QBManager

Complex Numbers and Quadratic Equation

117739 The locus of the point $z = x +$ iy satisfying $| \frac{z - 2 i}{z + 2 i} | = 1$ is

1

x

-axis

2

y

-axis

3

y = 2

4

x = 2

Explanation:

A Given,
$| \frac{z - 2 i}{z + 2 i} | = 1$
$| \frac{z - 2 i}{z + 2 i} | = 1$
$| \frac{x + i (y - 2) ∣}{| x + i (y + 2) |} | = 1$
$x^{2} + (y - 2)^{2} = x^{2} + (y + 2)^{2}$
$(y - 2)^{2} = (y + 2)^{2}$
$y^{2} + 4 - 4 y = y^{2} + 4 + 4 y$
$8 y = 0$
$y = 0, which is x -axis$ OrOr

AP EAMCET-2007

Complex Numbers and Quadratic Equation

117740 For $z \neq i$ , define $ω = \frac{z + i}{z - i}$ , then $| ω | < 1$ implies that

1

Re z > 0

2

Re z = 0

3

Im z = 0

4

Im z < 0

Explanation:

D Given,
$ω = \frac{z + i}{z - i} then | ω | < 1$
$= \frac{| z + i |}{| z - i |} < 1$
$= | \frac{x + i y + i}{x + i y - i} | < 1$
$= | x + i (y + 1) | < | x + i (y - 1) |$
$= x^{2} + (y + 1)^{2} < x^{2} + (y - 1)^{2}$
$= x^{2} + y^{2} + 1 + 2 y < x^{2} + y^{2} + 1 - 2 y$
$= 4 y < 0$
$= y < 0, \Rightarrow Im z < 0$

AMU-2004

Complex Numbers and Quadratic Equation

117741 If a complex number $z$ satisfies $| z^{2} - 1 | = | z |^{2} + 1$ , then $z$ lies on

1 The real axis

2 The imaginary axis

3

y = x

4 a circle

Explanation:

B Given,
$| z^{2} - 1 | = | z |^{2} + 1$
Put,
$z = x + i y$
$| x^{2} + i^{2} y^{2} + 2 x y i - 1 | = x^{2} + y^{2} + 1$
$| (x^{2} - y^{2} - 1) + 2 x y i | = x^{2} + y^{2} + 1$
$\sqrt{{(x^{2} - y^{2} - 1)}^{2} + 4 x^{2} y^{2}} = (x^{2} + y^{2} + 1)$
${(x^{2} - y^{2} - 1)}^{2} + 4 x^{2} y^{2} = {(x^{2} + y^{2} + 1)}^{2}$ Put $x = 0$ , $y$ has some real value which satisfy the equation.
$∴ z$ lies on imaginary axis.

AP EAMCET-2013

Complex Numbers and Quadratic Equation

117742 If a is a complex number and $b$ is a real number then the equation $\bar{a} + a + b = 0$ represents a

1 straight line

2 parabola

3 circle

4 hyperbola

Explanation:

A Given a is complex number and $b$ is real number.
Let, $a = x + i y$
Then, $\bar{a} + a + b = x - iy + x + iy + b = 0$
$2 x + b = 0$
$x = \frac{- b}{2}$ It is a straight line.

AP EAMCET-2001

Complex Numbers and Quadratic Equation

117743 $\frac{{(\sin \frac{π}{8} + i \cos \frac{π}{8})}^{8}}{{(\sin \frac{π}{8} - i \cos \frac{π}{8})}^{8}}$

1 -1

2 0

3 1

4

2 i

Explanation:

A Given,
$\frac{{(\sin \frac{π}{8} + i \cos \frac{π}{8})}^{8}}{{(\sin \frac{π}{8} - i \cos \frac{π}{8})}^{8}}$
$= \frac{\sin 8 \times \frac{π}{8} + i \cos 8 \times \frac{π}{8}}{\sin 8 \times \frac{π}{8} - i \cos 8 \times \frac{π}{8}}$
$= \frac{\sin π + i \cos π}{\sin π - i \cos π}$
$= \frac{0 + i (- 1)}{0 - i (- 1)} = \frac{- i}{i} = - 1$

