QBManager

Complex Numbers and Quadratic Equation

117491 The imaginary part of $i^{i}$ is

1 0

2 1

3 2

4 -1

Explanation:

A $a + i b = i^{i}$
Taking $\log$ both side -
$\log (a + ib) = \log (i^{i})$
$= i \log (i)$
$= i \log (e^{i π / 2}) {∵ e^{i π / 2} = \cos π / 2 + i \sin π / 2 = i}$
$= i \times i π / 2$
$= i^{2} π / 2$
$\log (a + i b) = - π / 2$
$a + i b = e^{- π / 2}$
On comparing imaginary part is zero.

Karnataka CET-2007

Complex Numbers and Quadratic Equation

117497 Number of solutions of the equation $z^{2} + | z |^{2} = 0$ is

1 1

2 2

3 3

4 infinitely many

Explanation:

D $W e h a v e, z^{2} + | z |^{2} = 0, z \neq 0$
$Let z = x + iy$
$\Rightarrow x^{2} - y^{2} + i 2 xy + x^{2} + y^{2} = 0$
$\Rightarrow 2 x^{2} + i 2 xy = 0 \Rightarrow 2 x (x + iy) = 0$
$\Rightarrow x = 0 or x + iy = 0 (not possible)$
$∴ x = 0 and z \neq 0$
So, y can have any real value.
Hence infinitely many solutions.

COMEDK-2018

Complex Numbers and Quadratic Equation

117499 Evaluate: $i^{141} + i^{142} + i^{143} + i^{144}$

1 -1

2 0

3 1

4 2

Explanation:

B $W e h a v e, i^{141} + i^{142} + i^{143} + i^{144}$
$= i^{140} [i + i^{2} + i^{3} + i^{4}]$
$= {(i^{4})}^{35} [i - 1 - i + 1]$
$= 0$
$= 0$

COMEDK-2020

Complex Numbers and Quadratic Equation

117500 The number of solutions of the equation $x^{2} - 5 | x | + 6 = 0$ is

1 4

2 3

3 2

4 1

Explanation:

A Given equation -
$x^{2} - 5 | x | + 6 = 0$
Then,
$x^{2} - 5 x + 6; x > 0$
$x^{2} + 5 x + 6; x < 0$
From the above equations roots are -
$x = (2, 3) & (- 2, - 3)$ Hence, Number of solution of given equation is4.

COMEDK-2019

Complex Numbers and Quadratic Equation

117491 The imaginary part of $i^{i}$ is

1 0

2 1

3 2

4 -1

Explanation:

A $a + i b = i^{i}$
Taking $\log$ both side -
$\log (a + ib) = \log (i^{i})$
$= i \log (i)$
$= i \log (e^{i π / 2}) {∵ e^{i π / 2} = \cos π / 2 + i \sin π / 2 = i}$
$= i \times i π / 2$
$= i^{2} π / 2$
$\log (a + i b) = - π / 2$
$a + i b = e^{- π / 2}$
On comparing imaginary part is zero.

Karnataka CET-2007

Complex Numbers and Quadratic Equation

117497 Number of solutions of the equation $z^{2} + | z |^{2} = 0$ is

1 1

2 2

3 3

4 infinitely many

Explanation:

D $W e h a v e, z^{2} + | z |^{2} = 0, z \neq 0$
$Let z = x + iy$
$\Rightarrow x^{2} - y^{2} + i 2 xy + x^{2} + y^{2} = 0$
$\Rightarrow 2 x^{2} + i 2 xy = 0 \Rightarrow 2 x (x + iy) = 0$
$\Rightarrow x = 0 or x + iy = 0 (not possible)$
$∴ x = 0 and z \neq 0$
So, y can have any real value.
Hence infinitely many solutions.

COMEDK-2018

Complex Numbers and Quadratic Equation

117499 Evaluate: $i^{141} + i^{142} + i^{143} + i^{144}$

1 -1

2 0

3 1

4 2

Explanation:

B $W e h a v e, i^{141} + i^{142} + i^{143} + i^{144}$
$= i^{140} [i + i^{2} + i^{3} + i^{4}]$
$= {(i^{4})}^{35} [i - 1 - i + 1]$
$= 0$
$= 0$

COMEDK-2020

Complex Numbers and Quadratic Equation

117500 The number of solutions of the equation $x^{2} - 5 | x | + 6 = 0$ is

1 4

2 3

3 2

4 1

Explanation:

A Given equation -
$x^{2} - 5 | x | + 6 = 0$
Then,
$x^{2} - 5 x + 6; x > 0$
$x^{2} + 5 x + 6; x < 0$
From the above equations roots are -
$x = (2, 3) & (- 2, - 3)$ Hence, Number of solution of given equation is4.

