QBManager

Complex Numbers and Quadratic Equation

117480 If $\frac{3}{2 + \cos θ + i \sin θ} = a + i b$ then $[(a - b)^{2} + b^{2}]$ is equal to

1 0

2 1

3 -1

4 2

Explanation:

Exp: (B):
Given, $\frac{3}{2 + \cos θ + i \sin θ} = a + i b$
Multiplying denominator and numerator by $2 + \cos θ - i$ $\sin θ$
$\frac{3}{2 + \cos θ + i \sin θ} \times \frac{2 + \cos θ - i \sin θ}{2 + \cos θ - i \sin θ} = a + i b$
$\frac{6 + 3 \cos θ - 3 i \sin θ}{(2 + \cos θ)^{2} - (i \sin θ)^{2}} = a + i b$
$\frac{6 + 3 \cos θ - 3 i \sin θ}{5 + 4 \cos θ} = a + i b$
$\frac{6 + 3 \cos θ}{5 + 4 \cos θ} - \frac{3 i \sin θ}{5 + 4 \cos θ} = a + i b$
equating real and imaginary part and we get
$a = \frac{6 + 3 \cos θ}{5 + 4 \cos θ}, b = \frac{- 3 \sin θ}{5 + 4 \cos θ}$
Now value of $[(a - 2)^{2} + b^{2}]$ is
$= {(\frac{6 + 3 \cos θ}{5 + 4 \cos θ} - 2)}^{2} + {(\frac{- 3 \sin θ}{5 + 4 \cos θ})}^{2}$
$= {(\frac{6 + 3 \cos θ - 10 - 8 \cos θ}{5 + 4 \cos θ})}^{2} + {(\frac{- 3 \sin θ}{5 + 4 \cos θ})}^{2}$
$= {\frac{(- 4 - 5 \cos θ)^{2}}{(5 + 4 \cos θ)^{2}} + \frac{9 \sin^{2} θ}{(5 + 4 \cos θ)^{2}}}$
$= \frac{16 + 25 \cos^{2} θ + 40 \cos θ + 9 \sin^{2} θ}{(5 + 4 \cos θ)^{2}}$
$= \frac{16 + 25 \cos^{2} θ + 40 \cos θ 9 (1 - \cos^{2} θ)}{{(5 + 4 \cos^{2} θ)}^{2}}$
$= \frac{(5 + 4 \cos θ)^{2}}{(5 + 4 \cos θ)^{2}} = 1$

BCECE-2009

Complex Numbers and Quadratic Equation

117483 The real value of $α$ for which the expression $\frac{1 - i \sin α}{1 + 2 i \sin α}$ is purely real, is :

1

(2 n + 1) \frac{π}{2}

2

(n + 1) \frac{π}{2}

3

n π

4 none of these

Explanation:

C Let complex number $(z) = \frac{1 - i \sin α}{1 + 2 i \sin α}$
Multiplying denominator \& Numerator by $1 - 2$ i sin $α$ Now,
$z = \frac{1 - i \sin α}{1 + 2 i \sin α} \times \frac{1 - 2 i \sin α}{1 - 2 i \sin α}$
$z = \frac{1 - 2 i \sin α - i \sin α + 2 i^{2} \sin^{2} α}{1 - 4 i^{2} \sin^{2} α}$
$z = \frac{1 - 2 \sin^{2} α - 3 i \sin α}{1 + 4 \sin^{2} α} {∵ i^{2} = - 1}$
$z = \frac{1 - 2 \sin^{2} α}{1 + 4 \sin^{2} α} - \frac{i 3 \sin α}{1 + 4 \sin^{2} α}$
$∵ z$ is purely real
$\frac{3 \sin α}{1 + 4 \sin^{2} α} = 0$
$3 \sin α = 0$
$\sin α = 0$
$α = 0, π, 2 π \dots \dots$
$α = n π where (n \in N)$

BCECE-2005

Complex Numbers and Quadratic Equation

117484 If the imaginary part of $\frac{2 z + 1}{i z + 1}$ is -2 , then the locus of the point represented by $z$ is a:

1 circle

2 straight line

3 parabola

4 none of these

Explanation:

(B) Let complex number $(z) = x + iy$
Thus,
$\frac{2 z + 1}{i z + 1} = \frac{2 (x + i y) + 1}{i (x + i y) + 1}$
$= \frac{2 x + 2 i y + 1}{i x + i^{2} y + 1} = \frac{2 x + 1 + 2 i y}{(1 - y) + i x}$
Multiplying denominator \& Numerator by $(1 - y) - ix$
$= \frac{2 x + 1 + 2 i y}{(1 - y) + i x} \times \frac{(1 - y) - i x}{(1 - y) - i x}$
$= \frac{2 x - 2 x y - 2 i x^{2} + 1 - y - i x + 2 i y - 2 i y^{2} - 2 i^{2} x y}{(1 - y)^{2} + x^{2}}$
$= \frac{2 x - 2 x y + 1 - y + 2 x y - 2 i x^{2} - i x + 2 i y - 2 i y^{2}}{(1 - y)^{2} + x^{2}}$
$\frac{(2 x - y + 1)}{(1 - y)^{2} + x^{2}} + \frac{i (- 2 x^{2} - x + 2 y - 2 y^{2})}{(1 - y)^{2} + x^{2}}$
$∵$ Given imaginary part is -2
$∴ \frac{- 2 x^{2} - x + 2 y - 2 y^{2}}{(1 - y)^{2} + x^{2}} = - 2$
$- 2 x^{2} - x + 2 y - 2 y^{2} = - 2 [(1 - y)^{2} + x^{2}]$
$- 2 x^{2} - x + 2 y - 2 y^{2} = - 2 - 2 y^{2} + 4 y - 2 x^{2}$
$x + 2 y = 2 {Which is represent a straight line}$

