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# Previous year Questions (2016-20) - Complex Numbers Notes | EduRev

## JEE : Previous year Questions (2016-20) - Complex Numbers Notes | EduRev

The document Previous year Questions (2016-20) - Complex Numbers Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 11.
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Q.1. If  where z = x + iy, then the point (x, y) lies on a    (2020)
(1) Circle whose centre is at
(2) Straight line whose slope is -2/3.
(3) Straight line whose slope is 3/2.
(4) Circle whose diameter is √5/2.
Ans.
(4)
Solution.
We have
...(1)

Therefore,
Hence, the diameter of circle is √5/2.

Q.2. If is a real number, then an argument of sinθ + i cosθ is    (2020)
(1) π - tan-1(4/3)
(2) π - tan-1(3/4)
(3) - tan-1(3/4)
(4) - tan-1(4/3)
Ans.
(1)
Solution.
We have

Given that the number is real, then
4 sinθ + 3 cosθ = 0 ⇒ tan θ = -3/4
Hence, the argument of sinθ + i cosθ is

Q.3. If z be a complex number satisfying |Re(z)| + |Im(z)| = 4, then |z| cannot be    (2020)
(1)
(2) √10
(3) √7
(4) √8
Ans.
(3)
Solution.
Let the complex number be z = x + iy. So,
|x|+|y| = 4   (1)
Now,

Minimum value of
Maximum value of |z| = 4
So, |z| ∈ (2√2, 4).

Hence, the value of |z| cannot be √7.

Q.4. Let  Then the sum of the elements in A is:   (2019)
(1) 5π/6
(2) π
(3) 3π/4
(4) 2π/3
Ans.
(4)
Solution.

Since, z is purely imaginary, then

Now, the sum of elements in A =

Q.5. Let z0 be a root of the quadratic equation, then arg z is equal    (2019)
(1) π/4
(2) π/6
(3) π/3
(4) 0
Ans.
(1)
Solution.
∵ z0 is a root of quadratic equation

Q.6. Let z1 and z2 be any two non-zero complex numbers such that  ,then:    (2019)
(1) Re(z) = 0
(2)
(3)
(4) Im(z) = 0
Ans.
(Bonus)
Solution.

Q.7. Let  If R(z) and I(z) respectively denote the real and imaginary parts of z, then:    (2019)
(1) I(z) = 0
(2) R(z) > 0 and I(z) > 0
(3) R(z) < 0 and I(z) > 0
(4) R(z) = -(3)
Ans.
(1)
Solution.

Q.8. Let  where x and y are real numbers then y - x equals:    (2019)
(1) 91
(2) -85
(3) 85
(4) -91
Ans.
(1)
Solution.

Q.9. Let z be a complex number such that |z| + z = 3 + i
Then |z| is equal to:   (2019)
(1)
(2) 5/3
(3)
(4) 5/4
Ans.
(2)
Solution.
Since, |z| + z = 3 + i
Let z = a + ib, then

Compare real and imaginary coefficients on both sides

Then,

Q.10.  is a purely imaginary number and |z| = 2, then a value of α is:    (2019)
(1) 2
(2) 1
(3) 1/2
(4)
Ans.
(1)
Solution.

∵ t is purely imaginary number.

Q.11. Let z1 and z2 be two complex numbers satisfying |z1| = 9 and |z2- 3 - 4i|=4. Then the minimum value of |z1 -z2| is:    (2019)
(1) 0
(2) √2
(3) 1
(4) 2
Ans.
(1)
Solution.
|Z1| = 9, |z2 - 3 - 4i| = 4
z1 lies on a circle with centre C1(0, 0) and radius r1 = 9
z2 lies on a circle with centre C2(3, 4) and radius r2 = 4
So, minimum value of |z1 -z2| is zero at point of contact (i.e. A)

Q.12. If  then (1 + iz + z5 + iz8)9 is equal to:
(1) 0
(2) 1
(3) (-1 + 2i)9
(4) -1
Ans.
(4)
Solution.

where ω is imaginary cube root of unity.

