JEE Advanced (Fill in the Blanks): Complex Numbers | Chapter-wise Tests for JEE Main & Advanced PDF Download

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Q. 1. If the expression            (1987 - 2 Marks)

is real, then the set of all possible values of x is ............

Ans. 

Sol.

But ATQ, I_m(z) = 0 (as z is real)

⇒ cos x = 1 ⇒ x = 2nπ and  

tan x  = 1 ⇒ x= nπ + π/4

∴ x = 2nπ, nπ + π/4

Q. 2. For any two complex numbers z₁, z₂ and any real number a and b. (1988 - 2 Marks) | az₁ – bz₂ |² + | bz₁ + az₂ |² = .............

Ans. (a² + b² )(| z₁ |²+ |z₂ |²)

Sol. 

Q. 3. If a, b, c, are the numbers between 0 and 1 such that the points z₁ = a + i, z₂ = 1 + bi and z₃ = 0 form an equilateral triangle, then a = .......and b = ........... (1989 - 2 Marks)

Ans. 

KEY CONCEPT : | z₁ - z₂ |= distance between two points represented by z₁ and z₂.

As z₁ = a + i, z₂ = 1+ bi and z₃ = 0 form an equilateral triangle, therefore

|z₁ – z₃| = |z₂ – z₃| = |z₁ – z₂|

| a + i | = | 1+ bi | = | ( a – 1) + i (1– b) |

⇒ a² + 1= 1+ b² = (a – 1)² + (1– b)²

⇒ a² = b² = a² + b² – 2a – 2b + 1

⇒ a = b ....(1)

(∴ a, b > 0       ∴ a≠ – b ) and

b² – 2a – 2b + 1= 0

or a² – 2a – 2b + 1 = 0 ....(2)

⇒ a² – 2a – 2a + 1= 0 [∴a = b]

⇒ a² – 4a + 1 = 0

Q. 4. ABCD is a rhombus. Its diagonals AC and BD intersect at the point M and satisfy BD = 2AC. If the points D and M represent the complex numbers 1 + i and 2 - i respectively, then A represents the complex number .........or.......... (1993 -  2 Marks)

Ans. 

Sol :

If we see the problem as in co-ordinate geometry we have D ≡ (1,1) and M≡ (2, – 1)

We know that diagonals of rhombus bisect each other at 90°

∴ AC passes through M and is ^ to BD

∴ Eq. of AC in symmetric form can be written as

Now for pt. A, as

Q. 5. Suppose Z₁, Z_2, Z₃ are the vertices of an equilateral triangle inscribed in the circle |Z| = 2. If Z₁ = 1 +  then Z₂ = ........, Z₃ = ............ (1994 -  2 Marks)

Ans. –2,  1 -

Sol : 

Let z₁, z₂,z₃ be the vertices A, B and C respectively of equilateral ΔABC, inscribed in a circle | z | = 2, centre (0, 0) rasius = 2

Q. 6. The value of the expression

where w is an imaginary cube root of unity, is..... (1996 - 2 Marks) 

Ans.  n (n - 1)(n² + 3n + 4)

Sol: 

rth term of the given series,

= r [(r +1) –ω](r +1) –w2 ]

= r [(r +1)² – (ω+ω² )(r +1) +ω³]

= r [(r +1)² – (-1)(r +1) +1]

= r [(r² + 3r + 3] = r³ + 3r² + 3r

Thus, sum of the given series,