JEE Advanced (Matrix Match): Complex Numbers | Chapter-wise Tests for JEE Main & Advanced PDF Download

DIRECTIONS (Q. 1 and 2) : Each question contains statements given in two columns, which have to be matched. The statements in Column-I are labelled A, B, C and D, while the statements in ColumnII are labelled p, q, r, s and t. Any given statement in Column-I can have correct matching with ONE OR MORE statement(s) in Column-II. The appropriate bubbles corresponding to the answers to these questions have to be darkened as illustrated in the following example : If the correct matches are A-p, s and t; B-q and r; C-p and q; and D-s then the correct darkening of bubbles will look like the given.

Q. 1. z ≠ 0 is a complex number (1992 -  2 Marks)

Column I                               Column II
 (A) Rez = 0                         (p) Rez² = 0
 (B) Argz =                       (q) Imz² = 0
                                              (r) Rez ² = Imz²

Ans. z ≠ 0 Let z = a + ib Re (z) = 0 ⇒ z = ib
⇒ z² = – b²
∴ Im (z)2 = 0
∴ (A) corresponds to (q)

Arg = ⇒ a = b ⇒ z = a + ia

z² = a² – a² + 2ia² ; z² = 2ia² ⇒ Re (z)² = 0
∴ (B) corresponds to (p).

Q. 2. Match the statements in Column I with those in Column II.  (2010) [Note : Here z takes values in the complex plane and Im z and Re z denote , respectively, the imaginary part and the real part of z.] 

Column  I	Column  II
The set of points z satisfying |z – i| z | | = |z + i | z || is contained in or equal to	(p) an ellipse with eccentricity
(B) The set of points z satisfying |z + 4 | + |z – 4 | = 10 is contained in or equal to	(q) the set of points z satisfying Im z = 0
(C) If | w | = 2, then the set of points z = w –  is contained in or equal to	(r) the set of points z satisfying |Im z | ≤ 1
(D) If | w | = 1, then the set of points z = w +    is contained in or equal to	(s) the set of points z satisfying | Re z | < 2
	(t) the set of points z satisfying | z | ≤ 3

Ans.  

⇒ z is equidistant from two points ( 0, |z|) and (0,– |z|) which lie on imaginary axis.

∴ z  must lie on real axis ⇒ Im ( z )=0 also |I_m(z)| ≤1

(B) → p

Sum of distances of z from two fined points (–4, 0) and (4, 0) is 10 which is greater than 8.
∴ z traces an ellipse with 2a = 10 and 2ae =8

⇒ e= 4/5

(C) → (p, s, t)

Let ω = 2(cosθ + i sinθ)

then z = ω -   (cosθ + i sinθ) - (cosθ + i sinθ)

⇒ x + iy

Here  

Also x = ⇒

Which is an ellipse with e =

(D) → (q,r, s,t)
Let _ω = cosθ + i sinq then z = 2 cosθ ⇒ Imz=0

Also z ≤ 3 and  | Im( z) |≤ 1, | R_e (z) |≤ 2

DIRECTIONS (Q. 3) : Following question has matching lists. The codes for the list have choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Q. 3. Let z_k =  + sin : k=1,2,......,9.                            (JEE Adv. 2014)

List-I	List-II
P. For each zk there exists as zj such that zk. zj = 1	1. True
Q. There exists a k ∈ {1, 2,...,9} such that z1.z = zk has no solution z in the set of complex numbers	2. False
R.  equals	3. 1
S.  equals	4. 2

      P    Q    R    S                                       P    Q    R    S
 (a) 1      2     4    3                             (b) 2      1     3    4 

(c) 1      2     3    4                              (d) 2      1     4    3

Ans. (c)

 (P) → (1) :, k = 1 to 9

Now z_k.z_j = 1 ⇒  =

We know if z_k is 10^th root of unity so will be

∴ For every z_k, there exist z_i =

Such that z_k . z _j = z_k.= 1

Hence the statement is true.

(Q) → (2) z₁ =z k ⇒    for z₁ ≠ 0

∴ We can always find a solution to z₁.z =z_k 

Hence the statement is false.

(R) → (3) : We know z¹⁰ - 1 = ( z - 1)( z - z₁ ) .... ( z-z₉ )

⇒ ( z - z₁)( z - z₂) .... ( z-z₉)

= 1 + z + z² + ... z⁹

For z = 1 we get

(1 - z₁) (1 - z₂) ..... (1 -z₉)= 10

(S) → (4) : 1, Z₁, Z₂, ... Z₉ are 10^th roots of unity.      

∴ Z¹⁰ – 1 = 0

From equation 1 + Z₁ + Z₂ + .... + Z₉ = 0

Re (1) + Re (Z₁) + Re (Z₂) + .... + Re(Z₉) = 0

⇒ Re (Z₁) + Re (Z₂) + ..... Re(Z₉) = – 1

Hence (c) is the correct option.