EAMCET-1994

Complex Numbers and Quadratic Equation

117739 The locus of the point $z = x +$ iy satisfying $| \frac{z - 2 i}{z + 2 i} | = 1$ is

1

x

-axis

2

y

-axis

3

y = 2

4

x = 2

Explanation:

A Given,
$| \frac{z - 2 i}{z + 2 i} | = 1$
$| \frac{z - 2 i}{z + 2 i} | = 1$
$| \frac{x + i (y - 2) ∣}{| x + i (y + 2) |} | = 1$
$x^{2} + (y - 2)^{2} = x^{2} + (y + 2)^{2}$
$(y - 2)^{2} = (y + 2)^{2}$
$y^{2} + 4 - 4 y = y^{2} + 4 + 4 y$
$8 y = 0$
$y = 0, which is x -axis$ OrOr

AP EAMCET-2007

Complex Numbers and Quadratic Equation

117740 For $z \neq i$ , define $ω = \frac{z + i}{z - i}$ , then $| ω | < 1$ implies that

1

Re z > 0

2

Re z = 0

3

Im z = 0

4

Im z < 0

Explanation:

D Given,
$ω = \frac{z + i}{z - i} then | ω | < 1$
$= \frac{| z + i |}{| z - i |} < 1$
$= | \frac{x + i y + i}{x + i y - i} | < 1$
$= | x + i (y + 1) | < | x + i (y - 1) |$
$= x^{2} + (y + 1)^{2} < x^{2} + (y - 1)^{2}$
$= x^{2} + y^{2} + 1 + 2 y < x^{2} + y^{2} + 1 - 2 y$
$= 4 y < 0$
$= y < 0, \Rightarrow Im z < 0$

AMU-2004

Complex Numbers and Quadratic Equation

117741 If a complex number $z$ satisfies $| z^{2} - 1 | = | z |^{2} + 1$ , then $z$ lies on

1 The real axis

2 The imaginary axis

3

y = x

4 a circle

Explanation:

B Given,
$| z^{2} - 1 | = | z |^{2} + 1$
Put,
$z = x + i y$
$| x^{2} + i^{2} y^{2} + 2 x y i - 1 | = x^{2} + y^{2} + 1$
$| (x^{2} - y^{2} - 1) + 2 x y i | = x^{2} + y^{2} + 1$
$\sqrt{{(x^{2} - y^{2} - 1)}^{2} + 4 x^{2} y^{2}} = (x^{2} + y^{2} + 1)$
${(x^{2} - y^{2} - 1)}^{2} + 4 x^{2} y^{2} = {(x^{2} + y^{2} + 1)}^{2}$ Put $x = 0$ , $y$ has some real value which satisfy the equation.
$∴ z$ lies on imaginary axis.

AP EAMCET-2013

Complex Numbers and Quadratic Equation

117742 If a is a complex number and $b$ is a real number then the equation $\bar{a} + a + b = 0$ represents a

1 straight line

2 parabola

3 circle

4 hyperbola

Explanation:

A Given a is complex number and $b$ is real number.
Let, $a = x + i y$
Then, $\bar{a} + a + b = x - iy + x + iy + b = 0$
$2 x + b = 0$
$x = \frac{- b}{2}$ It is a straight line.

AP EAMCET-2001

Complex Numbers and Quadratic Equation

117743 $\frac{{(\sin \frac{π}{8} + i \cos \frac{π}{8})}^{8}}{{(\sin \frac{π}{8} - i \cos \frac{π}{8})}^{8}}$

1 -1

2 0

3 1

4

2 i

Explanation:

A Given,
$\frac{{(\sin \frac{π}{8} + i \cos \frac{π}{8})}^{8}}{{(\sin \frac{π}{8} - i \cos \frac{π}{8})}^{8}}$
$= \frac{\sin 8 \times \frac{π}{8} + i \cos 8 \times \frac{π}{8}}{\sin 8 \times \frac{π}{8} - i \cos 8 \times \frac{π}{8}}$
$= \frac{\sin π + i \cos π}{\sin π - i \cos π}$
$= \frac{0 + i (- 1)}{0 - i (- 1)} = \frac{- i}{i} = - 1$

