JEE Advanced (Single Correct MCQs): Complex Numbers Chapter MCQs & solutions

(x – 1)³  +  8 = 0
⇒ (x – 1)³   = – 8 = (– 2)³
⇒ x – 1 = – 2       

 or   -2ω or -2ω²

⇒ x = -1, 1 - 2ω, 1 -2ω²

Now iⁿ = 1 ⇒ the smallest positive integral value of n should be 4.

ATQ | x + iy – 5i | = | x + iy + 5i |
⇒ | x + (y – 5) i | = | x + (y + 5) i |
⇒ x² + (y – 5)² = x² + (y + 5)²
⇒ x² + y² – 10y + 25 = x² + y² + 10y + 25
⇒ 20y = 0  ⇒ y = 0
∴ ‘a’ is the correct alternative.

⇒ Re(z) < 0 and Im(z) = 0

∴ (b) is the correct choice.

| z – 4 | <  | z – 2 |
⇒ | (x – 4) + iy | < | (x – 2) + iy |
⇒ (x – 4)² + y² < (x – 2)² + y²
⇒ – 8x + 16 < – 4x + 4    
⇒  4x – 12 > 0 ⇒ x  > 3 ⇒ Re (z) > 3

⇒ |1– iz | = | z – i | ⇒ | 1 – i (x + iy) | = | x + iy – i |
⇒ | (y + 1) – ix | = | x + i (y – 1) |
⇒ x² + (y + 1)² = x² + (y – 1)²
⇒ 4y = 0    ⇒ y = 0  ⇒ z lies on real axis

If vertices of a parallelogram are z₁, z₂, z₃, z₄ then as diagonals bisect each other

=  z₁+ z₃ = z₂+ z₄

Let ABC be the Δ with vertices a, b, c and PQR be the Δ with vertices u, v, w.
Then c = (1– r) a + rb.

⇒ c – a = r(b – a) ....(1)

⇒ w = (1– r) u + rv  ....(2)

From (1) and (2) and

⇒ ΔABC ~ΔPQR.

(1 + ω)⁷ = A +Bω
⇒ (- ω²)⁷ = A +Bω (∵1 + ω +ω² = 0)
⇒ - ω¹⁴ = A +Bω ⇒ - ω² = A +Bω (∵ ω³= 1)
⇒ 1 + ω = A +Bω ⇒ A = 1, B = 1

Q | z | = | ω | and argz = π - argω

Let

Given that  

⇒ z lies on perpendicular bisector of the line segment joining which is real ax is , being mirror images of  each other..

∴ Im(z) = 0.

If z = x then

∴  (c) is the correct option.

(1 + i) ⁿ1 + (1+ i3 )ⁿ1 + (1+ i5 )ⁿ2 + (1+i7 )ⁿ2

= (1 + i) ⁿ1 + (1 – i)ⁿ1 + (1 + i)ⁿ2+ (1 –i )ⁿ2

Using 

and

We get the given expression as

= real number irrespective the values of n₁ and n₂

∴  (d) is the most appropriate answer.

= 1+2w + 3(1+ w + w²) =

arg (z) < 0 (given) ⇒ arg (z) = – θ

Now

z = r cos(-θ) + i sin (-θ) = r [cos (θ) -i sin (θ)]
Again - z = -r [cos (θ) -i sin (θ)] = r [cos (π-θ) +i sin (π-θ)]
∴ arg (- z) = π -θ;
Thus arg (- z) - arg(z) =π - θ - (-θ) =π - θ +θ =π

| z₁ | = | z₂ | = | z₃| = 1  (given)

Now,

Similarly 

Now,

NOTE THIS STEP

Let z = (1)^{1/ n} = (cos ²kπ + i sin ²kπ)^1/n

 k, 0, 1, 2, .....,n -1.

Let 

and 

be the two values of z. s.t. they subtend ∠ of 90° at origin.

As k₁ and k₂ are integers and k₁ ≠ k_2.

