Class 11: Complex Numbers – Exercise 13.2 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Express the following complex numbers in the form $\displaystyle a + ib$

$\displaystyle \text{i) } (1+i)(1+2i) \hspace{1.0cm} \text{ii) } \frac{3+2i}{-2+i} \hspace{1.0cm} \text{iii) } \frac{1}{(2+i)^2} \hspace{1.0cm} \text{iv) } \frac{1-i}{1+i}$

$\displaystyle \text{v) } \frac{(2+i)^3}{2+3i} \hspace{1.0cm} \text{vi) } \frac{(1+i)(1+\sqrt{3} i)}{1-i} \hspace{1.0cm} \text{vii) } \frac{2+3i}{4+5i} \hspace{1.0cm} \text{viii) } \frac{(1-i)^3}{1-i^3}$

$\displaystyle \text{ix) } (1+2i)^{-3} \hspace{1.0cm} \text{x) } \frac{3-4i}{(4+2i)(1+i)}$

$\displaystyle \text{xi) } \Big( \frac{1}{1-4i} - \frac{2}{1+i} \Big) \Big( \frac{3-4i}{5+i} \Big) \hspace{1.0cm} \text{xii) } \frac{5+\sqrt{2} i}{1 - \sqrt{2} i}$

Answer:

$\displaystyle \text{i) } (1+i)(1+2i)$

$\displaystyle = 1 + 2i + i + 2i^2 = 1 + 3i - 2 = - 1 + 3i$

$\displaystyle \text{ii) } \frac{3+2i}{-2+i}$

$\displaystyle = \frac{3+2i}{-2+i} \times \frac{-2-i}{-2-i} = \frac{-6-3i-4i-2i^2}{4-i^2} = \frac{-6-7i+2}{4+1} = \frac{-4}{5} - i \big( \frac{7}{5} \big)$

$\displaystyle \text{iii) } \frac{1}{(2+i)^2}$

$\displaystyle = \frac{1}{4+i^2 + 4i} = \frac{1}{3+4i} \times \frac{3-4i}{3-4i} = \frac{3-4i}{9-16i^2} = \frac{3}{25} - i \big( \frac{4}{5} \big)$

$\displaystyle \text{iv) } \frac{1-i}{1+i}$

$\displaystyle = \frac{1-i}{1+i} \times \frac{1-i}{1-i} = \frac{1+i^2-2i}{1-i^2} = \frac{-2i}{2} = -i$

$\displaystyle \text{v) } \frac{(2+i)^3}{2+3i}$

$\displaystyle = \frac{( 4 + i^2 + 4 i )(2+i)}{2+3i} = \frac{8 + 2i^2 + 8i + 4i + i^3 + 4i^2}{2 + 3i} = \frac{2+11i}{2+3i} \times \frac{2-3i}{2-3i}$

$\displaystyle = \frac{4 - 6i + 22 i - 33i^2}{4 - 9i^2} = \frac{37+ 16i}{13} = \frac{37}{13} + i \big( \frac{16}{13} \big)$

$\displaystyle \text{vi) } \frac{(1+i)(1+\sqrt{3} i)}{1-i}$

$\displaystyle = \frac{(1+i)(1+\sqrt{3} i)}{1-i} \times \frac{1+i}{1-i} = \frac{(1+i^2+2i)(1+\sqrt{3}i}{1-i^2}$

$\displaystyle = \frac{2i ( 1 + \sqrt{3} i)}{2} = i(1 + \sqrt{3}i) = \sqrt{3} i^2 + i = - \sqrt{3}+ i$

$\displaystyle \text{vii) } \frac{2+3i}{4+5i}$

$\displaystyle = \frac{2+3i}{4+5i} \times \frac{4-5i}{4-5i} = \frac{8 - 10 i + 12 i - 15 i^2}{16 - 25 i^2} = \frac{23 + 2 i}{41} = \frac{23}{41} + i \big( \frac{2}{41} \big)$

$\displaystyle \text{viii) } \frac{(1-i)^3}{1-i^3}$

$\displaystyle = \frac{(1-i)^2 (1-i)}{1-i^3} = \frac{(1+i^2 - 2i)(1-i)}{1+i} = \frac{-2i(1-i)}{1+i} = \frac{-2(i - i^2)}{1+i} = \frac{-2(1+i)}{1+i} = - 2$

$\displaystyle \text{ix) } (1+2i)^{-3}$

$\displaystyle = \frac{1}{(1+2i)^2(1+2i)} = \frac{1}{(1+4i^2+4i)(1+2i)} = \frac{1}{(-3+4i)(1+2i)} = \frac{1}{-3+4i-6i+8^2}$

$\displaystyle = \frac{1}{-11-2i} = \frac{1}{-11-2i} \times \frac{11-2i}{11-2i} = \frac{-11+2i}{121+4} = \frac{-11}{125} + i \big( \frac{2}{125} \big)$

$\displaystyle \text{x) } \frac{3-4i}{(4+2i)(1+i)}$

$\displaystyle = \frac{3-4i}{4 - 2i + 4i - 2i^2} = \frac{3-4i}{6+2i} \times \frac{6-2i}{6-2i} = \frac{18 - 24i - 6i + 8i^2}{36 + 4}$

$\displaystyle = \frac{10-30i}{40} = \frac{1-3i}{4} = \frac{1}{4} + i \big( \frac{-3}{4} \big)$

