Very Short Answer on Complex Numbers Class 11 OP Malhotra Exe-9G ISC Maths Solutions Ch-9. In this article you would learn to solve all types questions 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 Very Short Answer ISC Maths Solutions Ch-9

Board	ICSE
Publications	S Chand
Subject	Maths
Class	11th
Chapter-9	Complex Numbers
Writer	OP Malhotra
Exe-9(G)	Revision and Self Assessment.

Very Short Answer on Complex Numbers

OP Malhotra ISC Class 11 Maths Solutions

Que-1: Evaluate :
(i) i⁻⁵⁰ + i⁶⁰
(ii) i⁴ + i⁵ + i⁶ + i⁷
(iii) {(1 − i)³}/(1 − i³)
(iv) {(1 + i)/(1 − i)}⁹⁶

Sol: (i) i^-50 + i⁶⁰
i^-50 = 1 / i⁵⁰ = 1 / i² = -1
i⁶⁰ = (i⁴)¹⁵ = 1
Answer: -1 + 1 = 0

(ii) i⁴ + i⁵ + i⁶ + i⁷
i⁴ = 1
i⁵ = i
i⁶ = -1
i⁷ = -i
Answer: 1 + i – 1 – i = 0

(iii) (1 − i)³ / (1 − i³)
i³ = -i
1 − i³ = 1 + i
(1 − i)³ = -2 − 2i
(-2 − 2i)/(1 + i)
= -2

(iv) ((1 + i)/(1 − i))⁹⁶
(1 + i)/(1 − i) = i
i⁹⁶ = (i⁴)²⁴ = 1

Que-2: If 5 + yi − 3i = x − 2i, find y.

Sol:  5 + (y − 3)i = x − 2i
Comparing imaginary parts:
y − 3 = −2
y = 1

Que-3: Write the additive inverse of
(i) −7 + 9i
(ii) 3/(1 − i)

Sol:  (i) Additive inverse of a complex number z is −z.
−(−7 + 9i) = 7 − 9i

(ii) Additive inverse = −3/(1 − i)

Que-4: If z = 2 + √3i, find the value of zz.

Sol:  z = 2 + √3i
z = 2 − √3i
zz = (2 + √3i)(2 − √3i)
= 4 + 3 = 7

Que-5:   If (1 + i)z = (1 − i) z, then show that x = −iz.

Sol:  Let z = x + iy
(1 + i)(x + iy) = (1 − i)(x − iy)
x − y + i(x + y) = x − y − i(x + y)
Comparing imaginary parts:
x + y = −(x + y)
⇒ x = −y

Que-6: Find the multiplicative inverse of (4 − 3i).

Sol:  1/(4 − 3i) × (4 + 3i)/(4 + 3i)
= (4 + 3i)/(16 + 9)
= (4 + 3i)/25

Que-7: Express in the form a + bi.
(i) i⁻³⁵
(ii) 5i (-3i/5)
(iii) i⁹ + i¹⁰ + i¹¹ + i¹²

Sol:  (i) i⁻³⁵ = 1/i³⁵
i³⁵ = i³ = −i
Therefore i⁻³⁵
= i

(ii) = −15i²/5
= −3(−1)
= 3

(iii) i⁹ = i
i¹⁰ = −1
i¹¹ = −i
i¹² = 1
Sum = i − 1 − i + 1
= 0

Que-8: If
{(1 − i)/(1+i)}¹⁰⁰
 =  a + bi, then find (a, b).

Sol:  (1−i)/(1+i) = -i
(-i)¹⁰⁰ = (i)¹⁰⁰ = 1
∴ a = 1 , b = 0

Que-9: If {(1 + i)²}/(2-i)
 = x + yi, then find the value of x + y.

Sol:  (1+i)² = 1 +2i + i² = 2i
2i/(2-i) × (2+i)/(2+i) = (4i +2i²)/5
= (-2 +4i)/5
x = -2/5 , y = 4/5
x + y = 2/5

Que-10: Find the modulus of (1+i)/(1-i)

Sol: | (1+i)/(1-i) | = |1+i| / |1-i|
= √2 / √2 = 1

Que-11: Find the conjugate of {(3-2i)(2+3i)}/{(1+2i)(2-i)}

Sol:  (3−2i)(2+3i) = 12 +5i
(1+2i)(2−i) = 4 +3i
(12+5i)/(4+3i) = (63−16i)/25
Conjugate = (63 +16i)/25

Que-12: Express in polar form:
(i) √9
(ii) (i²⁹)³
(iii) (1+3i)/(1-2i)

Sol:  (i) √9
= 3(cos0 + i sin0)

(ii) (i²⁹)³
= i⁸⁷ = -i
= 1 (cos270° + i sin270°)

(iii) (1+3i)/(1-2i)
= (-5+5i)/5 = -1 + i
r = √2 , θ = 135°
= √2 (cos135° + i sin135°)

Que-13: Find the argument of
(i) 1 + i
(ii) 1/(1-i)
(iii) (1+i)/(1-i)

Sol:  (i) 1 + i
|1+i| = √2
Arg(1+i) = π/4

(ii) 1/(1-i)
1/(1-i) = (1+i)/2
Arg = π/4

(iii) (1+i)/(1-i)
= i
Arg(i) = π/2

Que-14: (i) Write the principal argument of z = √3 + 3i.
(ii) The amplitude of −1 + √−3 is ______.

Sol:  (i) Principal argument of z = √3 + 3i
tanθ = 3/√3 = √3
θ = π/3

(ii) Amplitude of −1 + √−3
= −1 + i√3
Arg = 2π/3

Que-15: arg (z) + arg (z) where, (z ≠ 0) is ______.

Sol:  arg(z) + arg(z̅)
= 0

Que-16: If |z| = 4 and arg (z) = 5π/6, then z = ______.

Sol:  If |z| = 4 and arg(z) = 5π/6
z = 4[cos(5π/6) + i sin(5π/6)]

Que-17: If z = x + iy, then show that zz + 2 (z + z) + b = 0, where b ∈ R represents a circle.

Sol:  Given z = x + iy
z z̅ = x² + y²
z + z̅ = 2x
x² + y² + 4x + b = 0
This represents a circle.

Que-18: Find the number of non-zero integral solutions of the equation |1 − i|^x = z^x.

Sol:  |1 − i| = √2
Equation: (√2)^x = z^x
Non-zero integral solution: x = 0

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

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