Multiple Choice Questions on Complex Numbers Class 11 OP Malhotra Exe-9H ISC Maths Solutions Ch-9. In this article you would learn to solve all types mcqs on Complex Numbers. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11

Complex Numbers Class 11 OP Malhotra Multiple Choice Questions ISC Maths Solutions Ch-9

Board	ICSE
Publications	S Chand
Subject	Maths
Class	11th
Chapter-9	Complex Numbers
Writer	OP Malhotra
Exe-9(H)	Multiple Choice Questions.

Multiple Choice Questions on Complex Numbers

OP Malhotra ISC Class 11 Maths Solutions

Que-1: Find the value of 1 + i² + i³ − i⁶ + i⁵

(a) −1         (b) 2 − i           (c) 1          (d) 3

Sol: (c) 1
i² = −1, i³ = −i, i⁵ = i, i⁶ = −1
Substitute → 1 − 1 − i + 1 + i = 1

Que-2: If a + bi = c + di, then

(a) a² + c² = 0   
(b) b² + c² = 0   
(c) b² + d² = 0   
(d) a² + b² = c² + d²

Sol: (d) a² + b² = c² + d²
Equal complex numbers ⇒ a=c and b=d.

Que-3: In which quadrant of the complex plane, the point (1+2i)/(1-i) lies?

(a) fourth       (b) first         (c) second         (d) third

Sol: (c) second
(1+2i)/(1−i) = (−1 + 3i)/2
→ Real negative, Imaginary positive
⇒ Second quadrant.

Que-4: If z = {(1-i√3)/(1+i√3)}, then arg(z) is

(a) 60°       (b) 120°        (c) 240°        (d) 300°

Sol:  (c) 240°
After simplification z lies in third quadrant
⇒ argument = 240°.

Que-5: The imaginary part of {(1+i)/i(2i-1)} is

(a) −4/5        (b) −2/5         (c) 1/5         (d) 0

Sol: (a) -4/5
Simplifying the complex fraction gives imaginary part
= −4/5.

Que-6: The modulus of [{(1+i)/(1-i)} – {(1-i)/(1+i)}] is

(a) 2        (b) √2         (c) 4         (d) 8

Sol: (a) 2
(1+i)/(1−i) = i and (1−i)/(1+i) = −i
Expression = i − (−i)
= 2i → |2i| = 2

Que-7: The least positive integer n for which [(1+i√3)/(1-i√3)]^n  is

(a) 3    (b) 5    (c) 2    (d) 6

Sol: (a) 3
(1+i√3)/(1−i√3) = cos(120°)
Smallest n such that (cis120°)ⁿ=1
⇒ n = 3

Que-8: The value of {(1+i)/(1-i)}^1000 + {(1-i)/(1+i)}^200  is equal to

(a) 1 + i    (b) 2    (c) −i    (d) 1

Sol: (b) 2
(1+i)/(1−i) = i
i¹⁰⁰⁰=1 and (−i)²⁰⁰=1
Sum = 2

Que-9: If {(1+i)(2+3i)(3-4i)}/{(2-3i)(1-i)(3+4i)} = a + bi, then a² + b² =

(a) 132    (b) 25    (c) 144    (d) 1

Sol: (d) 1
Using modulus property: |z₁z₂/z₃| = |z₁||z₂|/|z₃|
Result modulus = 1
⇒ a²+b² =1

Que-10: If z = x + iy is a complex number such that |z| = Re(iz) + 1, then the locus of z is

(a) x² + y² = 1
(b) x² = 1 − 2y
(c) y² = 2x − 1
(d) x² = 2y − 1

Sol:  (b) x² = 1 − 2y
|z| = √(x²+y²) and Re(iz) = −y
√(x²+y²) = 1−y
⇒ x² = 1−2y

Que-11: If a complex number lies in the III quadrant find the quadrant in which its conjugate lies.

(a) I quadrant    (b) II quadrant    (c) III quadrant    (d) IV quadrant

Sol:  (b) II quadrant
If z = x + iy lies in III quadrant then x < 0, y < 0.
Conjugate = x − iy
⇒ x < 0, y > 0 → II quadrant.