$\displaystyle \text{xi) } \Big( \frac{1}{1-4i} - \frac{2}{1+i} \Big) \Big( \frac{3-4i}{5+i} \Big)$

$\displaystyle = \Big( \frac{1+i-2+8i}{1-4i+i-4i^2} \Big) \Big( \frac{3-4i}{5+i} \Big) = \Big( \frac{-1+9i}{5-3i} \Big) \Big( \frac{3-4i}{5+i} \Big) = \frac{-3+27i+4i-36i^2}{25-15i+5i-3i^2}$

$\displaystyle = \frac{33+ 31i}{28-10i} \times \frac{28+10i}{28+10i} = \frac{924+330i+868i-310}{784+100} = \frac{614+1198i}{884} = \frac{614}{884} + i \big( \frac{1198}{884} \big)$

$\displaystyle \text{xii) } \frac{5+\sqrt{2} i}{1 - \sqrt{2} i}$

$\displaystyle = \frac{5+\sqrt{2} i}{1 - \sqrt{2} i} \times \frac{1 + \sqrt{2} i}{1 + \sqrt{2} i} = \frac{5 + \sqrt{2} i + 5 \sqrt{2} i + 2 i^2}{1+2}$

$\displaystyle = \frac{3 + 6 \sqrt{2} i}{3} = 1 + 2\sqrt{2} i = 1 + i ( 2\sqrt{2})$

$\displaystyle \\$

Question 2: Find the real values of $\displaystyle x \text{ and } y$ , if

$\displaystyle \text{i) } (x+iy)(2-3i) = 4+i \hspace{1.0cm} \text{ii) } (3x-2iy)(2+i)^2 = 10(1+i)$

$\displaystyle \text{iii) } \frac{(1+i)x- 2i}{3+i} - \frac{(2-3i)y+i}{3-i} = i \hspace{1.0cm} \text{iv) } (1+i)(x+iy)= 2- 5i$

Answer:

$\displaystyle \text{i) } (x+iy)(2-3i) = 4+i$

$\displaystyle \Rightarrow 2x + i 2y - 3x i - 3y i^2 = 4 + i$

$\displaystyle \Rightarrow ( 2x + 3y) + i ( -3x + 2y) = 4 + i$

Therefore comparing, we get

$\displaystyle 2x + 3y = 4$ … … … … … i)

$\displaystyle -3x + 2y = 1$ … … … … … ii)

Solving i) and ii), multiplying i) by $\displaystyle 3$ and ii) by $\displaystyle 2$ and adding we get

$\displaystyle { \hspace{1.0cm} 6x + 9y = 12} \\ \underline { (+) - 6x + 4 y = 2 } \\ { \hspace{2.0cm}13 y = 14}$

$\displaystyle \Rightarrow y = \frac{14}{13}$

Now substituting in ii) we get

$\displaystyle 2x + 3 \big( \frac{14}{13} \big) = 4 \Rightarrow 2x = \frac{10}{13} \Rightarrow x = \frac{10}{26}$

Hence $\displaystyle x = \frac{10}{26}$ and $\displaystyle y = \frac{14}{13}$

$\displaystyle \text{ii) } (3x-2iy)(2+i)^2 = 10(1+i)$

$\displaystyle \Rightarrow ( 3x - 2iy) ( 4 + i^2 + 4i) = 10 + 10 i$

$\displaystyle \Rightarrow ( 3x - i 2y) ( 3 + 4i) = 10 + 10i$

$\displaystyle \Rightarrow 9x - i 6 y + i 12 x + 8 y = 10 + 10 i$

$\displaystyle \Rightarrow ( 9x + 8 y) + i ( 12 x - 6y) = 10 + 10 i$

Therefore comparing, we get

$\displaystyle 9x + 8y = 10$ … … … … … i)

$\displaystyle 12 x - 6y = 10$ … … … … … ii)

Solving i) and ii), multiplying i) by $\displaystyle 6$ and ii) by $\displaystyle 8$ and adding we get

$\displaystyle { \hspace{1.0cm} 54x + 48 y = 60 } \\ \underline { (+) 96x - 48 y = 80 } \\ { \hspace{2.0cm} 150x = 140}$

$\displaystyle \Rightarrow x = \frac{14}{15}$

Now substituting in i) we get

$\displaystyle 9 \big( \frac{14}{15} \big) + 8y = 10 \Rightarrow 8y = \frac{8}{5} \Rightarrow y = \frac{1}{5}$

Hence $\displaystyle x = \frac{14}{15}$ and $\displaystyle y = \frac{1}{5}$

$\displaystyle \text{iii) } \frac{(1+i)x- 2i}{3+i} - \frac{(2-3i)y+i}{3-i} = i$

$\displaystyle \Rightarrow \frac{[ (1+i) x - 2 i ] (3-i)+[(2-3i)y + 1](3+i)}{9+1} = i$

$\displaystyle \Rightarrow 3x + i 3x - 6i - i x - i^2 x + 2 i^2+ 6y - i 9y + 3 i + i 2y - i^2 3y + i^2 = 10i$

$\displaystyle \Rightarrow 4x + i ( 2x) - 6i - 2 + 9y + i ( -7y) + 3i - 1 = 10i$

$\displaystyle \Rightarrow ( 4x + 9y - 3) + i ( 2x - 7y - 3) = 10i$

$\displaystyle \Rightarrow 4x + 9y - 3 = 0 \Rightarrow 4x + 9y = 3$ … … … … … i)

