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This mock test of Complex Number MCQ Level - 2 for IIT JAM helps you for every IIT JAM entrance exam.
This contains 10 Multiple Choice Questions for IIT JAM Complex Number MCQ Level - 2 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The point of intersection the curves

arg ( z - i + 2 ) = and arg ( z + 4 - 3i ) = is given by

Solution:

⇒ y + x = - 1 ,x > - 4 , y < 3 ...(2)

so, there is no point of intersection.

The correct answer is: None of these

QUESTION: 2

If 1, α_{1},α_{2},α_{3}.........and α_{8} re nine, ninth roots of unity (taken in counter-

clockwise sequence) then | ( 2 - α_{1 }) (2 - α_{3}) ( 2 - α_{5 }) ( 2 - α_{7} ) | is equal to :

Solution:

( x - 1 ) ( x - α_{1}) ( x - α2) ...( x - α_{8}) ≡ x^{9} - 1

∴ ( 2 - α_{1} ) ( 2 - α_{2} )... ( 2 - α_{8} ) = 2^{9 }- 1

Now since 2 - α_{1 }and 2 - α_{8 }) are conjugates of each other

similarly

and

The correct answer is:

QUESTION: 3

If 'z' is complex number then the locus of ‘z’ satisfying the condition

|2z - 1| = | z - 1 | is

Solution:

or

Let us write z as x+iy

We then have ∣2z−1∣=∣z−1∣

⇒4x^{2}−8x+1+4y^{2}=x^{2}−2x+1+y^{2}

⇒3x^{2}+3y^{2}−6x=0 which is a circle.

So, locus of z is a circle.

QUESTION: 4

If t and c are two complex numbers such that | t | = 1 and z = z = x + iy , Locus of z is (where a, b are complex numbers)

Solution:

|t| = 1

The correct answer is: circle

QUESTION: 5

If ω is an imaginary cube root of unity, then (1+ ω - ω^{2} )^{7} is equal to

Solution:

T he correct answer is: -128 ω^{2}

QUESTION: 6

The principal argument of the complex number

Solution:

∴ argument

∴ principal argument is

The correct answer is:

QUESTION: 7

If a complex number z satisfies | 2z + 10 + 10i | __<__ 5 √3 - 5, then the least principal argument of z is.

Solution:

Point **B** has least principal argument

The correct answer is:

QUESTION: 8

Image of the point, whose affix is in the line (1 + i ) z + ( 1 - i ) is the point whose affix is :

Solution:

Equation of the line is 2x - 2y = 0

i.e. y = x

The correct answer is :

QUESTION: 9

Let S denote the set of all complex numbers z satisfying the inequality |z - 5i| __<__ 3.The complex numbers z in S having least positive argument is :

Solution:

See the figure

|z - 5i| __<__ 3

the point is ( 4 cosθ, 4 sinθ)

The correct answer is:

QUESTION: 10

Number of solution of the equation where z is a complex number is :

Solution:

Let

Since ‘r ’ cannot be zero

which will hold for

and 5 distinct value o f 'O'

Thus there are five solution.

The correct answer is: 5

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