EAMCET-1994

Complex Numbers and Quadratic Equation

117739 The locus of the point $z = x +$ iy satisfying $| \frac{z - 2 i}{z + 2 i} | = 1$ is

1

x

-axis

2

y

-axis

3

y = 2

4

x = 2

Explanation:

A Given,
$| \frac{z - 2 i}{z + 2 i} | = 1$
$| \frac{z - 2 i}{z + 2 i} | = 1$
$| \frac{x + i (y - 2) ∣}{| x + i (y + 2) |} | = 1$
$x^{2} + (y - 2)^{2} = x^{2} + (y + 2)^{2}$
$(y - 2)^{2} = (y + 2)^{2}$
$y^{2} + 4 - 4 y = y^{2} + 4 + 4 y$
$8 y = 0$
$y = 0, which is x -axis$ OrOr

AP EAMCET-2007

Complex Numbers and Quadratic Equation

117740 For $z \neq i$ , define $ω = \frac{z + i}{z - i}$ , then $| ω | < 1$ implies that

1

Re z > 0

2

Re z = 0

3

Im z = 0

4

Im z < 0

Explanation:

D Given,
$ω = \frac{z + i}{z - i} then | ω | < 1$
$= \frac{| z + i |}{| z - i |} < 1$
$= | \frac{x + i y + i}{x + i y - i} | < 1$
$= | x + i (y + 1) | < | x + i (y - 1) |$
$= x^{2} + (y + 1)^{2} < x^{2} + (y - 1)^{2}$
$= x^{2} + y^{2} + 1 + 2 y < x^{2} + y^{2} + 1 - 2 y$
$= 4 y < 0$
$= y < 0, \Rightarrow Im z < 0$

AMU-2004

Complex Numbers and Quadratic Equation

117741 If a complex number $z$ satisfies $| z^{2} - 1 | = | z |^{2} + 1$ , then $z$ lies on

1 The real axis

2 The imaginary axis

3

y = x

4 a circle

Explanation:

B Given,
$| z^{2} - 1 | = | z |^{2} + 1$
Put,
$z = x + i y$
$| x^{2} + i^{2} y^{2} + 2 x y i - 1 | = x^{2} + y^{2} + 1$
$| (x^{2} - y^{2} - 1) + 2 x y i | = x^{2} + y^{2} + 1$
$\sqrt{{(x^{2} - y^{2} - 1)}^{2} + 4 x^{2} y^{2}} = (x^{2} + y^{2} + 1)$
${(x^{2} - y^{2} - 1)}^{2} + 4 x^{2} y^{2} = {(x^{2} + y^{2} + 1)}^{2}$ Put $x = 0$ , $y$ has some real value which satisfy the equation.
$∴ z$ lies on imaginary axis.

AP EAMCET-2013

Complex Numbers and Quadratic Equation

117742 If a is a complex number and $b$ is a real number then the equation $\bar{a} + a + b = 0$ represents a

1 straight line

2 parabola

3 circle

4 hyperbola

Explanation:

A Given a is complex number and $b$ is real number.
Let, $a = x + i y$
Then, $\bar{a} + a + b = x - iy + x + iy + b = 0$
$2 x + b = 0$
$x = \frac{- b}{2}$ It is a straight line.

AP EAMCET-2001

Complex Numbers and Quadratic Equation

117743 $\frac{{(\sin \frac{π}{8} + i \cos \frac{π}{8})}^{8}}{{(\sin \frac{π}{8} - i \cos \frac{π}{8})}^{8}}$

1 -1

2 0

3 1

4

2 i

Explanation:

A Given,
$\frac{{(\sin \frac{π}{8} + i \cos \frac{π}{8})}^{8}}{{(\sin \frac{π}{8} - i \cos \frac{π}{8})}^{8}}$
$= \frac{\sin 8 \times \frac{π}{8} + i \cos 8 \times \frac{π}{8}}{\sin 8 \times \frac{π}{8} - i \cos 8 \times \frac{π}{8}}$
$= \frac{\sin π + i \cos π}{\sin π - i \cos π}$
$= \frac{0 + i (- 1)}{0 - i (- 1)} = \frac{- i}{i} = - 1$