$\displaystyle \Rightarrow 2x - 7y - 3 = 10 \Rightarrow 2x - 7y = 13$ … … … … … ii)

Solving i) and ii), multiplying i) by $\displaystyle 7$ and ii) by $\displaystyle 9$ and adding we get

$\displaystyle 46 x = 117 + 21 \Rightarrow x = \frac{138}{46} = 3$

Substituting in i) we get

$\displaystyle 4 ( 3) + 9y = 3 \Rightarrow y = -1$

Hence $\displaystyle x = 3$ and $\displaystyle y = -1$

$\displaystyle \text{iv) } (1+i)(x+iy)= 2- 5i$

$\displaystyle x + ix + iy + i^2 y = 2 - 5i$

$\displaystyle (x-y) + i( x+y) = 2 - 5i$

Comparing we get

$\displaystyle x-y = 2$ … … … … … i)

$\displaystyle x+y = -5$ … … … … … ii)

Adding i) and ii) we get

$\displaystyle 2x = - 3 \Rightarrow x = \frac{-3}{2}$

Substituting in i) we get

$\displaystyle y = x - 2 = \frac{-3}{2} - 2 = \frac{-7}{2}$

Hence $\displaystyle x = \frac{-3}{2}$ and $\displaystyle y = \frac{-7}{2}$

$\displaystyle \\$

Question 3: Find the conjugates of the following complex numbers:

$\displaystyle \text{i) } 4-5i \hspace{1.0cm} \text{ii) } \frac{1}{3+5i} \hspace{1.0cm} \text{iii) } \frac{1}{1+i}$

$\displaystyle \text{iv) } \frac{(3-i)^2}{2+i} \hspace{1.0cm} \text{v) } \frac{(1+i)(2+i)}{3+i} \hspace{1.0cm} \text{vi) } \frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$

Answer:

Note: $\displaystyle \text{If } z = x + iy$ , then conjugate of $\displaystyle z$ is $\displaystyle \overline{z} = x - iy$

$\displaystyle \text{i) } \text{If } z = 4-5i \Rightarrow \overline{z} = 4 + 5i$

$\displaystyle \text{ii) } \text{If } z = \frac{1}{3+5i} = \frac{1}{3+5i} \times \frac{3 - 5i}{3 - 5i} = \frac{3 - 5i}{34} = \frac{3}{34} -i \frac{5}{34}$

Therefore $\displaystyle \overline{z} = \frac{3}{34} + i \frac{5}{34}$

$\displaystyle \text{iii) } \text{If } z = \frac{1}{1+i} = \frac{1}{1+i} \times \frac{1-i}{1-i} = \frac{1-i}{2} = \frac{1}{2} - i \big( \frac{1}{2} \big)$

Therefore $\displaystyle \overline{z} = \frac{1}{2} + i \big( \frac{1}{2} \big)$

$\displaystyle \text{vi) } \text{If } z = \frac{(3-i)^2}{2+i} = \frac{9+i^2-6i}{2+i} = \frac{8-6i}{2+i} = \frac{8-6i}{2+i} \times \frac{2-i}{2-i}$

$\displaystyle = \frac{16-12i-8i-6}{4+1} = \frac{10-20i}{5} = 2-4i$

Therefore $\displaystyle \overline{z} = 2+4i$

$\displaystyle \text{v) } \text{If } z = \frac{(1+i)(2+i)}{3+i} = \frac{2+2i+i+i^2}{3+i} = \frac{1+3i}{3+i} = \frac{1+3i}{3+i} \times \frac{3-i}{3-i}$

$\displaystyle = \frac{3+9i - i-3i^2}{9+1} = \frac{6+8i}{10} = \frac{3}{4} + i \big( \frac{4}{5} \big)$

Therefore $\displaystyle \overline{z} = \frac{3}{4} - i \big( \frac{4}{5} \big)$

$\displaystyle \text{vi) } \text{If } z = \frac{(3-2i)(2+3i)}{(1+2i)(2-i)} = \frac{6-4i+9i-6i^2}{2+4i-i-2i^2} = \frac{12+5i}{4+3i} \times \frac{4-3i}{4-3i}$

$\displaystyle = \frac{48+20i - 36i - 15i^2}{16+9} = \frac{63}{25} - i \big( \frac{16}{25} \big)$

Therefore $\displaystyle \overline{z} = \frac{63}{25} + i \big( \frac{16}{25} \big)$

$\displaystyle \\$

Question 4: Find the multiplicative inverse of the following complex numbers:

$\displaystyle \text{i) } 1-i \hspace{1.0cm} \text{ii) } (1+i\sqrt{3})^2 \hspace{1.0cm} \text{iii) } 4-3i \hspace{1.0cm} \text{iv) } \sqrt{5}+3i$

Answer:

$\displaystyle \text{If } z = x + iy$ is the complex number, then multiplication inverse of $\displaystyle z$ is $\displaystyle z^{-1}$ or $\displaystyle \frac{1}{z}$

$\displaystyle \text{i) } z = 1-i$

$\displaystyle z^{-1} = \frac{1}{1-i} \times \frac{1+i}{1+i} = \frac{1+i}{1-i^2} = \frac{1}{2} + \frac{1}{2} i$

