Chapter-5: Complex Numbers and Quadratic Equations

Exercise: 5.1

Q.1: Express the following complex number in x + iy form; $$(4i)\;(-\frac{7}{4}i)$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

$$\\(4i)\;(-\frac{7}{4}i)$$ = -4 × $$\frac{7}{4}$$ × i × i = -7 i2 = -7(-1) [ Since, i2 = -1 ]

= 7

Q.2: Express the following complex number in x + iy form; $$i^{19} + i^{9}$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

$$i^{19} + i^{9}\;=i^{4\times2 + 1} + i^{4\times4 + 3}\\$$ = $$\\(i^{4})^{2}\cdot i + (i^{4})^{4}\cdot i^{3}\\$$

=$$\\ 1\times i + 1\times (-i)$$ [ Since, i4 = 1, i3 = -1 ]

= i + (-i)

= 0

Q.3: Express the following complex number in x + iy form; $$i^{-39}$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

$$i^{-39} = i^{-4\times 9 – 3}\\$$ $$\\= (i^{4})^{-9}\cdot i^{-3}$$

$$\\= (1)^{-9}\cdot i^{-3}\\$$ [ Since, i4 = 1 ]

$$\\=\frac{1}{i^{3}} = \frac{1}{-i}$$ [ Since, i3 = -i ]

$$\\= \frac{-1}{i}\times\frac{i}{i}\\$$

$$\\= \frac{-i}{i^{2}} = \frac{-i}{-1}$$ = i [ Since, i2 = -1 ]

Q.4: Express the following complex number in x + iy form; 5(6 + 6i) + i(6 + 6i)

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

5(6 + 6i) + i(6 + 6i) = 30 + 30i + 6i + 6i2

= 30 + 36i + 6(-1) [ Since, i2 = -1 ]

= 30 – 6 + 36i

= 24 + 36i

Q.5: Express the following complex number in x + iy form; (2 – 3i) – (-4 + 5i)

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

(2 – 3i) – (-4 + 5i) = 2 – 3i + 4 – 5i

= 6 – 8i

Q.6: Express the following complex number in x + iy form;

$$\left ( \frac{1}{4} + i\frac{3}{4} \right )- \left ( 6 + i\frac{4}{3} \right )$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

$$\\\left ( \frac{1}{4} + i\frac{3}{4} \right )- \left ( 6 + i\frac{4}{3} \right ) = \frac{1}{4} + i\frac{3}{4} – 6 – i\frac{4}{3}\\$$

=$$\\\left ( \frac{1}{4} – 6 \right )+ i\left ( \frac{3}{4} – \frac{4}{3} \right )\\$$

=$$\\\frac{-23}{4} + i(\frac{-7}{12})\\$$

=$$\boldsymbol{ \frac{-23}{4} – i\frac{7}{12}}$$

Q.7: Express the following complex number in x + iy form:

$$\left [ \left ( \frac{1}{3} + i\frac{7}{3} \right ) + \left ( 4 + i\frac{1}{3} \right ) \right ]- \left ( -\frac{4}{3} + i \right )$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

$$\left [ \left ( \frac{1}{3} + i\frac{7}{3} \right ) + \left ( 4 + i\frac{1}{3} \right ) \right ]- \left ( -\frac{4}{3} + i \right )\\$$

=$$\\\frac{1}{3} + i\frac{7}{3} + 4 + i\frac{1}{3} + \frac{4}{3} – i\\$$

=$$\\(\frac{1}{3} + \frac{4}{3} + 4) + i(\frac{7}{3} + \frac{1}{3} – 1)\\$$

=$$\\\boldsymbol{ \frac{17}{3} + i\frac{5}{3}}$$

Q.8: Express the following complex number in x + iy form: (1 – i)4

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

(1 – i)4 = $$[(1 – i)^{2}]^{2}\\$$

=$$[1 – 2i + i^{2}]^{2}$$=$$[1 – 1 – 2i]^{2}$$=$$(-2i)^{2}= (-2i)\times (-2i)= 4i^{2}$$ [ Since i2 = -1 ]

= -4

Q.9: Express the following complex number in ‘x + iy’ form; $$(\frac{1}{3} + 3i)^{3}$$

Sol:

=$$(\frac{1}{3} + 3i)^{3}:$$

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

=$$\\\frac{1}{27} + 27i^{3} + 3i(\frac{1}{3} + 3i)$$

=$$\\\frac{1}{27} + 27i^{3} + i + 9i^{2}$$

=$$\\\frac{1}{27} – 27i + i – 9$$ [Since, i3 = -i and i2 = -1]

