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NCERT Class 11 Mathematics Chapter 4 Complex Number and Quadratic Equation
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Complex Number and Quadratic Equation
Chapter – 4
| Exercise 4.1 |
Express each of the complex number given in the Exercises 1 to 10 in the form a + ib.
Ans:
= – 3i²
= -3(-1) [i² = – 1]
= 3
Q2: i⁹ + i¹⁹.
Ans: i⁹ + i¹⁹ = i⁴ˣ²⁺¹ + i⁴ˣ⁴⁺³
= (i⁴)² .i + (i⁴)⁴ .i³
= 1 x i + 1x (- i) [ i⁴ =1, i³ = -i]
= 0
Q3: i⁻³⁹.
Ans:
Q4. 3(7 + i7) + i(7 + i7).
Ans: 3(7+i7)+i(7+i7)= 21 + 21i + 7i + 7i²
= 21 + 28i + 7x(- 1) [∵i² = -1]
= 14 + 28i
Q5. (1 – i) -(-1+i6).
Ans : (1 – i)-(-1+i6)= 1 – i + 1 – 6i
= 2 – 7i
Q6:
Ans:
Q7
. 
Ans:
Q8. (1 – i)⁴.
Ans: (1 – i)⁴ = [(1 – i)²]²
= [I² + i² – 2i]²
= [1 – 1 – 2i]²
=(-2i)²
= (- 2i) × (- 2i)
= 4i² = – 4
[i² = – 1]
Q9:
Ans:
Q10
:
Ans:
Find the multiplicative inverse of each of the complex numbers given in the Exercises 11 to 13.
Q11: 4 – 3i.
Ans: Let z= 4 – 3i
Then, Z̅ = 4 + 3i and |z|² = 4² + (- 3)² = 16 + 9 = 25
Therefore, the multiplicative inverse of 4 – 3i is given by
Q12: √5 + 3i
Ans: Let z = √5 + 3i
Then, Z̅ = √5 – 3i and |z|² = (√5)² + 3² = 5 + 9 = 14
Therefore, the multiplicative inverse of √5 + 3i
is given by
Q13: – i.
Ans: Let z = – i
Then, Z̅ = i and |z|² = 1² = 1
Therefore, the multiplicative inverse of – i is given by
Q14: Express the following expression in the form of a + ib.
Ans:
| Miscellaneous Exercise on Chapter 4 |
Ans:
= [ -1-i ]³
=(-1)³ [1+I]³
= – [1³ + i³ + 3.1.i(1 + i)]
= – [1 – i³ + 3i – 3i²]
= – [1 – i + 3i – 3]
= – [- 2 + 2i]
= 2 – 2i
Q2: For any two complex numbers Z₁and Z₂ prove that Re (z₁ z₂) = Re z₁Re z₂ – Im z₁ Im Z₂
Ans: Let z₁ = x₁ + iy₁ and z₂ = x₂ + iy₂
∴ z₁ z₂ = (x₁ + iy₁)(x₂ + iy₂)
= x₁ (x₂ + iy₂) + iy₁(x₂ + iy₂)
= x₁ x₂ + ix₁ y₂ + iy₁x₂ + i² y₁y₂
= x₁x₂ + ix₁ y₂ + iy₁ x₂ – y₁y₂ [i²= – 1]
= (x₁x₂ – y₁y₂) + i(x₁y₂ + y₁x₂)
⇒Re(z₁z₂) = x₁x₂ – y₁y₂
⇒Re(z₁z₂) = Re z₁Rez₂ – Im z₁ Im z₂
Hence, proved.
Ans:
Ans:
Ans:
Ans:
7. Let z₁ = 2 – i, z₂= – 2 + i Find.
Ans: z₁ z₂ = (2 – i)(- 2 + i) = – 4 + 2i + 2i – i² = – 4 + 4i – (- 1) = – 3 + 4i
Z̅₁= 2 + i
On multiplying numerator and denominator by (2 – i), we obtain
Ans:
8. Find the real numbers x and y if (x – iy)(3 + 5i) is the conjugate of – 6 – 24i.
Ans: Let z = (x – iy)(3 + 5i)
z = 3x + 5xi – 3yi – 5yi² = 3x + 5xi – 3yi + 5y = (3x + 5y) + i(5x – 3y)
∴ Z̅ = (3x + 5y) – i(5x – 3y)
It is given that, Z̅ = – 6 – 24i
∴ (3x + 5y) – i(5x – 3y) = – 6 – 24i
Equating real and imaginary parts, we obtain
3x + 5y = – 6 ……(i)
5x – 3y = 24 ……(ii)
Multiplying equation (i) by 3 and equation (ii) by 5 and then adding them, we obtain
9x + 15y = – 18
Putting the value of x in equation (i), we obtain
3(3) + 5y = – 6
⇒ 5y = – 6 – 9 = – 15
⇒ y = – 3
Thus, the values of x and y are 3 and – 3 respectively.
Ans:
Ans: (x + iy)³ = u + iv
⇒x³ + (iy)³+ 3.x.iy(x + iy) = u + iv
⇒x³ + i³ y³ + 3x² yi + 3xy² i² = u + iv
⇒x³ – iy³ + 3x² yi – 3xy² = u + iv
⇒ (x³ – 3xy²) + i(3x² y – y³) = u + iv
On equating real and imaginary parts, we obtain
11. If a and ẞ are different complex numbers with |ẞ| = 1 , then find
Ans: For any complex number z, z z¹ = |z|²
It is given that |β| = 1 .
Then
|β|² = ββ’ = 1
Now, (β-α)/(1 – α’β)| = |(β-α)/(ββ’ – α’β)| [because 1 = ββ’ ]
= Ιβ – α| / |β| β’ – α’
= |β – α| / 1 β’ – α’| [because |β| = 1 ]
= |β – α| / | (β-α)’ |
= 1 (because |z| = |z’|, for any complex number z)
Therefore, |(β-α)/(1 – α’β)| = 1
12. Find the number of non-zero integral solutions of the equation |1 – i|ˣ = 2ˣ.
Ans: |1 – i|ˣ = 2ˣ
⇒ (√1² + (- 1)²)ˣ = 2ˣ
⇒ (√2)ˣ = 2ˣ
⇒ 2ˣ/² = 2ˣ
⇒ x/2 = x
⇒ x = 2x
⇒2x – x = 0
⇒ x = 0
Thus, 0 is the only integral solution of the given equation. Therefore, the number of non-zero integral solutions of the given equation is 0.
13. If (a + ib) (c + id) (e + if) (g + i h) = A + iB then show that (a² + b²) (c²+ d²)(e² + f²) (g² + h²) = A² + B².
Ans: (a + ib)(c + id)(e + if)(g + i h) = A + iB
∴ |(a + ib)(c + id)(e + if)(g + i h)| = |A + iB|
⇒|(a + ib)|x|(c + id)|x|(e + if)|x|(g + i h)| = |A + iB|
[|z₁z₂| = |z₁||z₂|]
⇒ √a² + b² × √c² + d² × √e² + f² × √g² + h² = √A² + B²
On squaring both sides, we obtain
(a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B²
Hence, proved.
Ans:
⇒iᵐ =1
∴ m = 4k where k is some integer.
Therefore, the least positive integer is 1.
Thus, the least positive integral value of m is 4(= 4 x1 ).
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