GSEB Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations Miscellaneous Exercise

Question 1.
Evaluate [i¹⁸ + (\(\frac{1}{i}\))²⁵]³.
Solution:
\(\frac{1}{i}\) = \(\frac{1}{i}\) × \(\frac{i}{i}\) = \(\frac{i}{i^{2}}\) = \(\frac{i}{- 1}\) = – i.
∴ [i¹⁸ + (\(\frac{1}{i}\))²⁵]³ = [i¹⁸ + (- i)²⁵]³ = [(i²)⁹ – i(i²)¹²]³
= [(- 1)⁹ – i( – 1)¹²]³
= [- 1 – i]³
= (- 1)³ + 3(- 1)².(- i) + 3(- 1)(- i)² + (- i)³
[∵ (a + b)³ = a³ + 3a²b + 3cb² + b³]
= – 1 – 3i + 3 – i³
= – 1 – 3i + 3 – i(- 1)
= – 1 – 3i + 3 + i = 2 – 2i.

Question 2.
For any two complex numbers z₁ and z₂, prove that Re(z₁z₂) = Rez₁ Rez₂ – Imz₁ Imz₂.
Solution:
Let z₁ = a + ib, where a – Rez₁; b = Im(Z₁)
and z₂ = c + id, where c – R(z₂), d – Im(z₂).
∴ z₁z₂ = (a + ib)(c + id) = ac – bd + i(ad + bc)
∴ Re(z₁z₂) = ac – bd
= Rez₁. Rez₂ – Imz₁Imz₂.

Question 3.
Reduce (\(\frac{1}{1 – 4i}\) – \(\frac{2}{1 + i}\)) (\(\frac{3 – 4i}{5 + i}\)) to the standard form.
Solution:

Question 4.
If x – iy = \(\sqrt{\frac{a-i}{a+i b}}\), prove that (x² + y²)² = \(\frac{a^{2}+b^{2}}{c^{2}+d^{2}}\)
Solution:

Question 5.
Convert the following in the polar form:
(i) \(\frac{1+7 i}{(2-i)^{2}}\)
(ii) \(\frac{1 + 3i}{1 – 2i}\)
Solution:

Put rcos θ = – 1 and rsin θ = 1.
Squaring and adding, we get
r² = 1 + 1 = 2 ⇒ r = \(\sqrt{2}\).
and cos θ = \(\frac{1}{\sqrt{2}}\), sin θ = \(\frac{1}{\sqrt{2}}\).
sin θ is + ve cos θ is – ve. Therefore, θ lies in II quadrant.
So, θ = π – \(\frac{π}{4}\) = \(\frac{3π}{4}\) [∵ sin θ = sin \(\frac{3π}{4}\) = \(\frac{1}{\sqrt{2}}\)]
∴ Polar form of \(\frac{1+7 i}{(2-i)^{2}}\) = \(\sqrt{2}\)(cos \(\frac{3π}{4}\) + isin \(\frac{3π}{4}\))

Solve each of the equations in questions 6 to 9:
6. 3x² – 4x + \(\frac{20}{3}\) = 0
7. x² – 2x + \(\frac{3}{2}\) = 0
8. 27x² – 10x + 1 = 0
9. 21x² – 28x + 10 = 0
Solutions to questions 6 to 9:
6. 3x² – 4x + \(\frac{20}{3}\) = 0
Multiplying by 3, we get
9x² – 12x + 20 = 0.
∴ a = 9, b = – 12, c = 20.
∴ b² – 4ac = (- 12)² – 4.9.20 = 144 – 720
= – 576.

7. x² – 2x + \(\frac{3}{2}\) = 0.
Multiplying by 2,
2x² – 4x + 3 = 0.
Here, a = 2, b = – 4, c = 3
∴ b² – 4ac = (- 4)² – 4.2.3 = 16 – 24 = – 8.

8. 27x² – 10x + 1 = 0
Here, a = 27, b = – 10, c = 1.
∴ b² – 4ac = (- 10)² – 4.27.1 = 100 – 108 = – 8.