Q.13. All the points in the set  lie on a:    (2019)
(1) straight line whose slope is 1.

(2) circle whose radius is 1.
(3) circle whose radius is √2 .
(4) straight line whose slope is -1.
Ans.
(2)
Solution.
Let z∈S then
Since, z is a complex number and let z = x + iy
Then,  (by rationalisation)

Then compare both sides
...(1)
...(2)
Now squaring and adding equations (1) and (2)

Q.14. Let z ∈ C be such that |z| < 1. If ω =  then:    (2019)
(1) 5 Re (ω) > 4
(2) 4 Im (ω) > 5
(3) 5 Re (ω) >1
(4) 5 Im (ω) < 1
Ans.
(3)
Solution.

Q.15. If a > 0 and z =  has magnitude  then  is equal to:    (2019)
(1)
(2)
(3)
(4)
Ans.
(1)
Solution.

Then, from equation (1),

Now, square on both side; we get

Q.16. If z and ω are two complex numbers such that |zω| = 1 and arg(z) - arg(ω) = π/2, then:    (2019)

Ans.
(3)
Solution:

Q.17. The equation  represents:    (2019)
(1) a circle of radius 1/2.
(2) the line through the origin with slope 1.
(3) a circle of radius 1.
(4) the line through the origin with slope -1.
Ans.
(2)
Solution.
Given equation is, |z - 1| = |z - i|

Hence, locus is straight line with slope 1.

Q.18. Let z ∈ C with Im(z) = 10 and it satisfies  for some natural number n. Then:    (2019)
(1) n = 20 and Re(z) = -10
(2) n = 40 and Re(z) = 10
(3) n = 40 and Re(z) = -10
(4) n = 20 and Re(z) = 10
Ans.
(3)
Solution.

On comparing real and imaginary parts,

Hence, Re(z) = -10

Q.19. If α ,β ∈ c are the distinct roots, of the equation x2 - x + 1 =0 , then α101 + β107 is equal to:    (2018)
(1) -1
(2) 0
(3) 1
(4) 2
Ans.
(3)
Solution. x2 - x+ 1 = 0

Q.20. The set of all α ∈ R, for which ω =is a purely imaginary number, for all z ∈ C Satifying |z| = 1 and Re z ≠ 1, is:    (2018)
(1) An empty set
(2)
(3) equal to R
(4) {0}
Ans.
(4)
Solution. As ω is purely imaginary

If Re(z) ≠ 1
then, α = 0

Q.21. The least positive integer n for which  is:    (2018)
(1) 2
(2) 5
(3) 6
(4) 3
Ans.
(4)
Solution.

Also,

∴ Least positive integer n is 3.

Q.22. Let z ∈ C, the set of complex numbers. Then the equation, 2|z + 3i| – |z – i| = 0  represents a circle with radius    (2017)
(1) a cirlce with radius 8/3
(2) an ellipse with length of minor axis 16/9
(3) an ellipse with length of major axis 16/3
(4) a circle with diameter 10/3
Ans.
(1)
Solution.

Q.23. The equation represents a part of a circle having radius equal to:    (2017)
(1) 1
(2) 2
(3) 3/4
(4) 1/2
Ans.
(3)
Solution. Let = x + y

Q.24. A value of θ for which is purely imaginary, is:    (2016)
(1) π/3
(2) π/6
(3) sin-1
(4) sin-1
Ans.
(4)
Solution.

To be purely imaginary if

Q.25. The point represented by 2 + i in the Argand plane moves 1 unit eastwards, then 2 units northwards and finally from there 2√2 units in the south- westwards direction. Then its new position in the Argand plane is at the point represented by:    (2016)
(1) 2 + 2i
(2) – 2 – 2i
(3) 1 + i
(4) – 1 – i
Ans.
(3)
Solution. Let P(2 + i)
By rotation theorem

Q.26. Let z = 1 + ai be a complex number, a > 0 such that z3 is a real number. Then the sum 1 + z + z2 +....+ z11 is equal to    (2016)

Ans.
(3)
Solution.

,

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