$\displaystyle \text{ii) } z = (1+i\sqrt{3})^2 = 1+3i^2+i (2\sqrt{3})= -2 +i (2\sqrt{3})$

$\displaystyle z^{-1} = \frac{1}{z} = \frac{1}{-2+i (2\sqrt{3})} \times \frac{-2-i (2\sqrt{3}) }{-2-i (2\sqrt{3}) } = \frac{-2-i (2\sqrt{3}) }{4 + 12} = \frac{-1}{8} -i \frac{\sqrt{3}}{6}$

$\displaystyle \text{iii) } z = 4-3i$

$\displaystyle z^{-1} = \frac{1}{z} = \frac{1}{4-3i} \times \frac{4+3i}{4+3i} = \frac{4+3i}{16+9} = \frac{4}{25} + i \big( \frac{3}{25} \big)$

$\displaystyle \text{iv) } z = \sqrt{5}+3i$

$\displaystyle z^{-1} = \frac{1}{z} = \frac{1}{\sqrt{5}+ 3i} \times \frac{\sqrt{5}- 3i}{\sqrt{5}-3i} = \frac{\sqrt{5}-3i}{5+9} = \frac{\sqrt{5}}{14} - \frac{3}{14} i$

$\displaystyle \\$

$\displaystyle \text{Question 5: If } z_1 = 2 - i, z_2 = 1+i , \text{ find } \Big| \frac{z_1+z_2+1}{z_1-z_2+1} \Big|$

Answer:

$\displaystyle \text{If } z = x + iy$ , then $\displaystyle |z| = \sqrt{x^2+y^2}$

$\displaystyle z_1 = 2 - i z_2 = 1 + i$

$\displaystyle \Rightarrow z_1 + z_2 + 1 = 2 - i + 1 + i + 1 = 4$

$\displaystyle \Rightarrow z_1 - z_2 + i = 2 - i - 1 - i + 1 = 1 - i$

$\displaystyle \Rightarrow \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} = \frac{4}{1-i} \times \frac{1+i}{1+i} = \frac{4+4i}{2} = 2+2i$

$\displaystyle \Rightarrow \Big| \frac{z_1 + z_2 + 1}{z_1 - z_2 + i} \Big| = \sqrt{2^2+2^2} = \sqrt{8} =2\sqrt{2}$

$\displaystyle \\$

Question 6: $\displaystyle \text{If } z_1 = 2-i, z_2 = -2+i$ , find $\displaystyle \text{i) } Re \Big( \frac{z_1z_2}{z_1} \Big) \hspace{1.0cm} \text{ii) } Im \Big( \frac{1}{z_1z_2} \Big)$

Answer:

$\displaystyle z_1 = 2-i, \hspace{1.0cm} z_2 = -2+i \hspace{1.0cm} \overline{z_1} = 2 + i$

$\displaystyle \text{i) } z_1z_2 = ( 2 - i) ( -2 + i) = -4 + 2i + 2i - i^2 = - 3 + 4i$

$\displaystyle \therefore \frac{z_1z_2}{\overline{z_1}} = \frac{-3+4i}{2+i} \times \frac{2-i}{2-i} = \frac{-6+8i+3i-4i^2}{4+1} = \frac{-2+5i}{5} = \frac{-2}{5} + i$

$\displaystyle \therefore Re \big( \frac{z_1z_2}{\overline{z_1}} \big) = \frac{-2}{5}$

$\displaystyle \text{ii) } z_1 = 2 - i \hspace{1.0cm} \overline{z_1} = 2+i$

$\displaystyle \therefore z_1 \overline{z_1} = (2-i)(2+i) = 4 + 1 = 5$

$\displaystyle \therefore \frac{1}{z_1 \overline{z_1}} = \frac{1}{5}$

$\displaystyle \therefore Im \big( \frac{1}{z_1 \overline{z_1}} \big) = 0$

$\displaystyle \\$

$\displaystyle \text{Question 7: Find the modulus of } \frac{1+i}{1-i} - \frac{1-i}{1+i}$

Answer:

$\displaystyle \frac{1+i}{1-i} - \frac{1-i}{1+i} = \frac{(1+i^2+2i)-(1+i^2-2i)}{1-i^2} = \frac{2i+2i}{2} = 2i$

$\displaystyle |2i| = \sqrt{2^2} = 2$

$\displaystyle \\$

$\displaystyle \text{Question 8: If } x+iy = \frac{a+ib}{a-ib} , \text{ prove that } x^2 + y^2 = 1$

Answer:

$\displaystyle x+iy = \frac{a+ib}{a-ib}$

$\displaystyle \overline{x+iy} = \Big( \overline{\frac{a+ib}{a-ib}} \Big) = \frac{\overline{a+ib}}{\overline{a-ib}}$

$\displaystyle x-iy = \frac{a-ib}{a+ib}$

$\displaystyle \therefore ( x +iy)(x-iy) = \frac{a+ib}{a-ib} \times \frac{a-ib}{a+ib}$

$\displaystyle \Rightarrow x^2 + y^2 = \frac{a^2 +b^2}{a^2 +b^2} = 1$

Hence proved.

Therefore the general solution is given by $\displaystyle \theta = 2 n \pi \pm \frac{\pi}{2} , n \in Z$

$\displaystyle \\$

Question 9: Find the least positive integral value of $\displaystyle n$ for which $\displaystyle \Big( \frac{1+i}{1-i} \Big)^n$ is real.