=$$\\(\frac{1}{27} – 9) + i(-27i + 1)\\$$ = $$\boldsymbol{\frac{-242}{27} -26i}$$

Q.10: Express the following complex number in ‘x + iy’ form: $$(-2 – i\frac{1}{3})^{3}$$

SoL: Based on formulae given in Complex Numbers and Quadratic Equations

$$(-2 – i\frac{1}{3})^{3}\;=\;(-1)^{3}\;(2 + i\frac{1}{3})^{3}$$ = $$-[2^{3} + (\frac{i}{3})^{3} +3(2)\;(\frac{i}{3})\;(2 + \frac{i}{3})]$$ = $$-[8 + \frac{i^{3}}{27} + 2i(2 + \frac{i}{3})]$$

=$$\\-[8 + \frac{i^{3}}{27} + 4i + \frac{2i^{2}}{3})]$$ = $$-[8 – \frac{i}{27} + 4i – \frac{2}{3}]\\$$ [ Since, i3 = -i and i2 = -1 ]

=$$[\frac{22}{3} + i\frac{107}{27}]\\$$=$$\boldsymbol{ -\frac{22}{1} – i\frac{107}{27}}$$

Q.11: Find the multiplicative inverse of the given complex number, 7 – 6i.

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming, z = 7 – 6i

Now, $$\overline{z} = 7 + 6i$$

And, |z|2 = (7)2 + (-6)2 = 49 + 36 = 85

Therefore, the multiplicative inverse of the given complex number is given by:

$$\\\boldsymbol{z^{-1} = \frac{\overline{z}}{|z|^{2}} \;= \frac{7 + 6i}{85} \;= \frac{7}{85} + \frac{6}{85}i}$$

Q.12: Find the multiplicative inverse of the given complex number, $$\sqrt{7} + 4i$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming, z = $$\sqrt{7} + 4i$$

Now, $$\overline{z} = \sqrt{7} + 4i$$

And, |z|2 = $$(\sqrt{7})^{2} + (4)^{2} = 23$$

Therefore, the multiplicative inverse of the given complex number is given by:

$$\\\boldsymbol{z^{-1} = \frac{\overline{z}}{|z|^{2}} \;= \frac{\sqrt{7} – 4i}{23} \;= \frac{\sqrt{7}}{23} – \frac{4}{23}i}$$

Q.13: Find the multiplicative inverse of the given complex number, – i

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming, z = – i

Now, $$\overline{z} = i$$

And, |z|2 = (-i)2 = 1

Therefore, the multiplicative inverse of the given complex number is given by:

$$\\z^{-1} = \frac{\overline{z}}{|z|^{2}} = \frac{i}{1}$$ = i

Q.14: Express the following complex number in ‘x + iy’ form: $$\frac{(3 + i\sqrt{5})(3 – i\sqrt{5})}{(\sqrt{3} + i\sqrt{2}) – (\sqrt{3} – i\sqrt{2})}$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

$$\boldsymbol{\frac{(3 + i\sqrt{5})(3 – i\sqrt{5})}{(\sqrt{3} + i\sqrt{2}) – (\sqrt{3} – i\sqrt{2})}}= \frac{(3)^{2} – (i\sqrt{5})^{2}}{\sqrt{3} + i\sqrt{2} – \sqrt{3} + i\sqrt{2}\\}$$

Now, by using (a + b) (a – b) = (a – b)2

$$= \frac{9 – 5i^{2}}{2\sqrt{2}\;i}= \frac{9 – 5\times (-1)}{2\sqrt{2}\;i}= \frac{9 + 5}{2\sqrt{2}\;i}\times\frac{i}{i}= \frac{14i}{2\sqrt{2}\;i^{2}}\\$$

$$\\\boldsymbol{= \frac{14i}{-2\sqrt{2}}= \frac{-7i}{\sqrt{2}}\times\frac{\sqrt{2}}{\sqrt{2}}=\frac{-7\sqrt{2}\;i}{2}}$$

Exercise: 5.2

Q.1: Find out the modulus and argument of the given complex number:

$$z = -1 -\;i\sqrt{3}$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming $$r\cos\Theta = -1 \;and\; r\sin\Theta = -\sqrt{3}$$

On squaring both sides and then adding them we will get:

$$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (-\sqrt{3})^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 3 \\ \Rightarrow r^{2}(1) = 4 \\ \Rightarrow r = \sqrt{4} = 2\\$$

Therefore, Modulus = 2 [ Since, r > 0 ]

Now, $$2\cos\Theta = -1 \;and\; 2\sin\Theta = \sqrt{3} \\ \Rightarrow \cos\Theta = \frac{-1}{2} \;and\; \sin\Theta = \frac{\sqrt{3}}{2}$$