BCECE-2004

Complex Numbers and Quadratic Equation

117485 The values of $x$ and $y$ satisfying the equation
$\frac{(1 + i) x - 2}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i are :$

1

x = - 1, x = 3

2

x = 3, y = - 1

3

x = 0, y = 1

4

x = 1, y = 0

Explanation:

B Given equation
$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$
$\frac{[(1 + i) x] (3 - i) - 2 i (3 - i) + [(2 - 3 i) y] (3 + i) + i (3 + i)}{9 - i^{2}} = i$
$\Rightarrow$
$\frac{3 x + 3 x i - i x - i^{2} x - 6 i + 2 i^{2} + 6 y - 9 i y + 2 i y - 3 i^{2} y + 3 i + i^{2}}{10} =$
$\Rightarrow 3 x + x - 2 + 6 y + 3 y - 1 + 2 xi - 3 i - 7 iy = 10 i$
$4 x + 9 y - 3 + 2 xi - 7 iy - 13 i = 0$
$4 x + 9 y - 3 + i (2 x - 7 y - 13) = 0$
$4 x + 9 y - 3 = 0 .....(i)$
$2 x - 7 y - 13 = 0 .....(ii)$
$23 y + 23 = 0$
$23 y = - 23$
$y = - 1$
$x = 3$
Now,
equation (ii) multiply by 2 and subtracted from equation (i)
$23 y + 23 = 0$
$23 y = - 23$
$y = - 1$ Putting the value of $y$ in equation (1) and we get,So value of $x = 3 & y = - 1$

BCECE-2003

Complex Numbers and Quadratic Equation

117486 The number of solutions of equation $z^{2} + \overset{―}{z} = 0$ , where $z \in C$ are

1 6

2 1

3 4

4 5

Explanation:

C The number of solution of equation
$z^{2} + \bar{z} = 0$
Let, $z = x +$ iy
$(x + i y)^{2} + (x - i y) = 0$
$x^{2} + i^{2} y^{2} + i 2 x y + x - i y = 0$
$x^{2} - y^{2} + x + i 2 x y - i y = 0$
$x^{2} - y^{2} + x + i (2 x y - y) = 0$
On comparing both real \& imaginary parts,
$x^{2} + x - y^{2} = 0$
$2 x y - y = 0$
$y (2 x - 1) = 0$
From equation (ii)
$y (2 x - 1) = 0$
$y = 0$
$x = \frac{1}{2}$
Putting $y = 0$ in equation (i) we get,
$x^{2} + x = 0$
$x (x + 1) = 0 \Rightarrow x = 0, - 1$
Now putting in equation (i) we get,
$\frac{1}{4} + \frac{1}{2} - y^{2} = 0$
$y^{2} = \frac{3}{4}$
$y = \pm \frac{\sqrt{3}}{2}$
Thus, $(x, y) = (0, 0) (- 1, 0), (\frac{1}{2}, \frac{\sqrt{3}}{2}) (\frac{1}{2}, \frac{- \sqrt{3}}{2})$
Hence, the number of solution of given equation is 4 .

Karnataka CET-2012

Complex Numbers and Quadratic Equation

117480 If $\frac{3}{2 + \cos θ + i \sin θ} = a + i b$ then $[(a - b)^{2} + b^{2}]$ is equal to

1 0

2 1

3 -1

4 2

Explanation:

Exp: (B):
Given, $\frac{3}{2 + \cos θ + i \sin θ} = a + i b$
Multiplying denominator and numerator by $2 + \cos θ - i$ $\sin θ$
$\frac{3}{2 + \cos θ + i \sin θ} \times \frac{2 + \cos θ - i \sin θ}{2 + \cos θ - i \sin θ} = a + i b$
$\frac{6 + 3 \cos θ - 3 i \sin θ}{(2 + \cos θ)^{2} - (i \sin θ)^{2}} = a + i b$
$\frac{6 + 3 \cos θ - 3 i \sin θ}{5 + 4 \cos θ} = a + i b$
$\frac{6 + 3 \cos θ}{5 + 4 \cos θ} - \frac{3 i \sin θ}{5 + 4 \cos θ} = a + i b$
equating real and imaginary part and we get
$a = \frac{6 + 3 \cos θ}{5 + 4 \cos θ}, b = \frac{- 3 \sin θ}{5 + 4 \cos θ}$
Now value of $[(a - 2)^{2} + b^{2}]$ is
$= {(\frac{6 + 3 \cos θ}{5 + 4 \cos θ} - 2)}^{2} + {(\frac{- 3 \sin θ}{5 + 4 \cos θ})}^{2}$
$= {(\frac{6 + 3 \cos θ - 10 - 8 \cos θ}{5 + 4 \cos θ})}^{2} + {(\frac{- 3 \sin θ}{5 + 4 \cos θ})}^{2}$
$= {\frac{(- 4 - 5 \cos θ)^{2}}{(5 + 4 \cos θ)^{2}} + \frac{9 \sin^{2} θ}{(5 + 4 \cos θ)^{2}}}$
$= \frac{16 + 25 \cos^{2} θ + 40 \cos θ + 9 \sin^{2} θ}{(5 + 4 \cos θ)^{2}}$
$= \frac{16 + 25 \cos^{2} θ + 40 \cos θ 9 (1 - \cos^{2} θ)}{{(5 + 4 \cos^{2} θ)}^{2}}$
$= \frac{(5 + 4 \cos θ)^{2}}{(5 + 4 \cos θ)^{2}} = 1$