Answer:

For $\displaystyle n = 1$

$\displaystyle \Big( \frac{1+i}{1-i} \Big)^1 = \frac{1+i^2+2i}{2} = i \text{ which is not real. }$

For $\displaystyle n = 2$

$\displaystyle \Big( \frac{1+i}{1-i} \Big)^2 = \frac{1+i^2+2i}{1+i^2-2i} = \frac{2i}{-2i} = -1 \text{ which is real. }$

Therefore the least positive integral value of $\displaystyle n = 2$

$\displaystyle \\$

$\displaystyle \text{Question 10: Find the real values of } \theta \text{ or which the complex number } \\ \\ \frac{1+ i \cos \theta}{1 - 2 i \cos \theta} \text{ is purely real. }$

Answer:

$\displaystyle \text{Let } z = \frac{1+ i \cos \theta}{1 - 2 i \cos \theta}$

$\displaystyle = \frac{1+ i \cos \theta}{1 - 2 i \cos \theta} \times \frac{1 + 2 i \cos \theta}{1 + 2 i \cos \theta}$

$\displaystyle = \frac{1+i \cos \theta + 2 i \cos \theta + 2 i \cos \theta + 2 i^2 \cos^2 \theta}{1+4 \cos^2 \theta}$

$\displaystyle = \frac{(1-2 \cos^2 \theta) + (i ( 3 \cos \theta)}{1+4 \cos^2 \theta}$

For $\displaystyle z$ to be purely Real, $\displaystyle Im (z) = 0$

$\displaystyle \therefore \frac{3 \cos \theta}{1+4 \cos^2 \theta} = 0$

$\displaystyle \Rightarrow \cos \theta = 0$

$\displaystyle \Rightarrow \theta = \frac{\pi}{2}$

$\displaystyle \\$

$\displaystyle \text{Question 11: Find the smallest positive integer value of n for which } \frac{(1+i)^n}{(1-i)^{n-2}} \\ \\ \text{ is a real number. }$

Answer:

$\displaystyle \frac{(1+i)^n}{(1-i)^{n-2}}$

$\displaystyle = \frac{ (1+i)^n }{ (1-i)^{n-2} } \times (1-i)^2$

$\displaystyle = \Big( \frac{1+i}{1-i} \times \frac{1+i}{1+i} \Big)^m \times ( 1+i^2 - 2i)$

$\displaystyle = \Big( \frac{1+i^2+ 2i}{1-i^2} \Big)^m \times ( 1-1 - 2i)$

$\displaystyle = \Big( \frac{1-1+ 2i}{1+1} \Big)^m \times ( - 2i)$

$\displaystyle = -2i (i^m)$

$\displaystyle = - 2 i^{m+1}$

For this to be real , the smallest positive value of $\displaystyle n$ will be $\displaystyle 1$ .

Thus $\displaystyle i^{1+1} = i^2 = - 1$ , which is real.

$\displaystyle \\$

$\displaystyle \text{Question 12: If } \Big( \frac{1+i}{1-i} \Big)^3 - \Big( \frac{1-i}{1+i} \Big)^3 = x+iy , \text{ find } ( x, y)$

Answer:

$\displaystyle \Big( \frac{1+i}{1-i} \Big)^3 - \Big( \frac{1+i}{1-i} \Big)^3 = x+iy$

$\displaystyle \Rightarrow \Big( \frac{1+i}{1-i} \times \frac{1+i}{1+i} \Big)^3 - \Big( \frac{1+i}{1-i} \times \frac{1-i}{1-i} \Big)^3 = x+iy$

$\displaystyle \Rightarrow \Big( \frac{2i}{2} \Big)^3 - \Big( \frac{-2i}{2} \Big)^3 = x+iy$

$\displaystyle \Rightarrow i^3 - ( -i)^3 = x + i y$

$\displaystyle \Rightarrow -i - i = x + iy$

$\displaystyle \Rightarrow -2i = x + iy$

$\displaystyle \Rightarrow x = 0$ and $\displaystyle y = - 2$

$\displaystyle \\$

$\displaystyle \text{Question 13: If } \frac{(1+i)^2}{2-i} = x+iy , \text{ find } x+y$

Answer:

$\displaystyle \frac{(1+i)^2}{2-i} = x + i y$

$\displaystyle \Rightarrow \frac{2i}{2-i} \times \frac{2+i}{2+i} = x+iy$

$\displaystyle \Rightarrow \frac{4i + 2i^2}{4+1} = x+iy$

$\displaystyle \Rightarrow \frac{-2+4i}{5} = x+iy$

$\displaystyle \Rightarrow x = \frac{-2}{5} , \ \ \ y = \frac{4}{5}$

$\displaystyle \therefore x + y = \frac{-2}{5} + \frac{4}{5} = \frac{2}{5}$

$\displaystyle \\$

$\displaystyle \text{Question 14: If } \Big( \frac{1-i}{1+ i} \Big)^{100} = a+ib , \text{ find } ( a, b)$

Answer:

$\displaystyle \Big( \frac{1-i}{1+ i} \Big)^{100} = a+ib$

$\displaystyle \Rightarrow \Big( \frac{1-i}{1+ i} \times \frac{1-i}{1-i} \Big)^{100} = a+ib$