EAMCET-1994

Complex Numbers and Quadratic Equation

117739 The locus of the point $z = x +$ iy satisfying $| \frac{z - 2 i}{z + 2 i} | = 1$ is

1

x

-axis

2

y

-axis

3

y = 2

4

x = 2

Explanation:

A Given,
$| \frac{z - 2 i}{z + 2 i} | = 1$
$| \frac{z - 2 i}{z + 2 i} | = 1$
$| \frac{x + i (y - 2) ∣}{| x + i (y + 2) |} | = 1$
$x^{2} + (y - 2)^{2} = x^{2} + (y + 2)^{2}$
$(y - 2)^{2} = (y + 2)^{2}$
$y^{2} + 4 - 4 y = y^{2} + 4 + 4 y$
$8 y = 0$
$y = 0, which is x -axis$ OrOr

AP EAMCET-2007

Complex Numbers and Quadratic Equation

117740 For $z \neq i$ , define $ω = \frac{z + i}{z - i}$ , then $| ω | < 1$ implies that

1

Re z > 0

2

Re z = 0

3

Im z = 0

4

Im z < 0

Explanation:

D Given,
$ω = \frac{z + i}{z - i} then | ω | < 1$
$= \frac{| z + i |}{| z - i |} < 1$
$= | \frac{x + i y + i}{x + i y - i} | < 1$
$= | x + i (y + 1) | < | x + i (y - 1) |$
$= x^{2} + (y + 1)^{2} < x^{2} + (y - 1)^{2}$
$= x^{2} + y^{2} + 1 + 2 y < x^{2} + y^{2} + 1 - 2 y$
$= 4 y < 0$
$= y < 0, \Rightarrow Im z < 0$

AMU-2004

Complex Numbers and Quadratic Equation

117741 If a complex number $z$ satisfies $| z^{2} - 1 | = | z |^{2} + 1$ , then $z$ lies on

1 The real axis

2 The imaginary axis

3

y = x

4 a circle

Explanation:

B Given,
$| z^{2} - 1 | = | z |^{2} + 1$
Put,
$z = x + i y$
$| x^{2} + i^{2} y^{2} + 2 x y i - 1 | = x^{2} + y^{2} + 1$
$| (x^{2} - y^{2} - 1) + 2 x y i | = x^{2} + y^{2} + 1$
$\sqrt{{(x^{2} - y^{2} - 1)}^{2} + 4 x^{2} y^{2}} = (x^{2} + y^{2} + 1)$
${(x^{2} - y^{2} - 1)}^{2} + 4 x^{2} y^{2} = {(x^{2} + y^{2} + 1)}^{2}$ Put $x = 0$ , $y$ has some real value which satisfy the equation.
$∴ z$ lies on imaginary axis.

AP EAMCET-2013

Complex Numbers and Quadratic Equation

117742 If a is a complex number and $b$ is a real number then the equation $\bar{a} + a + b = 0$ represents a

1 straight line

2 parabola

3 circle

4 hyperbola

Explanation:

A Given a is complex number and $b$ is real number.
Let, $a = x + i y$
Then, $\bar{a} + a + b = x - iy + x + iy + b = 0$
$2 x + b = 0$
$x = \frac{- b}{2}$ It is a straight line.

AP EAMCET-2001

Complex Numbers and Quadratic Equation

117743 $\frac{{(\sin \frac{π}{8} + i \cos \frac{π}{8})}^{8}}{{(\sin \frac{π}{8} - i \cos \frac{π}{8})}^{8}}$

1 -1

2 0

3 1

4

2 i

Explanation:

A Given,
$\frac{{(\sin \frac{π}{8} + i \cos \frac{π}{8})}^{8}}{{(\sin \frac{π}{8} - i \cos \frac{π}{8})}^{8}}$
$= \frac{\sin 8 \times \frac{π}{8} + i \cos 8 \times \frac{π}{8}}{\sin 8 \times \frac{π}{8} - i \cos 8 \times \frac{π}{8}}$
$= \frac{\sin π + i \cos π}{\sin π - i \cos π}$
$= \frac{0 + i (- 1)}{0 - i (- 1)} = \frac{- i}{i} = - 1$

