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# Fill in the Blanks: Complex Numbers | JEE Advanced Notes | Study Maths 35 Years JEE Main & Advanced Past year Papers - JEE

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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, Im(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, zand any real number a and b. (1988 - 2 Marks) | az1 – bz2 |2 + | bz1 + az2 |2 = .............

Ans. (a+ 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, Z2, Z3 are the vertices of an equilateral triangle inscribed in the circle |Z| = 2. If Z1 = 1 +  then Z2 = ........, Z= ............ (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)(n+ 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,

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## Maths 35 Years JEE Main & Advanced Past year Papers

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## Maths 35 Years JEE Main & Advanced Past year Papers

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