$\displaystyle \Rightarrow \Big( \frac{-2i}{2} \Big)^{100} = a+ib$

$\displaystyle \Rightarrow ( -i )^{100} = a+ib$

$\displaystyle \Rightarrow 1 = a + ib$

$\displaystyle \therefore a = 1 , b = 0$

$\displaystyle \therefore (a, b) = ( 1, 0)$

$\displaystyle \\$

$\displaystyle \text{Question 15: If } a = \cos \theta + i \sin \theta , \text{ find the value of } \frac{1+a}{1-a}$

Answer:

$\displaystyle a = \cos \theta + i \sin \theta$

$\displaystyle \frac{1+a}{1-a} = \frac{(1+\cos \theta) + i (\sin \theta) }{(1-\cos \theta) - i (\sin \theta) }$

$\displaystyle = \frac{(1+\cos \theta) + i (\sin \theta) }{(1-\cos \theta) - i (\sin \theta) } \times \frac{(1-\cos \theta) + i (\sin \theta)}{(1-\cos \theta) + i (\sin \theta)}$

$\displaystyle = \frac{(1+i \sin \theta)^2 - \cos^2 \theta}{(1-\cos \theta)^2 + \sin^2 \theta}$

$\displaystyle = \frac{1 - \sin^2 \theta + 2 i \sin \theta- \cos^2 \theta}{1+ \cos^2 \theta - 2 \cos \theta + \sin^2 \theta}$

$\displaystyle = \frac{2 i \sin \theta }{2 ( 1 - \cos \theta)}$

$\displaystyle = \frac{i \sin \theta}{1 - \cos \theta}$

$\displaystyle = i \frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \sin^2 \frac{\theta}{2}}$

$\displaystyle = i \cot^2 \frac{\theta}{2}$

$\displaystyle \\$

Question 16:

$\displaystyle \text{i) } 2x^3+2x^2-7x+72 , \text{ when } x = \frac{3-5i}{2}$

$\displaystyle \text{ii) } x^4-4x^3+4x^2+8x+44 , \text{ when } x= 3 + 2i$

$\displaystyle \text{iii) } x^4+4x^3+6x^2+4x+9 , \text{ when } x = - 1 +i \sqrt{2}$

$\displaystyle \text{iv) } x^6 + x^4 + x^2+1 , \text{ when } x = \frac{1+i}{\sqrt{2}}$

$\displaystyle \text{v) } 2x^4 + 5x^3 + 7x^2 - x +41 , \text{ when } x = - 2 - \sqrt{3}i$

Answer:

$\displaystyle \text{i) } x = \frac{3-5i}{2}$

$\displaystyle \Rightarrow 2x = 3 - 5i$

$\displaystyle \Rightarrow ( 2x - 3)^2 = ( -5 i)^2$

$\displaystyle \Rightarrow 4x^2 + 9 - 12 x = - 25$

$\displaystyle \Rightarrow 4x^2 - 12 x + 34 = 0$

$\displaystyle \Rightarrow 2x^2- 6x + 17 = 0$

$\displaystyle \Rightarrow 2x^3 + 2 x^2 - 7x + 72 = 0$

Now $\displaystyle 2x^3+2x^2-7x+72$

$\displaystyle = x ( 2x^2 - 6x + 17) + 6x^2 - 17x + 2x^2 - 7x + 72$

$\displaystyle = x(0) + 8x^2 - 24x + 72$

$\displaystyle = 4 ( 2x^2 - 6x +17) + 4$

$\displaystyle = 4(0) + 4 = 4$

$\displaystyle \text{ii) } x = 3 + 2i$

$\displaystyle \Rightarrow (x-3)^2 = (2i)^2$

$\displaystyle \Rightarrow x^2 + 9 - 6x = - 4$

$\displaystyle \Rightarrow x^2 - 6x + 13 = 0$

Now, $\displaystyle x^4-4x^3+4x^2+8x+44$

$\displaystyle = x^2( x^2 - 6x + 13) + 6x^2 - 13 x^2 - 4x^3 + 4x^2 + 8x + 144$

$\displaystyle = x^2 (0) + 2x^3 - 9x^2 + 8x + 44$

$\displaystyle = 2x ( x^2 - 6x + 13) + 12x^2 - 26 x - 9x^2 +8x+ 44$

$\displaystyle = 2x (0) + 3x^2 - 18x +44$

$\displaystyle = 3 ( x^2 - 6x + 13) + 5$

$\displaystyle = 3 ( 0) + 5 = 5$

$\displaystyle \text{iii) } x = - 1 + i \sqrt{2}$

$\displaystyle \Rightarrow ( x + 1) ^2 = ( i \sqrt{2})^2$

$\displaystyle \Rightarrow x^2 + 1 + 2 x = - 2$

$\displaystyle \Rightarrow x^2 + 2x + 3 = 0$

Now $\displaystyle x^4+4x^3+6x^2+4x+9$

$\displaystyle = x^2 ( x^2 + 2x + 3) + 2x^3 + 3x^2 + 4x + 9$

$\displaystyle = x (0) + 2x ( x^2 + 2x + 3) - ( x^2 - 2x + 9)$

$\displaystyle = 2x (0) - ( x^2 + 2x + 3) + 3 + 9$

$\displaystyle = 12$

$\displaystyle \text{iv) } x= \frac{1+i}{\sqrt{2}}$

$\displaystyle (\sqrt{2} x)^2 = ( 1 + i)^2$

$\displaystyle \Rightarrow 2x^2 = 2i$

$\displaystyle \Rightarrow x^2 = i$

$\displaystyle \Rightarrow x^4 = - 1$

$\displaystyle \Rightarrow x^4 + 1 = 0$

Now, $\displaystyle x^6 + x^4 + x^2+1$

$\displaystyle = x^2 ( x^4 + 1) + ( x^4 +1 )$

$\displaystyle = x^2 ( 0) + (0) = 0$

$\displaystyle \text{v) } x = - 2 - \sqrt{3} i$

$\displaystyle x^2 = ( -2 - \sqrt{3} i)^2 = 1 + 4 \sqrt{3} i$

$\displaystyle x^3 = ( 1+4\sqrt{3} i)( - 2 - \sqrt{3} i) = 10 - 9 \sqrt{3} i$

$\displaystyle x^4 = ( 10 - 9 \sqrt{3} i) ( - 2 - \sqrt{3} i) = - 20 + 18 \sqrt{3} i - 10 \sqrt{3} i + 27 i^2 = -47 + 8 \sqrt{3} i$

Now, $\displaystyle 2x^4 + 5x^3 + 7x^2 - x +41$

$\displaystyle = 2 ( -47 + 8\sqrt{3} i) + 5 ( 10 - 9 \sqrt{3} i) + 7 ( 1 + 4 \sqrt{3} i) - ( -2 - \sqrt{3} i) + 41$

$\displaystyle = - 94 + 16 \sqrt{3} i + 50 - 45 \sqrt{3} i + 7 + 28 \sqrt{3} i + 2 + \sqrt{3} i + 41$

$\displaystyle = ( - 94 + 50 + 7 + 2 + 41) + ( 16 \sqrt{3} i - 45 \sqrt{3} i + 28 \sqrt{3} i + \sqrt{3} i)$

$\displaystyle = 6+0 =6$

$\displaystyle \\$

$\displaystyle \text{Question 17: For a positive integer, find the value of } (1-i)^n \Big( 1 - \frac{1}{i} \Big)^n$

Answer:

$\displaystyle (1-i)^n \big( 1 - \frac{1}{i} \big)^n = \Big[ (1-i) \big( 1 - \frac{1}{i} \big) \Big]^n = \Big[ \frac{(1-i)(i-1)}{i} \Big]^n = \Big[ \frac{i+1-1+i}{i} \Big]^n = 2^n$

$\displaystyle \\$

Question 18: $\displaystyle \text{If } (1 +i) z = (1-i) \overline{z}$ , then show that $\displaystyle z = -i \overline{z}$

Answer:

$\displaystyle (1+i) z = ( 1- i) \overline{z}$

$\displaystyle \Rightarrow z = \Big( \frac{1-i}{1+i} \Big) \overline{z}$

$\displaystyle \Rightarrow z = \Big( \frac{1-i}{1+i} \times \frac{1-i}{1-i} \Big) \overline{z}$

$\displaystyle \Rightarrow z = \Big( \frac{-2i}{2} \Big) \overline{z}$

$\displaystyle \Rightarrow z = -i \overline{z}$

Hence proved.

$\displaystyle \\$

Question 19: Solve the system of equations $\displaystyle Re(z^2) = 0 , |z|= 2$ .

Answer:

$\displaystyle Re(z^2) = 0 , |z| = 2$

Let $\displaystyle z = x + iy$

$\displaystyle z^2 = ( x + iy)^2 = ( x^2 - y^2)+ i ( 2xy)$

$\displaystyle Re(z^2) = 0 \Rightarrow x^2 - y^2 = 0$ … … … … … i)

$\displaystyle |z| = 2$

$\displaystyle \Rightarrow \sqrt{x^2+y^2} = 2$

$\displaystyle \Rightarrow x^2 + y^2 = 4$ … … … … … ii)

Adding i) and ii) we get

$\displaystyle 2x^2 = 4 \Rightarrow x^2 = 2 \Rightarrow x = \pm \sqrt{2}$

$\displaystyle \therefore y = \pm \sqrt{2}$

Hence $\displaystyle x + iy = \pm \sqrt{2} \pm i \sqrt{2}$

For positive sign for $\displaystyle x : z = \sqrt{2} \pm i \sqrt{2} = \sqrt{2} ( 1 \pm i)$

For negative sign for $\displaystyle x : z = - \sqrt{2} \pm i \sqrt{2} = \sqrt{2} ( - 1 \pm i )$

$\displaystyle \\$

$\displaystyle \text{Question 20: If } \frac{z-1}{z+1} \text{ is purely imaginary number } (z \neq -1) , \text{ find the value of } |z|$

Answer:

Let $\displaystyle z = x + iy$

$\displaystyle \frac{z-1}{z+1} = \frac{x + iy-1}{x + iy+1} = \frac{(x-1) + iy}{(x +1) + iy}$

$\displaystyle = \frac{(x-1) + iy}{(x +1) + iy} \times \frac{(x+1) - iy}{(x +1) - iy}$

$\displaystyle = \frac{[(x-1)+iy][(x+1)-iy]}{(x+1)^2 + y^2}$

$\displaystyle = \frac{x^2 -1 + i ( yx+y) - i ( xy - y) + y^2}{(x+1)^2 + y^2}$

$\displaystyle = \frac{(x^2 + y^2 -1) + i ( yx + y - xy + y)}{(x+1)^2 + y^2}$

$\displaystyle = \frac{(x^2 + y^2 -1) }{(x+1)^2 + y^2} + i \frac{ 2y}{(x+1)^2 + y^2}$