EAMCET-1994

Complex Numbers and Quadratic Equation

117739 The locus of the point $z = x +$ iy satisfying $| \frac{z - 2 i}{z + 2 i} | = 1$ is

1

x

-axis

2

y

-axis

3

y = 2

4

x = 2

Explanation:

A Given,
$| \frac{z - 2 i}{z + 2 i} | = 1$
$| \frac{z - 2 i}{z + 2 i} | = 1$
$| \frac{x + i (y - 2) ∣}{| x + i (y + 2) |} | = 1$
$x^{2} + (y - 2)^{2} = x^{2} + (y + 2)^{2}$
$(y - 2)^{2} = (y + 2)^{2}$
$y^{2} + 4 - 4 y = y^{2} + 4 + 4 y$
$8 y = 0$
$y = 0, which is x -axis$ OrOr

AP EAMCET-2007

Complex Numbers and Quadratic Equation

117740 For $z \neq i$ , define $ω = \frac{z + i}{z - i}$ , then $| ω | < 1$ implies that

1

Re z > 0

2

Re z = 0

3

Im z = 0

4

Im z < 0

Explanation:

D Given,
$ω = \frac{z + i}{z - i} then | ω | < 1$
$= \frac{| z + i |}{| z - i |} < 1$
$= | \frac{x + i y + i}{x + i y - i} | < 1$
$= | x + i (y + 1) | < | x + i (y - 1) |$
$= x^{2} + (y + 1)^{2} < x^{2} + (y - 1)^{2}$
$= x^{2} + y^{2} + 1 + 2 y < x^{2} + y^{2} + 1 - 2 y$
$= 4 y < 0$
$= y < 0, \Rightarrow Im z < 0$

AMU-2004

Complex Numbers and Quadratic Equation

117741 If a complex number $z$ satisfies $| z^{2} - 1 | = | z |^{2} + 1$ , then $z$ lies on

1 The real axis

2 The imaginary axis

3

y = x

4 a circle

Explanation:

B Given,
$| z^{2} - 1 | = | z |^{2} + 1$
Put,
$z = x + i y$
$| x^{2} + i^{2} y^{2} + 2 x y i - 1 | = x^{2} + y^{2} + 1$
$| (x^{2} - y^{2} - 1) + 2 x y i | = x^{2} + y^{2} + 1$
$\sqrt{{(x^{2} - y^{2} - 1)}^{2} + 4 x^{2} y^{2}} = (x^{2} + y^{2} + 1)$
${(x^{2} - y^{2} - 1)}^{2} + 4 x^{2} y^{2} = {(x^{2} + y^{2} + 1)}^{2}$ Put $x = 0$ , $y$ has some real value which satisfy the equation.
$∴ z$ lies on imaginary axis.

AP EAMCET-2013

Complex Numbers and Quadratic Equation

117742 If a is a complex number and $b$ is a real number then the equation $\bar{a} + a + b = 0$ represents a

1 straight line

2 parabola

3 circle

4 hyperbola

Explanation:

A Given a is complex number and $b$ is real number.
Let, $a = x + i y$
Then, $\bar{a} + a + b = x - iy + x + iy + b = 0$
$2 x + b = 0$
$x = \frac{- b}{2}$ It is a straight line.

AP EAMCET-2001

Complex Numbers and Quadratic Equation

117743 $\frac{{(\sin \frac{π}{8} + i \cos \frac{π}{8})}^{8}}{{(\sin \frac{π}{8} - i \cos \frac{π}{8})}^{8}}$

1 -1

2 0

3 1

4

2 i

Explanation:

A Given,
$\frac{{(\sin \frac{π}{8} + i \cos \frac{π}{8})}^{8}}{{(\sin \frac{π}{8} - i \cos \frac{π}{8})}^{8}}$
$= \frac{\sin 8 \times \frac{π}{8} + i \cos 8 \times \frac{π}{8}}{\sin 8 \times \frac{π}{8} - i \cos 8 \times \frac{π}{8}}$
$= \frac{\sin π + i \cos π}{\sin π - i \cos π}$
$= \frac{0 + i (- 1)}{0 - i (- 1)} = \frac{- i}{i} = - 1$

EAMCET-1994

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