Instructions
- Each question carry 2 Marks
- Total 40 marks test, 20 questions
- Time- 90 minutes
- Answer key will be provided after the exam get over
Q1) The value of √(-121) is
(a) -11i
(b) 11i
(c) -12i
(d) 12i
Q2) The value of √(-441) is
(a) 21i
(b) -21i
(c) ±21i
(d) All of these
Q3) The value of √(-25) + 3√(-4) + 2√(-9) is
(a) 13i
(b) -13i
(c) 17i
(d) -17i
Q4) If z lies on |z| = 1, then 2/z lies on
(a) a circle
(b) an ellipse
(c) a straight line
(d) a parabola
Q5) If ω is an imaginary cube root of unity, then (1 + ω – ω²)7 equals
(a) 128 ω
(b) -128 ω
(c) 128 ω²
(d) -128 ω²
Q6) The least value of n for which {(1 + i)/(1 – i)}n is real, is
(a) 1
(b) 2
(c) 3
(d) 4
Q7) Let z be a complex number such that |z| = 4 and arg(z) = 5π/6, then z =
(a) -2√3 + 2i
(b) 2√3 + 2i
(c) 2√3 – 2i
(d) -√3 + i
Q8) The value of i-999 is
(a) 1
(b) -1
(c) i
(d) -i
Q9) Let z1 and z2 be two roots of the equation z² + az + b = 0, z being complex. Further assume that the origin, z1 and z1 form an equilateral triangle. Then
(a) a² = b
(b) a² = 2b
(c) a² = 3b
(d) a² = 4b
Q10) The complex numbers sin x + i cos 2x are conjugate to each other for
(a) x = nπ
(b) x = 0
(c) x = (n + 1/2) π
(d) no value of x
Q11) The curve represented by Im(z²) = k, where k is a non-zero real number, is
(a) a pair of striaght line
(b) an ellipse
(c) a parabola
(d) a hyperbola
Q12) The value of x and y if (3y – 2) + i(7 – 2x) = 0
(a) x = 7/2, y = 2/3
(b) x = 2/7, y = 2/3
(c) x = 7/2, y = 3/2
(d) x = 2/7, y = 3/2
Q13) Find real θ such that (3 + 2i × sin θ)/(1 – 2i × sin θ) is imaginary
(a) θ = nπ ± π/2 where n is an integer
(b) θ = nπ ± π/3 where n is an integer
(c) θ = nπ ± π/4 where n is an integer
(d) None of these
Q14) If {(1 + i)/(1 – i)}n = 1 then the least value of n is
(a) 1
(b) 2
(c) 3
(d) 4
Q15) If arg (z) < 0, then arg (-z) – arg (z) =
(a) π
(b) -π
(c) -π/2
(d) π/2
Q16) if x + 1/x = 1 find the value of x2000 + 1/x2000 is
(a) 0
(b) 1
(c) -1
(d) None of these
Q17) The value of √(-169) is
(a) 13i
(b) -13i
(c) ±13i
(d) None of these
Q18) If the cube roots of unity are 1, ω, ω², then the roots of the equation (x – 1)³ + 8 = 0 are
(a) -1, -1 + 2ω, – 1 – 2ω²
(b) – 1, -1, – 1
(c) – 1, 1 – 2ω, 1 – 2ω²
(d) – 1, 1 + 2ω, 1 + 2ω²
Q19) (1 – w + w²)×(1 – w² + w4)×(1 – w4 + w8) × …………… to 2n factors is equal to
(a) 2n
(b) 22n
(c) 23n
(d) 24n
Q20) The modulus of 5 + 4i is
(a) 41
(b) -41
(c) √41
(d) -√41
No comments:
Post a Comment