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 [i^{18} + (\(\frac{1}{i}\))^{25}]^{3}.

Solution:

\(\frac{1}{i}\) = \(\frac{1}{i}\) × \(\frac{i}{i}\) = \(\frac{i}{i^{2}}\) = \(\frac{i}{- 1}\) = – i.

∴ [i^{18} + (\(\frac{1}{i}\))^{25}]^{3} = [i^{18} + (- i)^{25}]^{3} = [(i^{2})^{9} – i(i^{2})^{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} = a^{3} + 3a^{2}b + 3cb^{2} + b^{3}]

= – 1 – 3i + 3 – i^{3}

= – 1 – 3i + 3 – i(- 1)

= – 1 – 3i + 3 + i = 2 – 2i.

Question 2.

For any two complex numbers z_{1} and z_{2}, prove that Re(z_{1}z_{2}) = Rez_{1} Rez_{2} – Imz_{1} Imz_{2}.

Solution:

Let z_{1} = a + ib, where a – Rez_{1}; b = Im(Z_{1})

and z_{2} = c + id, where c – R(z_{2}), d – Im(z_{2}).

∴ z_{1}z_{2} = (a + ib)(c + id) = ac – bd + i(ad + bc)

∴ Re(z_{1}z_{2}) = ac – bd

= Rez_{1}. Rez_{2} – Imz_{1}Imz_{2}.

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 (x^{2} + y^{2})^{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

r^{2} = 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. 3x^{2} – 4x + \(\frac{20}{3}\) = 0

7. x^{2} – 2x + \(\frac{3}{2}\) = 0

8. 27x^{2} – 10x + 1 = 0

9. 21x^{2} – 28x + 10 = 0

Solutions to questions 6 to 9:

6. 3x^{2} – 4x + \(\frac{20}{3}\) = 0

Multiplying by 3, we get

9x^{2} – 12x + 20 = 0.

∴ a = 9, b = – 12, c = 20.

∴ b^{2} – 4ac = (- 12)^{2} – 4.9.20 = 144 – 720

= – 576.

7. x^{2} – 2x + \(\frac{3}{2}\) = 0.

Multiplying by 2,

2x^{2} – 4x + 3 = 0.

Here, a = 2, b = – 4, c = 3

∴ b^{2} – 4ac = (- 4)^{2} – 4.2.3 = 16 – 24 = – 8.

8. 27x^{2} – 10x + 1 = 0

Here, a = 27, b = – 10, c = 1.

∴ b^{2} – 4ac = (- 10)^{2} – 4.27.1 = 100 – 108 = – 8.

9. 21x^{2} – 28x + 10 = 0.

Here, a = 21, b = – 28, c = 10

∴ b^{2} – 4ac = (- 28)^{2} – 4.21 × 10.

= 784 – 840 = – 56

10. If z_{1} = 2 – i and z_{2} = 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 a^{2} + b^{2} = \(\frac{\left(x^{2}+1\right)^{2}}{\left(2 x^{2}+1\right)^{2}}\).

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 z_{1} = 2 – i and z_{2} = – 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 – 5yi^{2} – 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(x^{2} – y^{2}).

Solution:

(x + iy)^{3} = x^{3} + 3x^{2}iy + 3x.(iy)^{2} + (iy)^{2}

= x^{3} + 3x^{2}yi – 3xy^{2} – iy^{3}

∴ (x^{3} – 3xy^{2}) + (3x^{2}y – y^{3})i = u + iv.

Equating real and imaginary parts, we get

x^{3} – 3xy^{2} = u ∴ x^{2} – 3y^{2} = \(\frac{u}{x}\) ……………….. (1)

3x^{2} – y^{3} = v ∴ 3x^{2} – y^{2} = \(\frac{v}{y}\) ………………… (2)

Adding (1) and (2), 4x^{2} – 4y^{2} = \(\frac{u}{x}\) + \(\frac{v}{y}\)

∴ \(\frac{u}{x}\) + \(\frac{v}{y}\) = 4(x^{2} – y^{2}).

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 (a^{2} + b^{2})(c^{2} + d^{2})(c^{2} + f^{2}) + (g^{2} + h^{2}) = A^{2} + B^{2}.

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 (a^{2} – i^{2}b^{2})(c^{2} – i^{2}d^{2})(e^{2} – i^{2}f^{2})(g^{2} – i^{2}h^{2}) = A^{2} – i^{2}B^{2}

(a^{2} + b^{2})(c^{2} + d^{2})(e^{2} + f^{2})(g^{2} + h^{2}) = A^{2} + B^{2}.

(a^{2} + b^{2})(c^{2} + d^{2})(e^{2} + f^{2})(g^{2} + h^{2}) = A^{2} + B^{2}. [∵ i^{2} = – 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.