∴ n = 4k, k ∈ I

⇒ arg (cos(- π /3) + i sin (-π /3))
⇒ angle between z₁ – z₃  and  z₂– z₃  is 60°.

and 

NOTE THIS STEP
⇒ The Δ  with vertices z₁, z₂ and z₃ is isosceles with vertical ∠60° . Hence rest of the two angles should also be 60° each.
⇒  Req. Δ is an equilateral Δ.

| z₁ |=12  ⇒ z₁ lies on a circle with centre (0, 0) and radius 12 units, and | z₂ – 3 – 4i | = 5
⇒ z₂ lies on a circle with centre (3, 4) and radius 5 units.

From fig. it is clear that | z₁– z₂ | i.e., distance between z₁ and z₂ will be min when they lie at A and B resp. i.e., O,C, B, A are collinear  as shown.
Then z₁– z₂ = AB = OA – OB = 12 – 2(5) = 2.
As above is the min, value, we must have | z₁– z₂|  ≥ 2.

Given that | z | = 1 and

Now we know that  

          (for  | z | = 1)

⇒ Re (ω)=0

(1 +ω²)ⁿ = (1 +ω⁴)ⁿ
⇒ (-ω)ⁿ = (1+ ω)ⁿ = (-ω²)ⁿ ⇒ ωⁿ = 1 ⇒ n =3

Here we observe that.
AB = AC = AD = 2
∴ BCD is an arc of a circle with centre at A and radius 2.
Shaded region is outer (exterior) part of this sector ABCDA.
∴ For any pt. z on arc BCD we should have | z – (– 1) | = 2
and for shaded region, | z + 1| > 2 ....(i)
For shaded region we also have -p /4 < arg ( z + 1) <p /4
or | arg (z  + 1) | <p /4 ...(ii)

Combining (i) and (ii), (a)  is the correct option

Given that a, b, c are integers not all equal, w is cube root of unity ≠ 1,  then

| a + bω +cω²|

∴ The min value is obtained when any two are zero and third is a minimum magnitude integer i.e. 1.
Thus b = c = 0, a = 1 gives us the minimum value 1

Operating R₁ + R₂ + R₃, we get

 is purely real

⇒  | z |²  = 1 (∵ ω = α + iβ and β ≠ 0)

⇒  | z | = 1  also given z ≠ 1

∴ The required set is {z : | z | =1, z ≠ 1}

= 3ω (ω- 1)

Given | z | = 1 and z ≠ ± 1

To find locus of

We have 

purely imaginary number

∴ ω must lie on y – axis.

The initial position of point is Z₀ = 1+ 2i
∴ Z₁ = (1 + 5) + (2 + 3) i = 6 + 5i
Now Z₁ is moved through a distance of 2 units in the direction

∴ It becomes 

Now OZ₁'  is rotated through an angle  inanticlockwise direction, therefore Z₂ = iZ₁' = - 6+7i

z = cosθ + i sinθ

= cos(2m - 1)θ + i sin(2m - 1)θ

= sinθ + sin 3θ + sin 5θ + ......+ upto15 terms

Given z = x + iy where x and y are integer

Also 

⇒ ( x² +y²) ( x² -y²)= 175
⇒ ( x² +y²) ( x² -y²) = 25x7 ...(i)
or(x² + y²) (x² – y²) = 35 × 5 ...(ii)
∵ x and y are integers,
∴ x² +y²= 25 and x² - y² = 7       [From eq (i)]
⇒ x² = 16 and y 2 = 9 ⇒ x = ± 4 and y =±3
∴ Vertices of rectangle are (4, 3) , (4, – 3), (– 4, – 3) , (– 4, 3).
So, area of rectangle = 8 × 6 = 48 sq. units Now from eq. (ii)
or x² + y² = 35 and x² – y² = 5
⇒ x² = 20, which is not possible for any integral value of x

∵ Im (z) ≠ 0  
⇒ z is non real and equation z² + z + (1 -a )=0 will have non real roots, if D < 0

⇒ 1 – 4(1 – a) < 0 ⇒ 4a < 3 ⇒ a < 

 ∴ a can not take the value