COMEDK-2019

Complex Numbers and Quadratic Equation

117491 The imaginary part of $i^{i}$ is

1 0

2 1

3 2

4 -1

Explanation:

A $a + i b = i^{i}$
Taking $\log$ both side -
$\log (a + ib) = \log (i^{i})$
$= i \log (i)$
$= i \log (e^{i π / 2}) {∵ e^{i π / 2} = \cos π / 2 + i \sin π / 2 = i}$
$= i \times i π / 2$
$= i^{2} π / 2$
$\log (a + i b) = - π / 2$
$a + i b = e^{- π / 2}$
On comparing imaginary part is zero.

Karnataka CET-2007

Complex Numbers and Quadratic Equation

117497 Number of solutions of the equation $z^{2} + | z |^{2} = 0$ is

1 1

2 2

3 3

4 infinitely many

Explanation:

D $W e h a v e, z^{2} + | z |^{2} = 0, z \neq 0$
$Let z = x + iy$
$\Rightarrow x^{2} - y^{2} + i 2 xy + x^{2} + y^{2} = 0$
$\Rightarrow 2 x^{2} + i 2 xy = 0 \Rightarrow 2 x (x + iy) = 0$
$\Rightarrow x = 0 or x + iy = 0 (not possible)$
$∴ x = 0 and z \neq 0$
So, y can have any real value.
Hence infinitely many solutions.

COMEDK-2018

Complex Numbers and Quadratic Equation

117499 Evaluate: $i^{141} + i^{142} + i^{143} + i^{144}$

1 -1

2 0

3 1

4 2

Explanation:

B $W e h a v e, i^{141} + i^{142} + i^{143} + i^{144}$
$= i^{140} [i + i^{2} + i^{3} + i^{4}]$
$= {(i^{4})}^{35} [i - 1 - i + 1]$
$= 0$
$= 0$

COMEDK-2020

Complex Numbers and Quadratic Equation

117500 The number of solutions of the equation $x^{2} - 5 | x | + 6 = 0$ is

1 4

2 3

3 2

4 1

Explanation:

A Given equation -
$x^{2} - 5 | x | + 6 = 0$
Then,
$x^{2} - 5 x + 6; x > 0$
$x^{2} + 5 x + 6; x < 0$
From the above equations roots are -
$x = (2, 3) & (- 2, - 3)$ Hence, Number of solution of given equation is4.

COMEDK-2019

Complex Numbers and Quadratic Equation

117491 The imaginary part of $i^{i}$ is

1 0

2 1

3 2

4 -1

Explanation:

A $a + i b = i^{i}$
Taking $\log$ both side -
$\log (a + ib) = \log (i^{i})$
$= i \log (i)$
$= i \log (e^{i π / 2}) {∵ e^{i π / 2} = \cos π / 2 + i \sin π / 2 = i}$
$= i \times i π / 2$
$= i^{2} π / 2$
$\log (a + i b) = - π / 2$
$a + i b = e^{- π / 2}$
On comparing imaginary part is zero.

Karnataka CET-2007

Complex Numbers and Quadratic Equation

117497 Number of solutions of the equation $z^{2} + | z |^{2} = 0$ is

1 1

2 2

3 3

4 infinitely many

Explanation:

D $W e h a v e, z^{2} + | z |^{2} = 0, z \neq 0$
$Let z = x + iy$
$\Rightarrow x^{2} - y^{2} + i 2 xy + x^{2} + y^{2} = 0$
$\Rightarrow 2 x^{2} + i 2 xy = 0 \Rightarrow 2 x (x + iy) = 0$
$\Rightarrow x = 0 or x + iy = 0 (not possible)$
$∴ x = 0 and z \neq 0$
So, y can have any real value.
Hence infinitely many solutions.

COMEDK-2018

Complex Numbers and Quadratic Equation

117499 Evaluate: $i^{141} + i^{142} + i^{143} + i^{144}$

1 -1

2 0

3 1

4 2

Explanation:

B $W e h a v e, i^{141} + i^{142} + i^{143} + i^{144}$
$= i^{140} [i + i^{2} + i^{3} + i^{4}]$
$= {(i^{4})}^{35} [i - 1 - i + 1]$
$= 0$
$= 0$

COMEDK-2020

Complex Numbers and Quadratic Equation

117500 The number of solutions of the equation $x^{2} - 5 | x | + 6 = 0$ is

1 4

2 3

3 2

4 1

Explanation:

A Given equation -
$x^{2} - 5 | x | + 6 = 0$
Then,
$x^{2} - 5 x + 6; x > 0$
$x^{2} + 5 x + 6; x < 0$
From the above equations roots are -
$x = (2, 3) & (- 2, - 3)$ Hence, Number of solution of given equation is4.

COMEDK-2019