Since, the values of both $$\cos\Theta \;and\; \sin\Theta$$are negative and they both are negative in 3rd quadrant

Therefore, Argument = $$-(\pi – \frac{\pi}{3}) = \frac{-2\pi}{3}$$

Q.2: Find out the modulus and argument of the given complex number

$$z = -\sqrt{3} + i$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming $$r\cos\Theta = -\sqrt{3} \;and\; r\sin\Theta = 1$$

On squaring both sides and then adding them we will get:

$$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-\sqrt{3})^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 3 + 1 \\ \Rightarrow r^{2}(1) = 4 \\ \Rightarrow r = \sqrt{4} = 2\\$$

Therefore, Modulus = 2 [ Since, r > 0 ]

Now, $$2\cos\Theta = -\sqrt{3} \;\;and\;\; 2\sin\Theta = 1 \\ \Rightarrow \cos\Theta = \frac{-\sqrt{3}}{2} \;and\; \sin\Theta = \frac{1}{2}$$

Here, $$\Theta$$ lies in the 2nd quadrant.

Therefore, Argument = $$(\pi – \frac{\pi}{6}) = \frac{5\pi}{6}$$

Q.3: Convert the following complex number in polar form: 1 – i

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming $$r\cos\Theta = 1 \;and\; r\sin\Theta = -1$$

On squaring both sides and then adding them we will get:

$$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (1)^{2} + (-1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}\\$$

Therefore, Modulus = $$\sqrt{2}$$ [ Since, r > 0 ]

Now, $$\sqrt{2}\cos\Theta = 1 \;and\; \sqrt{2}\sin\Theta = -1$$

$$\Rightarrow \cos\Theta = \frac{1}{\sqrt{2}} \;and\; \sin\Theta = \frac{-1}{\sqrt{2}}\\$$

Here, $$\Theta$$ lies in 4th quadrant.

Therefore, $$\Theta$$ = $$(- \frac{\pi}{4})$$

Therefore, the required polar form is:

$$1 – i = r\cos\Theta + ir\sin\Theta = \sqrt{2}\cos(-\frac{\pi}{4}) + i\sqrt{2}\sin(-\frac{\pi}{4})\\$$

Therefore, $$\\\boldsymbol{1-i=\sqrt{2}\left [ \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) \right ]}$$

Q.4: Convert the following complex number in polar form; -1 + i

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming $$r\cos\Theta = -1 \;and\; r\sin\Theta = 1$$

On squaring both sides and then adding them we will get:

$$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}$$

Therefore, Modulus = $$\sqrt{2}$$ [ Since, r > 0 ]

Now,$$\sqrt{2}\cos\Theta = -1 \;and\; \sqrt{2}\sin\Theta = 1$$

$$\Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;and\; \sin\Theta = \frac{1}{\sqrt{2}}\\$$

Here, $$\Theta$$ lies in 2nd quadrant.

Therefore, $$\Theta$$ = $$(\pi- \frac{\pi}{4}) = \frac{3\pi}{4}$$

So, the required polar form is:

$$-1 + i = r\cos\Theta + ir\sin\Theta = \sqrt{2}\cos(\frac{3\pi}{4}) + i\sqrt{2}\;\sin(\frac{3\pi}{4})\\$$

Therefore, -1 + i = $$\\\mathbf{\sqrt{2}\left [ \cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4}) \right ]}$$

Q.5: Convert the following complex number in polar form; -3

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming $$r\cos\Theta = -3 \;and\; r\sin\Theta = 0$$

On squaring both sides and then adding them we will get:

$$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-3)^{2} + (0)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 9 \\ \Rightarrow r^{2}(1) = 9 \\ \Rightarrow r = \sqrt{9} = 3\\$$

Therefore, Modulus = 3 [ Since, r > 0]

Now, $$3\cos\Theta = -3 \;and\; 3\sin\Theta = 0 \\ \Rightarrow \cos\Theta = -1 \;and\; \sin\Theta = 0$$

Therefore, $$\Theta = \pi$$

So, the required polar form is:

$$-3 = r\cos\Theta + ir\sin\Theta = 3\cos(\pi) + i.\;3\sin(\pi)\\$$

Therefore, $$\boldsymbol{-3= 3[\cos(\pi) + i\sin(\pi)]}$$

Q.6: Convert the following complex number in polar form; -1 – i

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming $$r\cos\Theta = -1 \;and\; r\sin\Theta = -1$$

On squaring both sides and then adding them we will get:

$$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (-1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}$$

