{2}}{(1^{2} - i^{2})}]}^{n} \times (1 - i)^{2}$
$= \frac{(1 - 1 + 2 i)^{n}}{2^{n}} \times (- 2 i)$
$= - \frac{(2 i)^{n + 1}}{2^{n}} = \frac{2^{n + 1}}{2^{n}} [- 1 (i)^{n + 1}]$
$= - 2 (i)^{n + 1} = 2$
For least positive integer, $n + 1 = 2$
$n = 1$

Karnataka CET-2010

Complex Numbers and Quadratic Equation

117992 If $x + i y = (- 1 + i \sqrt{3})^{2010}$ , then $x =$

1

2^{2010}

2

- 2^{2010}

3 -1

4 1

Explanation:

(A) $Given,$
$x + i y = (- 1 + i \sqrt{3})^{2010}$
$= {[\frac{2 (- 1 + i \sqrt{3})}{2}]}^{2010}$
$= 2^{2010} {[\frac{- 1 + i \sqrt{3}}{2}]}^{2010}$
$x + iy = 2^{2010} ω^{2010} [∵ ω = \frac{- 1 + i \sqrt{3}}{2}]$
$= 2^{2010} \times {(ω^{3})}^{670}$
$= 2^{2010} \times (1)^{670}$
$x + iy = 2^{2010} + i \times 0$
$So, = 2^{2010}, y = 0$
So, $x = 2^{2010}, y = 0$

Karnataka CET-2010

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