BCECE-2009

Complex Numbers and Quadratic Equation

117483 The real value of $α$ for which the expression $\frac{1 - i \sin α}{1 + 2 i \sin α}$ is purely real, is :

1

(2 n + 1) \frac{π}{2}

2

(n + 1) \frac{π}{2}

3

n π

4 none of these

Explanation:

C Let complex number $(z) = \frac{1 - i \sin α}{1 + 2 i \sin α}$
Multiplying denominator \& Numerator by $1 - 2$ i sin $α$ Now,
$z = \frac{1 - i \sin α}{1 + 2 i \sin α} \times \frac{1 - 2 i \sin α}{1 - 2 i \sin α}$
$z = \frac{1 - 2 i \sin α - i \sin α + 2 i^{2} \sin^{2} α}{1 - 4 i^{2} \sin^{2} α}$
$z = \frac{1 - 2 \sin^{2} α - 3 i \sin α}{1 + 4 \sin^{2} α} {∵ i^{2} = - 1}$
$z = \frac{1 - 2 \sin^{2} α}{1 + 4 \sin^{2} α} - \frac{i 3 \sin α}{1 + 4 \sin^{2} α}$
$∵ z$ is purely real
$\frac{3 \sin α}{1 + 4 \sin^{2} α} = 0$
$3 \sin α = 0$
$\sin α = 0$
$α = 0, π, 2 π \dots \dots$
$α = n π where (n \in N)$

BCECE-2005

Complex Numbers and Quadratic Equation

117484 If the imaginary part of $\frac{2 z + 1}{i z + 1}$ is -2 , then the locus of the point represented by $z$ is a:

1 circle

2 straight line

3 parabola

4 none of these

Explanation:

(B) Let complex number $(z) = x + iy$
Thus,
$\frac{2 z + 1}{i z + 1} = \frac{2 (x + i y) + 1}{i (x + i y) + 1}$
$= \frac{2 x + 2 i y + 1}{i x + i^{2} y + 1} = \frac{2 x + 1 + 2 i y}{(1 - y) + i x}$
Multiplying denominator \& Numerator by $(1 - y) - ix$
$= \frac{2 x + 1 + 2 i y}{(1 - y) + i x} \times \frac{(1 - y) - i x}{(1 - y) - i x}$
$= \frac{2 x - 2 x y - 2 i x^{2} + 1 - y - i x + 2 i y - 2 i y^{2} - 2 i^{2} x y}{(1 - y)^{2} + x^{2}}$
$= \frac{2 x - 2 x y + 1 - y + 2 x y - 2 i x^{2} - i x + 2 i y - 2 i y^{2}}{(1 - y)^{2} + x^{2}}$
$\frac{(2 x - y + 1)}{(1 - y)^{2} + x^{2}} + \frac{i (- 2 x^{2} - x + 2 y - 2 y^{2})}{(1 - y)^{2} + x^{2}}$
$∵$ Given imaginary part is -2
$∴ \frac{- 2 x^{2} - x + 2 y - 2 y^{2}}{(1 - y)^{2} + x^{2}} = - 2$
$- 2 x^{2} - x + 2 y - 2 y^{2} = - 2 [(1 - y)^{2} + x^{2}]$
$- 2 x^{2} - x + 2 y - 2 y^{2} = - 2 - 2 y^{2} + 4 y - 2 x^{2}$
$x + 2 y = 2 {Which is represent a straight line}$

BCECE-2004

Complex Numbers and Quadratic Equation

117485 The values of $x$ and $y$ satisfying the equation
$\frac{(1 + i) x - 2}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i are :$

1

x = - 1, x = 3

2

x = 3, y = - 1

3

x = 0, y = 1

4

x = 1, y = 0

Explanation:

B Given equation
$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$
$\frac{[(1 + i) x] (3 - i) - 2 i (3 - i) + [(2 - 3 i) y] (3 + i) + i (3 + i)}{9 - i^{2}} = i$
$\Rightarrow$
$\frac{3 x + 3 x i - i x - i^{2} x - 6 i + 2 i^{2} + 6 y - 9 i y + 2 i y - 3 i^{2} y + 3 i + i^{2}}{10} =$
$\Rightarrow 3 x + x - 2 + 6 y + 3 y - 1 + 2 xi - 3 i - 7 iy = 10 i$
$4 x + 9 y - 3 + 2 xi - 7 iy - 13 i = 0$
$4 x + 9 y - 3 + i (2 x - 7 y - 13) = 0$
$4 x + 9 y - 3 = 0 .....(i)$
$2 x - 7 y - 13 = 0 .....(ii)$
$23 y + 23 = 0$
$23 y = - 23$
$y = - 1$
$x = 3$
Now,
equation (ii) multiply by 2 and subtracted from equation (i)
$23 y + 23 = 0$
$23 y = - 23$
$y = - 1$ Putting the value of $y$ in equation (1) and we get,So value of $x = 3 & y = - 1$

BCECE-2003

Complex Numbers and Quadratic Equation

117486 The number of solutions of equation $z^{2} + \overset{―}{z} = 0$ , where $z \in C$ are

1 6

2 1

3 4

4 5

Explanation:

C The number of solution of equation
$z^{2} + \bar{z} = 0$
Let, $z = x +$ iy
$(x + i y)^{2} + (x - i y) = 0$
$x^{2} + i^{2} y^{2} + i 2 x y + x - i y = 0$
$x^{2} - y^{2} + x + i 2 x y - i y = 0$
$x^{2} - y^{2} + x + i (2 x y - y) = 0$
On comparing both real \& imaginary parts,
$x^{2} + x - y^{2} = 0$
$2 x y - y = 0$
$y (2 x - 1) = 0$
From equation (ii)
$y (2 x - 1) = 0$
$y = 0$
$x = \frac{1}{2}$
Putting $y = 0$ in equation (i) we get,
$x^{2} + x = 0$
$x (x + 1) = 0 \Rightarrow x = 0, - 1$
Now putting in equation (i) we get,
$\frac{1}{4} + \frac{1}{2} - y^{2} = 0$
$y^{2} = \frac{3}{4}$
$y = \pm \frac{\sqrt{3}}{2}$
Thus, $(x, y) = (0, 0) (- 1, 0), (\frac{1}{2}, \frac{\sqrt{3}}{2}) (\frac{1}{2}, \frac{- \sqrt{3}}{2})$
Hence, the number of solution of given equation is 4 .

Karnataka CET-2012

Complex Numbers and Quadratic Equation

117480 If $\frac{3}{2 + \cos θ + i \sin θ} = a + i b$ then $[(a - b)^{2} + b^{2}]$ is equal to

1 0

2 1

3 -1

4 2

Explanation:

Exp: (B):
Given, $\frac{3}{2 + \cos θ + i \sin θ} = a + i b$
Multiplying denominator and numerator by $2 + \cos θ - i$ $\sin θ$
$\frac{3}{2 + \cos θ + i \sin θ} \times \frac{2 + \cos θ - i \sin θ}{2 + \cos θ - i \sin θ} = a + i b$
$\frac{6 + 3 \cos θ - 3 i \sin θ}{(2 + \cos θ)^{2} - (i \sin θ)^{2}} = a + i b$
$\frac{6 + 3 \cos θ - 3 i \sin θ}{5 + 4 \cos θ} = a + i b$
$\frac{6 + 3 \cos θ}{5 + 4 \cos θ} - \frac{3 i \sin θ}{5 + 4 \cos θ} = a + i b$
equating real and imaginary part and we get
$a = \frac{6 + 3 \cos θ}{5 + 4 \cos θ}, b = \frac{- 3 \sin θ}{5 + 4 \cos θ}$
Now value of $[(a - 2)^{2} + b^{2}]$ is
$= {(\frac{6 + 3 \cos θ}{5 + 4 \cos θ} - 2)}^{2} + {(\frac{- 3 \sin θ}{5 + 4 \cos θ})}^{2}$
$= {(\frac{6 + 3 \cos θ - 10 - 8 \cos θ}{5 + 4 \cos θ})}^{2} + {(\frac{- 3 \sin θ}{5 + 4 \cos θ})}^{2}$
$= {\frac{(- 4 - 5 \cos θ)^{2}}{(5 + 4 \cos θ)^{2}} + \frac{9 \sin^{2} θ}{(5 + 4 \cos θ)^{2}}}$
$= \frac{16 + 25 \cos^{2} θ + 40 \cos θ + 9 \sin^{2} θ}{(5 + 4 \cos θ)^{2}}$
$= \frac{16 + 25 \cos^{2} θ + 40 \cos θ 9 (1 - \cos^{2} θ)}{{(5 + 4 \cos^{2} θ)}^{2}}$
$= \frac{(5 + 4 \cos θ)^{2}}{(5 + 4 \cos θ)^{2}} = 1$

BCECE-2009

Complex Numbers and Quadratic Equation

117483 The real value of $α$ for which the expression $\frac{1 - i \sin α}{1 + 2 i \sin α}$ is purely real, is :

1

(2 n + 1) \frac{π}{2}

2

(n + 1) \frac{π}{2}

3

n π

4 none of these

Explanation:

C Let complex number $(z) = \frac{1 - i \sin α}{1 + 2 i \sin α}$
Multiplying denominator \& Numerator by $1 - 2$ i sin $α$ Now,
$z = \frac{1 - i \sin α}{1 + 2 i \sin α} \times \frac{1 - 2 i \sin α}{1 - 2 i \sin α}$
$z = \frac{1 - 2 i \sin α - i \sin α + 2 i^{2} \sin^{2} α}{1 - 4 i^{2} \sin^{2} α}$
$z = \frac{1 - 2 \sin^{2} α - 3 i \sin α}{1 + 4 \sin^{2} α} {∵ i^{2} = - 1}$
$z = \frac{1 - 2 \sin^{2} α}{1 + 4 \sin^{2} α} - \frac{i 3 \sin α}{1 + 4 \sin^{2} α}$
$∵ z$ is purely real
$\frac{3 \sin α}{1 + 4 \sin^{2} α} = 0$
$3 \sin α = 0$
$\sin α = 0$
$α = 0, π, 2 π \dots \dots$
$α = n π where (n \in N)$

BCECE-2005

Complex Numbers and Quadratic Equation

117484 If the imaginary part of $\frac{2 z + 1}{i z + 1}$ is -2 , then the locus of the point represented by $z$ is a:

1 circle

2 straight line

3 parabola

4 none of these

Explanation:

(B) Let complex number $(z) = x + iy$
Thus,
$\frac{2 z + 1}{i z + 1} = \frac{2 (x + i y) + 1}{i (x + i y) + 1}$
$= \frac{2 x + 2 i y + 1}{i x + i^{2} y + 1} = \frac{2 x + 1 + 2 i y}{(1 - y) + i x}$
Multiplying denominator \& Numerator by $(1 - y) - ix$
$= \frac{2 x + 1 + 2 i y}{(1 - y) + i x} \times \frac{(1 - y) - i x}{(1 - y) - i x}$
$= \frac{2 x - 2 x y - 2 i x^{2} + 1 - y - i x + 2 i y - 2 i y^{2} - 2 i^{2} x y}{(1 - y)^{2} + x^{2}}$
$= \frac{2 x - 2 x y + 1 - y + 2 x y - 2 i x^{2} - i x + 2 i y - 2 i y^{2}}{(1 - y)^{2} + x^{2}}$
$\frac{(2 x - y + 1)}{(1 - y)^{2} + x^{2}} + \frac{i (- 2 x^{2} - x + 2 y - 2 y^{2})}{(1 - y)^{2} + x^{2}}$
$∵$ Given imaginary part is -2
$∴ \frac{- 2 x^{2} - x + 2 y - 2 y^{2}}{(1 - y)^{2} + x^{2}} = - 2$
$- 2 x^{2} - x + 2 y - 2 y^{2} = - 2 [(1 - y)^{2} + x^{2}]$
$- 2 x^{2} - x + 2 y - 2 y^{2} = - 2 - 2 y^{2} + 4 y - 2 x^{2}$
$x + 2 y = 2 {Which is represent a straight line}$

BCECE-2004

Complex Numbers and Quadratic Equation

117485 The values of $x$ and $y$ satisfying the equation
$\frac{(1 + i) x - 2}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i are :$

1

x = - 1, x = 3

2

x = 3, y = - 1

3

x = 0, y = 1

4

x = 1, y = 0

Explanation:

B Given equation
$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$
$\frac{[(1 + i) x] (3 - i) - 2 i (3 - i) + [(2 - 3 i) y] (3 + i) + i (3 + i)}{9 - i^{2}} = i$
$\Rightarrow$
$\frac{3 x + 3 x i - i x - i^{2} x - 6 i + 2 i^{2} + 6 y - 9 i y + 2 i y - 3 i^{2} y + 3 i + i^{2}}{10} =$
$\Rightarrow 3 x + x - 2 + 6 y + 3 y - 1 + 2 xi - 3 i - 7 iy = 10 i$
$4 x + 9 y - 3 + 2 xi - 7 iy - 13 i = 0$
$4 x + 9 y - 3 + i (2 x - 7 y - 13) = 0$
$4 x + 9 y - 3 = 0 .....(i)$
$2 x - 7 y - 13 = 0 .....(ii)$
$23 y + 23 = 0$
$23 y = - 23$
$y = - 1$
$x = 3$
Now,
equation (ii) multiply by 2 and subtracted from equation (i)
$23 y + 23 = 0$
$23 y = - 23$
$y = - 1$ Putting the value of $y$ in equation (1) and we get,So value of $x = 3 & y = - 1$

BCECE-2003

Complex Numbers and Quadratic Equation

117486 The number of solutions of equation $z^{2} + \overset{―}{z} = 0$ , where $z \in C$ are

1 6

2 1

3 4

4 5

Explanation:

C The number of solution of equation
$z^{2} + \bar{z} = 0$
Let, $z = x +$ iy
$(x + i y)^{2} + (x - i y) = 0$
$x^{2} + i^{2} y^{2} + i 2 x y + x - i y = 0$
$x^{2} - y^{2} + x + i 2 x y - i y = 0$
$x^{2} - y^{2} + x + i (2 x y - y) = 0$
On comparing both real \& imaginary parts,
$x^{2} + x - y^{2} = 0$
$2 x y - y = 0$
$y (2 x - 1) = 0$
From equation (ii)
$y (2 x - 1) = 0$
$y = 0$
$x = \frac{1}{2}$
Putting $y = 0$ in equation (i) we get,
$x^{2} + x = 0$
$x (x + 1) = 0 \Rightarrow x = 0, - 1$
Now putting in equation (i) we get,
$\frac{1}{4} + \frac{1}{2} - y^{2} = 0$
$y^{2} = \frac{3}{4}$
$y = \pm \frac{\sqrt{3}}{2}$
Thus, $(x, y) = (0, 0) (- 1, 0), (\frac{1}{2}, \frac{\sqrt{3}}{2}) (\frac{1}{2}, \frac{- \sqrt{3}}{2})$
Hence, the number of solution of given equation is 4 .

Karnataka CET-2012

Complex Numbers and Quadratic Equation

117480 If $\frac{3}{2 + \cos θ + i \sin θ} = a + i b$ then $[(a - b)^{2} + b^{2}]$ is equal to

1 0

2 1

3 -1

4 2

Explanation:

Exp: (B):
Given, $\frac{3}{2 + \cos θ + i \sin θ} = a + i b$
Multiplying denominator and numerator by $2 + \cos θ - i$ $\sin θ$
$\frac{3}{2 + \cos θ + i \sin θ} \times \frac{2 + \cos θ - i \sin θ}{2 + \cos θ - i \sin θ} = a + i b$
$\frac{6 + 3 \cos θ - 3 i \sin θ}{(2 + \cos θ)^{2} - (i \sin θ)^{2}} = a + i b$
$\frac{6 + 3 \cos θ - 3 i \sin θ}{5 + 4 \cos θ} = a + i b$
$\frac{6 + 3 \cos θ}{5 + 4 \cos θ} - \frac{3 i \sin θ}{5 + 4 \cos θ} = a + i b$
equating real and imaginary part and we get
$a = \frac{6 + 3 \cos θ}{5 + 4 \cos θ}, b = \frac{- 3 \sin θ}{5 + 4 \cos θ}$
Now value of $[(a - 2)^{2} + b^{2}]$ is
$= {(\frac{6 + 3 \cos θ}{5 + 4 \cos θ} - 2)}^{2} + {(\frac{- 3 \sin θ}{5 + 4 \cos θ})}^{2}$
$= {(\frac{6 + 3 \cos θ - 10 - 8 \cos θ}{5 + 4 \cos θ})}^{2} + {(\frac{- 3 \sin θ}{5 + 4 \cos θ})}^{2}$
$= {\frac{(- 4 - 5 \cos θ)^{2}}{(5 + 4 \cos θ)^{2}} + \frac{9 \sin^{2} θ}{(5 + 4 \cos θ)^{2}}}$
$= \frac{16 + 25 \cos^{2} θ + 40 \cos θ + 9 \sin^{2} θ}{(5 + 4 \cos θ)^{2}}$
$= \frac{16 + 25 \cos^{2} θ + 40 \cos θ 9 (1 - \cos^{2} θ)}{{(5 + 4 \cos^{2} θ)}^{2}}$
$= \frac{(5 + 4 \cos θ)^{2}}{(5 + 4 \cos θ)^{2}} = 1$

BCECE-2009

Complex Numbers and Quadratic Equation

117483 The real value of $α$ for which the expression $\frac{1 - i \sin α}{1 + 2 i \sin α}$ is purely real, is :

1

(2 n + 1) \frac{π}{2}

2

(n + 1) \frac{π}{2}

3

n π

4 none of these

Explanation:

C Let complex number $(z) = \frac{1 - i \sin α}{1 + 2 i \sin α}$
Multiplying denominator \& Numerator by $1 - 2$ i sin $α$ Now,
$z = \frac{1 - i \sin α}{1 + 2 i \sin α} \times \frac{1 - 2 i \sin α}{1 - 2 i \sin α}$
$z = \frac{1 - 2 i \sin α - i \sin α + 2 i^{2} \sin^{2} α}{1 - 4 i^{2} \sin^{2} α}$
$z = \frac{1 - 2 \sin^{2} α - 3 i \sin α}{1 + 4 \sin^{2} α} {∵ i^{2} = - 1}$
$z = \frac{1 - 2 \sin^{2} α}{1 + 4 \sin^{2} α} - \frac{i 3 \sin α}{1 + 4 \sin^{2} α}$
$∵ z$ is purely real
$\frac{3 \sin α}{1 + 4 \sin^{2} α} = 0$
$3 \sin α = 0$
$\sin α = 0$
$α = 0, π, 2 π \dots \dots$
$α = n π where (n \in N)$

BCECE-2005

Complex Numbers and Quadratic Equation

117484 If the imaginary part of $\frac{2 z + 1}{i z + 1}$ is -2 , then the locus of the point represented by $z$ is a:

1 circle

2 straight line

3 parabola

4 none of these

Explanation:

(B) Let complex number $(z) = x + iy$
Thus,
$\frac{2 z + 1}{i z + 1} = \frac{2 (x + i y) + 1}{i (x + i y) + 1}$
$= \frac{2 x + 2 i y + 1}{i x + i^{2} y + 1} = \frac{2 x + 1 + 2 i y}{(1 - y) + i x}$
Multiplying denominator \& Numerator by $(1 - y) - ix$
$= \frac{2 x + 1 + 2 i y}{(1 - y) + i x} \times \frac{(1 - y) - i x}{(1 - y) - i x}$
$= \frac{2 x - 2 x y - 2 i x^{2} + 1 - y - i x + 2 i y - 2 i y^{2} - 2 i^{2} x y}{(1 - y)^{2} + x^{2}}$
$= \frac{2 x - 2 x y + 1 - y + 2 x y - 2 i x^{2} - i x + 2 i y - 2 i y^{2}}{(1 - y)^{2} + x^{2}}$
$\frac{(2 x - y + 1)}{(1 - y)^{2} + x^{2}} + \frac{i (- 2 x^{2} - x + 2 y - 2 y^{2})}{(1 - y)^{2} + x^{2}}$
$∵$ Given imaginary part is -2
$∴ \frac{- 2 x^{2} - x + 2 y - 2 y^{2}}{(1 - y)^{2} + x^{2}} = - 2$
$- 2 x^{2} - x + 2 y - 2 y^{2} = - 2 [(1 - y)^{2} + x^{2}]$
$- 2 x^{2} - x + 2 y - 2 y^{2} = - 2 - 2 y^{2} + 4 y - 2 x^{2}$
$x + 2 y = 2 {Which is represent a straight line}$

BCECE-2004

Complex Numbers and Quadratic Equation

117485 The values of $x$ and $y$ satisfying the equation
$\frac{(1 + i) x - 2}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i are :$

1

x = - 1, x = 3

2

x = 3, y = - 1

3

x = 0, y = 1

4

x = 1, y = 0

Explanation:

B Given equation
$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$
$\frac{[(1 + i) x] (3 - i) - 2 i (3 - i) + [(2 - 3 i) y] (3 + i) + i (3 + i)}{9 - i^{2}} = i$
$\Rightarrow$
$\frac{3 x + 3 x i - i x - i^{2} x - 6 i + 2 i^{2} + 6 y - 9 i y + 2 i y - 3 i^{2} y + 3 i + i^{2}}{10} =$
$\Rightarrow 3 x + x - 2 + 6 y + 3 y - 1 + 2 xi - 3 i - 7 iy = 10 i$
$4 x + 9 y - 3 + 2 xi - 7 iy - 13 i = 0$
$4 x + 9 y - 3 + i (2 x - 7 y - 13) = 0$
$4 x + 9 y - 3 = 0 .....(i)$
$2 x - 7 y - 13 = 0 .....(ii)$
$23 y + 23 = 0$
$23 y = - 23$
$y = - 1$
$x = 3$
Now,
equation (ii) multiply by 2 and subtracted from equation (i)
$23 y + 23 = 0$
$23 y = - 23$
$y = - 1$ Putting the value of $y$ in equation (1) and we get,So value of $x = 3 & y = - 1$