Therefore, Modulus = $$\sqrt{2}$$ [ Since, r > 0 ]

Now, $$\sqrt{2}\cos\Theta$$=$$-1 \;\;and\;\; \sqrt{2}\sin\Theta = -1$$

$$\Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;\;and\;\; \sin\Theta = \frac{-1}{\sqrt{2}}$$

Q.7: Convert the following complex number in polar form: $$z = \sqrt{3} + i$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

$$z = \sqrt{3} + i$$

Assuming $$r\cos\Theta = \sqrt{3} \;and\; r\sin\Theta = 1$$

On squaring both sides and then adding them we will get:

$$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (\sqrt{3})^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 3 + 1 \\ \Rightarrow r^{2}(1) = 4 \\ \Rightarrow r = \sqrt{4} = 2\\$$

Therefore, Modulus = 2 [Since, r > 0]

Now,

$$2\cos\Theta = \sqrt{3} \;and\; 2\sin\Theta = 1 \\ \Rightarrow \cos\Theta = \frac{\sqrt{3}}{2} \;and\; \sin\Theta = \frac{1}{2}$$

Here, $$\Theta$$ lies in 1st quadrant.

Therefore, $$\Theta$$ = $$(\frac{\pi}{6})$$

So, the required polar form is:

$$\sqrt{3} + i = r\cos\Theta + ir\sin\Theta = 2\cos(\frac{\pi}{6}) + i2\sin(\frac{\pi}{6})\\$$

Therefore, $$\\\boldsymbol{\sqrt{3} + i = 2\left [ \cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6}) \right ]}$$

Q.8: Convert the following complex number in polar form: i

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming $$r\cos\Theta = 0 \;and\; r\sin\Theta = 1$$

On squaring both sides and then adding them we will get:

$$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (0)^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 \\ \Rightarrow r^{2}(1) = 1 \\ \Rightarrow r = \sqrt{1} = 1\\$$

Therefore, Modulus = 1 [ Since, as r > 0)]

Now, $$\cos\Theta = 0 \;and\; 3\sin\Theta = 1$$

Therefore, $$\Theta = \frac{\pi}{2}$$

So, the required polar form is:

$$\boldsymbol{i= \cos(\frac{\pi}{2}) + i\sin(\frac{\pi}{2})}$$

Miscellaneous Exercise

Q-1: Evaluate the following:

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

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

$$[i^{18} + (\frac{1}{i})^{25}]^{3}$$ = $$[i^{4\times4 + 2} + \frac{1}{i^{4\times6 + 1}}]^{3}$$ = $$[(i^{4})^{4}\cdot i^{2} + \frac{1}{(i^{6})^{4}\cdot i}]^{3}\\$$ = $$[i^{2} + \frac{1}{i}]^{3}$$ [Since, i4 = 1]

=$$[-1 + \frac{1}{i}\times\frac{i}{i}]^{3} = [-1 + \frac{i}{i^{2}}]^{3} = [-1 – i]^{3}$$

=$$(-1)^{3}[1 + i]^{3} = -[1 + i^{3} + 3i(i + 1)]$$

=$$-[1 – i^{3} + 3i + 3i^{2}]$$

= – [1 – i + 3i – 3] = – [-2 + 2i]

= 2 – 2i

Q-2: For the two given complex number z1 and z2 below prove that

Re(z1z2) = Re z1Re z2 – Im z1 Im z2

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming z1 = a1 + ib1 and z2 = a2 + ib2

Therefore, z1z2 = (a1 + ib1) ( a2 + ib2)

= a1(a2 + ib2) + ib2(a2 + ib2)

= a1a2 + ia1b2 + ib1a2 + i2b1b2

= a1a2 + ia1b2 + ib1a2 – b1b2

= (a1a2 – b1b2) + i(a1b2 + b1a2)

Re(z1z2) = a1a2 – b1b2

Re(z1z2) = Rez1 Rez2 – Imz1 Imz2

Hence, Proved

Q-3: Express $$\left ( \frac{1}{1 – 4i} – \frac{2}{1 + i} \right )\;\left ( \frac{3 – 4i}{5 + i} \right )$$ in the standard form i.e. a + ib.

