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# Matrix-Match Type Questions: Complex Numbers | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

## Document Description: Matrix-Match Type Questions: Complex Numbers | JEE Advanced for JEE 2022 is part of Maths 35 Years JEE Main & Advanced Past year Papers preparation. The notes and questions for Matrix-Match Type Questions: Complex Numbers | JEE Advanced have been prepared according to the JEE exam syllabus. Information about Matrix-Match Type Questions: Complex Numbers | JEE Advanced covers topics like and Matrix-Match Type Questions: Complex Numbers | JEE Advanced Example, for JEE 2022 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Matrix-Match Type Questions: Complex Numbers | JEE Advanced.

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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

z= a2 – a2 + 2ia2 ; z= 2ia⇒ Re (z)= 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 pointsz = 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 pointsz = 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 |Im(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, | Re (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 zk =  + 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 = zkhas 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 zk.z= 1 ⇒  =

We know if zis 10th root of unity so will be

∴ For every zk, there exist zi =

Such that zk . z j = zk.= 1

Hence the statement is true.

(Q) → (2) z=z k ⇒    for z1 ≠ 0

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

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 + z+ ... z9

For z = 1 we get

(1 - z1) (1 - z2) ..... (1 -z9)= 10

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

∴ Z10 – 1 = 0

From equation 1 + Z1 + Z+ .... + Z= 0

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

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

Hence (c) is the correct option.

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