If it is purely imaginary, $\displaystyle Re(z) = 0$

$\displaystyle \Rightarrow x^2 + y^2 - 1 = 0$

$\displaystyle \Rightarrow x^2 + y^2 = 1$

$\displaystyle \Rightarrow \sqrt{x^2 + y^2} = 1$

$\displaystyle \Rightarrow |z| = 1$

$\displaystyle \\$

Question 21: $\displaystyle \text{If } z_1$ is a complex number other than $\displaystyle - 1$ such that $\displaystyle |z_1|= 1$ and $\displaystyle z_2 = \frac{z_1-1}{z_1+1} \text{ then show that the real parts of } z_2 \text{ is zero. }$

Answer:

Let $\displaystyle z_1 = x_1 + i y_2$ and $\displaystyle z_2 = x_ 2 + i y_2$

$\displaystyle |z| = 1 \Rightarrow {x_1}^2 + {y_1}^2 = 1$ … … … … … i)

$\displaystyle z_2 = \frac{z_1 - 1}{z_1 + 1}$

$\displaystyle \Rightarrow z_2 = \frac{(x_1-1)+iy_1}{(x_1+1)+iy_1}$

$\displaystyle = \frac{(x_1-1)+iy_1}{(x_1+1)+iy_1} \times \frac{(x_1+1)- iy_1}{(x_1+1)-iy_1}$

$\displaystyle = \frac{ ({x_1}^2-1) + i ( y_1x_1+y_1) - i ( y_1x_1-y_1)+{y_1}^2}{(x_1+1)^2+{y_1}^2}$

$\displaystyle = \frac{( {x_1}^2+{y_1}^2-1)+i ( 2y_1)}{(x_1+1)^2+{y_1}^2}$

$\displaystyle =0 + i \frac{2y_1}{(x_1+1)^2+{y_1}^2}$

Since there is no real part, $\displaystyle z_2$ is purely an imaginary numbers

$\displaystyle \Rightarrow x_2 = 0$

$\displaystyle \\$

Question 22: $\displaystyle \text{If } |z+1|= z + 2 (1+i)$ , find $\displaystyle z$

Answer:

Let $\displaystyle z = x + i y$

$\displaystyle |z+1| = z + 2 ( 1 + i)$

$\displaystyle \Rightarrow |x+1+iy| = ( x+iy) + 2 ( 1+i)$

$\displaystyle \Rightarrow \sqrt{(x+1)^2 + y^2} = (x+2) + i ( y + 2)$

Comparing, $\displaystyle y + 2 = 0 \Rightarrow y = - 2$

$\displaystyle \therefore \sqrt{ (x+1)^2+y^2 } = (x+2)$

$\displaystyle \Rightarrow (x+1)^2 + 4 = ( x+2)^2$

$\displaystyle \Rightarrow x^2 + 1 + 2x + 4 = ( x+2)^2$

$\displaystyle \Rightarrow 5-4 = 2x$

$\displaystyle \Rightarrow x = \frac{1}{2}$

$\displaystyle \therefore z = \frac{1}{2} + i (-2)$

$\displaystyle \\$

Question 23: Solve the equation $\displaystyle |z| = z+1+2i$

Answer:

Let $\displaystyle z = x + iy$

$\displaystyle |z| = z + 1 + 2i$

$\displaystyle |x+iy| = x + iy + 1 + 2i$

$\displaystyle \sqrt{x^2 + y^2} = (x+1) + i ( y+2)$

Comparing, $\displaystyle y+2 = 0 \Rightarrow y = -2$

and $\displaystyle \sqrt{x^2 + y^2} = (x+1)$

$\displaystyle \Rightarrow x^2 + y^2 = x^2 + 1 + 2x$

$\displaystyle \Rightarrow 2x = y^2 - 1 = ( -2)^2 - 1 = 3$

$\displaystyle \Rightarrow x = \frac{3}{2}$

$\displaystyle \therefore z = \frac{3}{2} - 2i$

$\displaystyle \\$

Question 24: What is the smallest positive integer $\displaystyle n$ for which $\displaystyle (1+i)^{2n}= ( 1- i)^{2n} \ ?$

Answer:

$\displaystyle (1+i)^{2n}= ( 1- i)^{2n}$

$\displaystyle \Rightarrow [(1+i)^{2}]^n= [( 1- i)^{2}]^n$

$\displaystyle \Rightarrow (2i)^n = ( -2i)^n$

$\displaystyle \Rightarrow (2i)^n = ( -1)^n (2i)^n$

$\displaystyle \Rightarrow ( -1 )^n = 1$

$\displaystyle \therefore n$ is a multiple of $\displaystyle 2$ .

Hence the smallest positive number $\displaystyle n$ for which $\displaystyle (1+i)^{2n} = (1-i)^{2n}$ is $\displaystyle 2$

$\displaystyle \\$

Question 25: $\displaystyle \text{If } z_1, z_2, z_3$ , are complex numbers such that, $\displaystyle |z_1|= |z_2|=|z_3|= \Big| \frac{1}{z_1} + \frac{1}{z_2} + \frac{1}{z_3} \Big| = 1 , \text{ then find the value of } |z_1+z_2+z_3|$ .