BCECE-2003

Complex Numbers and Quadratic Equation

117486 The number of solutions of equation $z^{2} + \overset{―}{z} = 0$ , where $z \in C$ are

1 6

2 1

3 4

4 5

Explanation:

C The number of solution of equation
$z^{2} + \bar{z} = 0$
Let, $z = x +$ iy
$(x + i y)^{2} + (x - i y) = 0$
$x^{2} + i^{2} y^{2} + i 2 x y + x - i y = 0$
$x^{2} - y^{2} + x + i 2 x y - i y = 0$
$x^{2} - y^{2} + x + i (2 x y - y) = 0$
On comparing both real \& imaginary parts,
$x^{2} + x - y^{2} = 0$
$2 x y - y = 0$
$y (2 x - 1) = 0$
From equation (ii)
$y (2 x - 1) = 0$
$y = 0$
$x = \frac{1}{2}$
Putting $y = 0$ in equation (i) we get,
$x^{2} + x = 0$
$x (x + 1) = 0 \Rightarrow x = 0, - 1$
Now putting in equation (i) we get,
$\frac{1}{4} + \frac{1}{2} - y^{2} = 0$
$y^{2} = \frac{3}{4}$
$y = \pm \frac{\sqrt{3}}{2}$
Thus, $(x, y) = (0, 0) (- 1, 0), (\frac{1}{2}, \frac{\sqrt{3}}{2}) (\frac{1}{2}, \frac{- \sqrt{3}}{2})$
Hence, the number of solution of given equation is 4 .

Karnataka CET-2012

Complex Numbers and Quadratic Equation

117480 If $\frac{3}{2 + \cos θ + i \sin θ} = a + i b$ then $[(a - b)^{2} + b^{2}]$ is equal to

1 0

2 1

3 -1

4 2

Explanation:

Exp: (B):
Given, $\frac{3}{2 + \cos θ + i \sin θ} = a + i b$
Multiplying denominator and numerator by $2 + \cos θ - i$ $\sin θ$
$\frac{3}{2 + \cos θ + i \sin θ} \times \frac{2 + \cos θ - i \sin θ}{2 + \cos θ - i \sin θ} = a + i b$
$\frac{6 + 3 \cos θ - 3 i \sin θ}{(2 + \cos θ)^{2} - (i \sin θ)^{2}} = a + i b$
$\frac{6 + 3 \cos θ - 3 i \sin θ}{5 + 4 \cos θ} = a + i b$
$\frac{6 + 3 \cos θ}{5 + 4 \cos θ} - \frac{3 i \sin θ}{5 + 4 \cos θ} = a + i b$
equating real and imaginary part and we get
$a = \frac{6 + 3 \cos θ}{5 + 4 \cos θ}, b = \frac{- 3 \sin θ}{5 + 4 \cos θ}$
Now value of $[(a - 2)^{2} + b^{2}]$ is
$= {(\frac{6 + 3 \cos θ}{5 + 4 \cos θ} - 2)}^{2} + {(\frac{- 3 \sin θ}{5 + 4 \cos θ})}^{2}$
$= {(\frac{6 + 3 \cos θ - 10 - 8 \cos θ}{5 + 4 \cos θ})}^{2} + {(\frac{- 3 \sin θ}{5 + 4 \cos θ})}^{2}$
$= {\frac{(- 4 - 5 \cos θ)^{2}}{(5 + 4 \cos θ)^{2}} + \frac{9 \sin^{2} θ}{(5 + 4 \cos θ)^{2}}}$
$= \frac{16 + 25 \cos^{2} θ + 40 \cos θ + 9 \sin^{2} θ}{(5 + 4 \cos θ)^{2}}$
$= \frac{16 + 25 \cos^{2} θ + 40 \cos θ 9 (1 - \cos^{2} θ)}{{(5 + 4 \cos^{2} θ)}^{2}}$
$= \frac{(5 + 4 \cos θ)^{2}}{(5 + 4 \cos θ)^{2}} = 1$

BCECE-2009

Complex Numbers and Quadratic Equation

117483 The real value of $α$ for which the expression $\frac{1 - i \sin α}{1 + 2 i \sin α}$ is purely real, is :

1

(2 n + 1) \frac{π}{2}

2

(n + 1) \frac{π}{2}

3

n π

4 none of these

Explanation:

C Let complex number $(z) = \frac{1 - i \sin α}{1 + 2 i \sin α}$
Multiplying denominator \& Numerator by $1 - 2$ i sin $α$ Now,
$z = \frac{1 - i \sin α}{1 + 2 i \sin α} \times \frac{1 - 2 i \sin α}{1 - 2 i \sin α}$
$z = \frac{1 - 2 i \sin α - i \sin α + 2 i^{2} \sin^{2} α}{1 - 4 i^{2} \sin^{2} α}$
$z = \frac{1 - 2 \sin^{2} α - 3 i \sin α}{1 + 4 \sin^{2} α} {∵ i^{2} = - 1}$
$z = \frac{1 - 2 \sin^{2} α}{1 + 4 \sin^{2} α} - \frac{i 3 \sin α}{1 + 4 \sin^{2} α}$
$∵ z$ is purely real
$\frac{3 \sin α}{1 + 4 \sin^{2} α} = 0$
$3 \sin α = 0$
$\sin α = 0$
$α = 0, π, 2 π \dots \dots$
$α = n π where (n \in N)$

