1 Crore+ students have signed up on EduRev. Have you? 
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 z_{1}, z_{2 }and any real number a and b. (1988  2 Marks)  az_{1} – bz_{2} ^{2} +  bz_{1} + az_{2} ^{2} = .............
Ans. (a^{2 }+ b^{2} )( z_{1} ^{2}+ z_{2} ^{2})
Sol.
Q. 3. If a, b, c, are the numbers between 0 and 1 such that the points z_{1} = a + i, z_{2} = 1 + bi and z_{3} = 0 form an equilateral triangle, then a = .......and b = ........... (1989  2 Marks)
Ans.
KEY CONCEPT :  z_{1}  z_{2} = distance between two points represented by z_{1} and z_{2}.
As z_{1} = a + i, z_{2} = 1+ bi and z_{3} = 0 form an equilateral triangle, therefore
z_{1} – z_{3} = z_{2} – z_{3} = z_{1} – z_{2}
 a + i  =  1+ bi  =  ( a – 1) + i (1– b) 
⇒ a^{2} + 1= 1+ b^{2} = (a – 1)^{2} + (1– b)^{2}
⇒ a^{2} = b^{2} = a^{2} + b^{2} – 2a – 2b + 1
⇒ a = b ....(1)
(∴ a, b > 0 ∴ a≠ – b ) and
b^{2} – 2a – 2b + 1= 0
or a^{2} – 2a – 2b + 1 = 0 ....(2)
⇒ a^{2} – 2a – 2a + 1= 0 [∴a = b]
⇒ a^{2} – 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 coordinate 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 Z_{1}, Z_{2,} Z_{3} are the vertices of an equilateral triangle inscribed in the circle Z = 2. If Z_{1} = 1 + then Z_{2} = ........, Z_{3 }= ............ (1994  2 Marks)
Ans. –2, 1 
Sol :
Let z_{1}, z_{2},z_{3} 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^{2 }+ 3n + 4)
Sol:
rth term of the given series,
= r [(r +1) –ω](r +1) –w2 ]
= r [(r +1)^{2} – (ω+ω^{2} )(r +1) +ω^{3}]
= r [(r +1)^{2} – (1)(r +1) +1]
= r [(r^{2} + 3r + 3] = r^{3} + 3r^{2} + 3r
Thus, sum of the given series,
132 docs70 tests
