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QUESTION: 1

The equation represents | z + 3 | - | z - 5 | = 0

Solution:

, it is bisector of line segment joining (-3.,0) and (5,0)

QUESTION: 2

Point z with in the complex plane lies on

Solution:

I z - 4 I = | z + 4|

If ( z - 4 ) = k |z + 4| k ≠ 1

(x - y)^{2} + y^{2} = ± (x+y)^{2} + y^{2} then it is circle

but now x = 0

x = 0 hence it is y - axis.

QUESTION: 3

The locus of z satisfying the condition is

Solution:

it is a circle

QUESTION: 4

The inequality | z - 4 | < | z - 2 | region given by

Solution:

| z - 4 | < | z - 2 |

=> | x - 4 + iy | < | x + 2 - iy |

=> (x - y)^{2} + y^{2} < (x - 2)^{2} + y^{2}

=> x^{2}- 8 x + 16 < x^{2 }- 4 x +4

=> - 4 x < -12

=> x > 3

=> real value of z should be > 3

QUESTION: 5

Which of the following is correct

Solution:

The concept of inequality does not hold in the set of complex number

QUESTION: 6

If z = x - iy and w = (1 - iz)/( z - l ) , then | w | = 1 implies that in the complex plane, z iies on

Solution:

Let. z = x + iy

and |w| = 1

squaring both side

Hence z = x + iy = x + 0

which is real

QUESTION: 7

If the complex number z_{1}, z_{2}, z_{3} are in A.P. then they lies on a

Solution:

z_{1}, z_{2}, z_{3}, are in AP, so they lie on a straight line (by mid point method)

QUESTION: 8

The value of is

Solution:

is

= -i (sum of complex roots of 1)

= i

QUESTION: 9

The point z_{1 }, z_{2}, z_{3}, z_{4 }in the complex plane are the vertices of a parallelogram taken in order if and only if.

Solution:

z_{1}, z_{2}, z_{3}, z_{4}, are the complex numbers which are the vertices of a parallelogram taken in order if

Diagonal of a parelielogam bisect each other at midpoint of AC and BD

∴

z_{1} + z_{3 }= z_{2 }+ z_{4}

QUESTION: 10

If (1, ω, ω^{2})^{5} are the cube root of unity, then (1 - ω + ω^{2})^{5} + (1 + ω - ω^{2})^{5} =

Solution:

1, ω, ω^{2} are cube roots of one (unity).

QUESTION: 11

If a, b are the complex cube roots of unity and x = a + b, y = ωa + bω^{2}, z = aω^{2} - bω, then xyz =

Solution:

If a and b are cube roots of unity

x = a + b , y = ωa + bω^{2}

QUESTION: 12

If 1, ω, ω^{2} ... ω^{n-1 }are the n^{th }roots of unity then (1 - ω) (1 - ω^{2}) ... (1 — ω^{n-2}) is equal to

Solution:

If 1, ω, ω^{2}..., ω^{n-1} are the n^{th} root of unity then

QUESTION: 13

If ω, ω^{2} are the complex root of unity, then [(a + bω - cω^{2})/(c + aω + bω^{2})] + [(a + bω+ aω^{2})/ (b + cω + aω^{2})]

Solution:

if ω_{1}, ω_{2}, are the complex roots of unity

= (ω^{2}) + (ω)

QUESTION: 14

If cube root of unity are 1, ω, ω^{2}, then the roots of the equation (x - 1)^{3} + 8 = 0 are

Solution:

1 ,ω, ω^{2 }are root of cube root of unity , then

(x - 1)^{3} + 8 = 0

hence required roots are - 1 , 1 - 2w , 1 - 2w

QUESTION: 15

For any complex number z, the minimum value of |z| + | z - 1 | is

Solution:

Value of |z| + |z - 1| is at least |1 - 0 |

so minimum value of |z| + |z - 1| is 1

QUESTION: 16

Multiplying a complex number z by 1 + i rotates the radius vector of z by an angle of

Solution:

tan θ = 1/1

θ = π/4

QUESTION: 17

If |z_{1 }+ z_{2}| = |z_{1}| + |z_{2}| , then arg z_{1} - arg z_{2 }is equal to

Solution:

QUESTION: 18

The area of the triangle formed by the complex number z, iz, z + iz in the argand plane is

Solution:

Area, AABC =

QUESTION: 19

Let z be a complex number. Then the angle between z and iz is

Solution:

As e^{π/2} is i so, iz is obtained by rotating z by an angle of π/2. so , angle between z and iz is π/2.

QUESTION: 20

If z_{1}, z_{2}, z_{3} are vertices of equilateral triangle with z_{o} as centroid, then z_{1}^{2} + z_{2}^{2} + z_{3}^{2} is equal to

Solution:

A, B, C are equilateral A and P be the centroid of ΔABC.

since AB = BC = CA = 2a

Δ AOC

sin 60 = AO/CA

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