JEE Advanced (One or More Correct Option): Complex Numbers | Chapter-wise Tests for JEE Main & Advanced PDF Download

Q.1. If z₁ and z₂ are two complex numbers such that |z₁ + z₂|² = |z₁ + z₂|² then
(a) z₁/z₂ is purely real
(b) z₁/z₂ is purely imaginary
(c)
(d)

Correct Answer is options (b, c)

=
We have,

So, z₁/z_{2 is purely imaginary.}

Q.2. If (a cos θ₁, a sin θ₁), (a cos θ₂, a sin θ₂), (a cos θ₃, a sin θ₃) represents the vertices of an equilateral triangle inscribed in a circle, then
(a) cos θ₁ + cos θ₂ + cos θ₃ = 0
(b) sin θ₁ + sin θ₂ + sin θ₃ = 0
(c) tan θ₁ + tan θ₂ + tan θ₃ = 0
(d) cot θ₁ + cot θ₂ + cot θ₃ = 0

Correct Answer is options (a, b)
Choices (A) and (B) are true since origin is centre of the given circle and in an equilateral triangle centroid and circumcentre are same point.
⇒

Q.3. If  and z₁, z₂ , z₃ are non-zero complex numbers, then
(a)  is purely real
(b) z₂/z₁ is purely real
(c)  is purely real
(d)  is purely real

Correct Answer is options (a, b)
We have
 is purely real
⇒  is purely real ⇒ z₂/z₁ is purely real

Q.4. Let Z₁ and Z₂ be complex numbers such that Z₁ ≠ Z2 and |Z₁| = |Z₂| .  If Z₁ has positive real part and Z₂ has negative imaginary part then  may be
(a) Zero
(b) real and positive
(c) real and negative
(d) purely imaginary

Correct Answer is options (a, d)
Given, |z₁| = |z₂| ,
Now,
=
=
As, we know
∴
∴
Which is purely imaginary or zero. Therefore, (A) and (D) are correct answers.

Q.5. If z ≠ 0 is a complex number, then z, iz, -z and -iz are the vertices of a
(b) square
(b) rectangle
(c) rhombus
(d) parallelogram, which is not a rectangle 

Correct Answer is options (a, b, c)
Suppose, z, iz, -z and -iz are represented by A, B, C and D in the complex plane. We have mid-point of AC is  . and mid-point of BD is

Thus, the diagonals bisect each other.
Also,  AC =| -z - z |= 2 | z | and   BD =| -iz - iz |= 2 | -i || z |= 2 | z |
∴ AC = BD
Finally,  and
Hence, ABCD is a square.

Q.6. If 1, ω, ω², &. ω^n-1 are the n^th roots of unity, then
(a) 2ⁿ -1
(b) ⁿC₁ + ⁿC₂ + .... + ⁿC_n-1 + ⁿC_n
(c)
(d) none of these

Correct Answer is options (a, b)
We have
⇒
Putting z = 2, we get

Q.7. If ω ≠ 1 is a cube root of unity and , then  
(a) A is singular
(b) |A| = 0
(c) A is symmetric
(d) A is skew-symmetric

Correct Answer is options (a, b)
We have

=
 (taking ω common from C₂)
Thus,  |A| = 0 and hence A is singular.  

Q.8. Let z₁, z₂, z₃ be the vertices of a triangle ABC. Then which of the following statements is correct?
(a) If , where , then ABC is an equilateral triangle.
(b) If ABC is an equilateral triangle then , where
(c) If  then the triangle ABC is equilateral
(d) If |z₁| = |z₂ |= |z₃| and z₁ + z₂ + z₃ = 0 , then the triangle ABC is equilateral.

Correct Answer is options (a, b, c, d)
A necessary and sufficient condition for a triangle having vertices z₁, z₂ and z₃ to form an equilateral triangle is  
(A) and (B) will follow by performing some algebraic jugglery on the known condition given above.
To prove (D) note that z₁ + z₂ + z₃ = 0 can be changed to

Q.9. If z₁, z₂, z₃, z₄ are the vertices of a square in that order, then
(a) z₁ + z₃ = z₂ + z₄
(b)
(c)
(d)  is a real number

Correct Answer is options (a, b, c)
Let the four points represented by z₁, z₂, z₃ and z₄ be A, B, C and D respectively.
Since ABCD is a square, the mid point AC = the mid point of BD.
⇒  
Also,  AB = BC = CD = DA.
⇒
Since diagonals of the square ABCD are equal
AC = BD   or   |z₁ – z₃| = |z₂ – z₄|
Also, since AC ⊥ BD, (z₁- z3)/(z₂ - z₄) is purely imaginary.

Q.10. If z lies on the circle centered at origin and if area of the triangle, whose vertices are z, ωz and z + ωz, (ω being an imaginary cube root of unity), is 4√3 sq. units. Then radius of the circle is
(a) 1 unit
(b) 2 units
(c) 3 units
(d) 4 units

Correct Answer is option (d)

Area of the triangle =
⇒ a² = 16
⇒ a = 4