BCECE-2005

Complex Numbers and Quadratic Equation

117484 If the imaginary part of $\frac{2 z + 1}{i z + 1}$ is -2 , then the locus of the point represented by $z$ is a:

1 circle

2 straight line

3 parabola

4 none of these

Explanation:

(B) Let complex number $(z) = x + iy$
Thus,
$\frac{2 z + 1}{i z + 1} = \frac{2 (x + i y) + 1}{i (x + i y) + 1}$
$= \frac{2 x + 2 i y + 1}{i x + i^{2} y + 1} = \frac{2 x + 1 + 2 i y}{(1 - y) + i x}$
Multiplying denominator \& Numerator by $(1 - y) - ix$
$= \frac{2 x + 1 + 2 i y}{(1 - y) + i x} \times \frac{(1 - y) - i x}{(1 - y) - i x}$
$= \frac{2 x - 2 x y - 2 i x^{2} + 1 - y - i x + 2 i y - 2 i y^{2} - 2 i^{2} x y}{(1 - y)^{2} + x^{2}}$
$= \frac{2 x - 2 x y + 1 - y + 2 x y - 2 i x^{2} - i x + 2 i y - 2 i y^{2}}{(1 - y)^{2} + x^{2}}$
$\frac{(2 x - y + 1)}{(1 - y)^{2} + x^{2}} + \frac{i (- 2 x^{2} - x + 2 y - 2 y^{2})}{(1 - y)^{2} + x^{2}}$
$∵$ Given imaginary part is -2
$∴ \frac{- 2 x^{2} - x + 2 y - 2 y^{2}}{(1 - y)^{2} + x^{2}} = - 2$
$- 2 x^{2} - x + 2 y - 2 y^{2} = - 2 [(1 - y)^{2} + x^{2}]$
$- 2 x^{2} - x + 2 y - 2 y^{2} = - 2 - 2 y^{2} + 4 y - 2 x^{2}$
$x + 2 y = 2 {Which is represent a straight line}$

BCECE-2004

Complex Numbers and Quadratic Equation

117485 The values of $x$ and $y$ satisfying the equation
$\frac{(1 + i) x - 2}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i are :$

1

x = - 1, x = 3

2

x = 3, y = - 1

3

x = 0, y = 1

4

x = 1, y = 0

Explanation:

B Given equation
$\frac{(1 + i) x - 2 i}{3 + i} + \frac{(2 - 3 i) y + i}{3 - i} = i$
$\frac{[(1 + i) x] (3 - i) - 2 i (3 - i) + [(2 - 3 i) y] (3 + i) + i (3 + i)}{9 - i^{2}} = i$
$\Rightarrow$
$\frac{3 x + 3 x i - i x - i^{2} x - 6 i + 2 i^{2} + 6 y - 9 i y + 2 i y - 3 i^{2} y + 3 i + i^{2}}{10} =$
$\Rightarrow 3 x + x - 2 + 6 y + 3 y - 1 + 2 xi - 3 i - 7 iy = 10 i$
$4 x + 9 y - 3 + 2 xi - 7 iy - 13 i = 0$
$4 x + 9 y - 3 + i (2 x - 7 y - 13) = 0$
$4 x + 9 y - 3 = 0 .....(i)$
$2 x - 7 y - 13 = 0 .....(ii)$
$23 y + 23 = 0$
$23 y = - 23$
$y = - 1$
$x = 3$
Now,
equation (ii) multiply by 2 and subtracted from equation (i)
$23 y + 23 = 0$
$23 y = - 23$
$y = - 1$ Putting the value of $y$ in equation (1) and we get,So value of $x = 3 & y = - 1$

BCECE-2003

Complex Numbers and Quadratic Equation

117486 The number of solutions of equation $z^{2} + \overset{―}{z} = 0$ , where $z \in C$ are

1 6

2 1

3 4

4 5

Explanation:

C The number of solution of equation
$z^{2} + \bar{z} = 0$
Let, $z = x +$ iy
$(x + i y)^{2} + (x - i y) = 0$
$x^{2} + i^{2} y^{2} + i 2 x y + x - i y = 0$
$x^{2} - y^{2} + x + i 2 x y - i y = 0$
$x^{2} - y^{2} + x + i (2 x y - y) = 0$
On comparing both real \& imaginary parts,
$x^{2} + x - y^{2} = 0$
$2 x y - y = 0$
$y (2 x - 1) = 0$
From equation (ii)
$y (2 x - 1) = 0$
$y = 0$
$x = \frac{1}{2}$
Putting $y = 0$ in equation (i) we get,
$x^{2} + x = 0$
$x (x + 1) = 0 \Rightarrow x = 0, - 1$
Now putting in equation (i) we get,
$\frac{1}{4} + \frac{1}{2} - y^{2} = 0$
$y^{2} = \frac{3}{4}$
$y = \pm \frac{\sqrt{3}}{2}$
Thus, $(x, y) = (0, 0) (- 1, 0), (\frac{1}{2}, \frac{\sqrt{3}}{2}) (\frac{1}{2}, \frac{- \sqrt{3}}{2})$
Hence, the number of solution of given equation is 4 .

Karnataka CET-2012

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