Que-12: The complex number z which satisfies |(i + z)/(i − z)| = 1 lies on

(a) circle x² + y² = 1
(b) the x-axis
(c) the y-axis
(d) the line x + y = 1

Sol:  (b) the x-axis
|(i+z)/(i−z)| = 1
⇒ |i+z| = |i−z|
Points equidistant from i and −i lie on the x-axis.

Que-13: If z is a complex number such that z = − z

(a) z is purely real
(b) z is purely imaginary
(c) z is any complex number
(d) real part of z is same as its imaginary part

Sol: (b) z is purely imaginary
Let z = x + iy → z̅ = x − iy
x + iy = −x + iy
⇒ x = 0 → purely imaginary.

Que-14: If w is a cube root of unity, then the value
(1 − w + w²)⁵ + 5(1 + w − w²)⁵ is

(a) 30    (b) 32    (c) 2    (d) None of these

Sol: (b) 32
Using cube root property: w³ = 1 and 1 + w + w² = 0.
After simplification
⇒ value = 32.

Que-15: If w is a complex cube root of unity, then the value of sin { (w¹⁰ + w²³)π − π/6 } is

(a) 1/√2        (b) √3/2         (c) −1/2         (d) 1/2

Sol: (d) 1/2
Since w³ = 1 → w¹⁰ = w, w²³ = w².
w + w² = −1.
So sin(−π − π/6)
= sin(−7π/6) = 1/2.

Que-16: If √i = x + iy, then the positive value of x

(a) 1    (b) 0    (c) 2    (d) 1/√2

Sol: (d) 1/√2
Since √i = (1+i)/√2
⇒ x = 1/√2

Que-17: Arg (2i) is

(a) π/2    (b) −π/2    (c) −π    (d) π

Sol: (a) π/2
2i lies on positive imaginary axis
⇒ argument = π/2

Que-18: If the conjugate of a complex number z is {1/(i-1)}, then z is

(a) 1/(i − 1)   
(b) 1/(i + 1)   
(c) −1/(i − 1)   
(d) −1/(i + 1)   
(e) 1/i

Sol: (d) −1/(i + 1)
Since conjugate of z = 1/(i−1),
therefore z = −1/(i+1).

Que-19: The real value of θ for which {(1 + i cosθ)/(1 – 2i cosθ)}is a real number is

(a) nπ + π/4
(b) nπ + (−1)ⁿ π/4
(c) 2nπ ± π/2
(d) None of these

Sol: (c) 2nπ ± π/2
For real number ⇒ imaginary part = 0
⇒ cosθ = 0
⇒ θ = 2nπ ± π/2.

Que-20: The amplitude of the complex number 1 + sin α − i cos α is

(a) π/4
(b) α − π/4
(c) α/2 − π/4
(d) π/4 − α

Sol: (c) α/2 − π/4
Using argument formula of complex numbers.

Que-21: sin x + i cos 2x and cos x − i sin 2x are conjugate to each other for

(a) x = nπ
(b) x = (n + 1/2)π/2
(c) x = 0
(d) No value of x

Sol: (c) x = 0
For conjugate numbers imaginary parts must be opposite.

Que-22: If α = cos θ + i sin θ, then (1 + α) / (1 − α) equals

(a) i cot(θ/2)
(b) i tan(θ/2)
(c) cot(θ/2)
(d) tan θ

Sol: (a) i cot(θ/2)
Using Euler form and half angle identities.

Que-23: If z is a complex number of unit modulus and argument θ, then arg ((1 + z) / (1 + z̄)) equals

(a) π/2 − θ        (b) θ          (c) π − θ           (d) −θ

Sol: (b) θ
Using properties of arguments of complex numbers.

Que-24: The locus of the point z in the argand plane for which |z + 1|² + |z − 1|² = 4 is a

(a) straight line
(b) pair of straight lines
(c) parabola
(d) circle

Sol:  (d) circle
Simplifying gives x² + y² = constant (circle).

–: End Complex Numbers Class 11 OP Malhotra Exe-9H ISC Maths Ch-9 Solutions :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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