9. 21x² – 28x + 10 = 0.
Here, a = 21, b = – 28, c = 10
∴ b² – 4ac = (- 28)² – 4.21 × 10.
= 784 – 840 = – 56

10. If z₁ = 2 – i and z₂ = 1 + i, find |\(\frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+i}\)|.
Solution:

11. If (a + b) = \(\frac{(x+i)^{2}}{2 x^{2}+1}\), prove that a² + b² = \(\frac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}}\).
Solution:
(a + ib) = \(\frac{(x+i)^{2}}{2 x^{2}+1}\) ………………… (1)
Replacing i by – i, we get
a – ib = \(\frac{(x-i)^{2}}{2 x^{2}+1}\) …………………….. (2)
Multiplying (1) and (2), we get

12. Let z₁ = 2 – i and z₂ = – 2 + i. Find:
(i) Re\(\left(\frac{z_{1} z_{2}}{z_{1}}\right)\)
(ii) Im\(\left(\frac{1}{z z_{1}}\right)\)
Solution:

13. Find the modulus and argument of the complex number \(\frac{1 + 2i}{1 – 3i}\).
Solution:

Put r cos θ = – \(\frac{1}{2}\) and r sin θ = \(\frac{1}{2}\).

sin θ is +ve and cos θ is -ve.
∴ θ = π – \(\frac{π}{4}\) = \(\frac{3π}{4}\).
∴ |\(\frac{1 + 2i}{1 – 3i}\)| = \(\frac{1}{\sqrt{2}}\) and arg (\(\frac{1 + 2i}{1 – 3i}\)) = \(\frac{3π}{4}\).

14. Find the real number x and y, if (x – iy)(3 + 5i) is conjugate of – 6 – 24i.
Solution:
Conjugate of – 6 – 24i is – 6 + 24i …………………. (1)
Also, (x – iy)(3 + 5i) = 3x – 5yi² – 3yi + 5xi
= (3x + 5y) + (5x – 3y)i …………………… (2)
Equating (1) and (2), we get
3x + 5y + (5x – 3y)i = – 6 + 24i
∴ 3x + 5y = – 6
and 5x – 3y = 24
or 3x + 5y + 6 = 0
and 5x – 3y – 24 = 0.

∴ x = 3, y = – 3.

15. Find the modulus of \(\frac{1+i}{1-i}\) = \(\frac{1-i}{1+i}\).
Solution:

16. If (x + iy)³ = u + iv, then show that \(\frac{u}{x}\) + \(\frac{u}{y}\) = 4(x² – y²).
Solution:
(x + iy)³ = x³ + 3x²iy + 3x.(iy)² + (iy)²
= x³ + 3x²yi – 3xy² – iy³
∴ (x³ – 3xy²) + (3x²y – y³)i = u + iv.
Equating real and imaginary parts, we get
x³ – 3xy² = u ∴ x² – 3y² = \(\frac{u}{x}\) ……………….. (1)
3x² – y³ = v ∴ 3x² – y² = \(\frac{v}{y}\) ………………… (2)
Adding (1) and (2), 4x² – 4y² = \(\frac{u}{x}\) + \(\frac{v}{y}\)
∴ \(\frac{u}{x}\) + \(\frac{v}{y}\) = 4(x² – y²).

17. If α and β are different complex numbers with |β| = 1, then find |\(\frac{\beta-\alpha}{1-\alpha \beta}\)|.
Solution:

18. Find the number of non-zero integral solutions of the equation |1 – i|^x = 2^x.
Solution:
|1 – i| = \(\sqrt{1^{2}+(-1)^{2}}\) = \(\sqrt{2}\) = 2^1/2.
∴ |1 – i|^x = (2^1/2)^x = 2^x/2.
Since |1 – i|^x = 2^x, therefore
2^x/2 = 2^x.
∴ \(\frac{x}{2}\) = x is admissible except x = 0.
⇒ There is no non-zero integral solution. Its solution is x = 0.

19. If (a + ib)(c + id)(e + if)(g + ih) = A + iB, then show that (a² + b²)(c² + d²)(c² + f²) + (g² + h²) = A² + B².
Solution:
(a + ib)(c + id)(e + if)(g + ih) = A + iB ………………. (1)
Replacing i by – i, we get
(a – ib)(c – id)(e – if)(g – ib) = A – iB …………………… (2)
Multiplying (1) and (2), we get
[(a + ib )(a – ib)] [(c + id)(c – id)] [(e + if)(e – if)] [(g + ih)(g – ih)] = (A + iB)(A – iB)
or (a² – i²b²)(c² – i²d²)(e² – i²f²)(g² – i²h²) = A² – i²B²
(a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B².
(a² + b²)(c² + d²)(e² + f²)(g² + h²) = A² + B². [∵ i² = – 1]

20. If (\(\frac{1+i}{1-i}\))^m = 1, then find the least integral value of m?
Solution:

⇒ m is a multiple of 4.
∴ Least value of m = 4.