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

$$\left [ \frac{1}{1 – 4i} – \frac{2}{1 + i} \right ]\;\left [ \frac{3 – 4i}{5 + i} \right ] = \left [ \frac{(1 + i) -2(1 – 4i)}{(1- 4i)(1 + i)} \right ]\left [ \frac{3 – 4i}{5 + i} \right ]$$ $$= \left [ \frac{1 + i – 2 + 8i}{1 + i – 4i -4i^{2}} \right ]\left [ \frac{3 – 4i}{5 + i} \right ] = \left [ \frac{-1 + 9i}{5 – 3i} \right ]\left [ \frac{3 – 4i}{5 + i} \right ]$$ $$= \left [ \frac{-3 + 4i + 27i – 36i^{2}}{25 + 5i – 15i – 3i^{2}} \right ] = \frac{33 + 31i}{28 – 10i} = \frac{33 + 31i}{2(14 – 5i)}$$ $$= \frac{33 + 31i}{2(14 – 5i)}\times\frac{14 + 5i}{14 + 5i}$$

Now, on multiplying and dividing by (14 + 5i) we will get:

$$=\frac{462 + 165i + 434i + 155i^{2}}{2[(14)^{2} – (5i)^{2}]} = \frac{307 + 599i}{2(196 – 25i^{2})}$$ $$=\frac{307 + 599i}{2(221)} = \frac{307 + 599i}{442}$$ $$=\frac{307}{442} + i\frac{599}{442}$$

Q-4: If $$a- ib =\sqrt{\frac{x \;-\; iy}{u \;- \;iv}}$$ then prove that $$(a^{2} + b^{2})^{2} = \frac{x^{2} \;+ \;y^{2}}{u^{2} \;+\; v^{2}}$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

$$a – ib =\sqrt{\frac{x \;-\; iy}{u \;- \;iv}}$$ $$= \sqrt{\frac{x – iy}{u – iv}\times\frac{x + iy}{u + iv}}\\$$

Now, on multiplying and dividing by (u + iv) we will get:

=$$\\\sqrt{\frac{(xu\; +\; yv) \;+\; i\;(xv \;- \;yu)}{u^{2} \;+\; v^{2}}}$$

Therefore, $$(a – ib)^{2} = \frac{(xu \;+\; yv) \;+ \;i(xv\; – \;yu)}{u^{2}\; + \;v^{2}}\\$$

$$\\\Rightarrow a^{2} – 2iab – b^{2} = \frac{(xu \;+ \;yv) \;+\;i(xv \;-\; yu)}{u^{2} \;+\; v^{2}}$$

On comparing both sides, we will get:

$$a^{2} – b^{2} = \frac{xu \;+\; yv}{u^{2} \;+\; v^{2}}\;and\; -2ab = \frac{xv\; -\; yu}{u^{2} + v^{2}}$$……………..(a)

(a2 + b2)2 = (a2 – b2)2 + 4a2b2

=$$\\(\frac{xu \;+\; yv}{u^{2} \;+\; v^{2}})^{2} +(\frac{xv \;- \;u}{u^{2}\; + \;v^{2}})^{2}\\$$

Now, by using equation (a) we will get:

$$= \frac{x^{2}u^{2}\; +\; y^{2}v^{2} \;+\; 2xyuv \;+\; x^{2}v^{2}\; +\; y^{2}u^{2} \;- \;2xyuv}{(u^{2} \;+\; v^{2})^{2}}$$ $$=\frac{x^{2}u^{2}\; +\; y^{2}v^{2}\; +\; x^{2}v^{2} \;+\; y^{2}u^{2}}{(u^{2} \;+\; v^{2})^{2}}$$ $$= \frac{x^{2}(u^{2}\; +\; v^{2}) \;+ \;y^{2}(u^{2}\; +\; v^{2})^{2}}{(u^{2}\; + \;v^{2})^{2}}$$ $$= \frac{(x^{2} \;+ \;y^{2})\;(u^{2} \;+\; v^{2})}{(u^{2} \;+ \;v^{2})^{2}} = \frac{x^{2} \;+ \;y^{2}}{u^{2}\; +\; v^{2}}\\$$

Hence, proved

Q-5) Convert the following complex number in polar form:

1: $$\frac{1 + 7i}{(2 – i)^{2}}\\$$

2:$$\\\frac{1 + 3i}{1 – 2i}$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

1: Here, z = $$\frac{1 + 7i}{(2 – i)^{2}}$$

$$\frac{1 + 7i}{(2 – i)^{2}} = \frac{1 + 7i}{4 + i^{2} – 4i} = \frac{1 + 7i}{4 – 1 – 4i}\\$$

=$$\\\frac{1 + 7i}{3 – 4i}\times\frac{3 + 4i}{3 + 4i} = \frac{3 + 4i + 21i + 28i^{2}}{3^{2} + 4^{2}}\\$$

=$$\\\frac{3 + 4i + 21i – 28}{3^{2} + 4^{2}} = \frac{-25 + 25i}{25}$$ = -i + 1

Assuming $$r\cos\Theta = -1 \;and\; r\sin\Theta = 1$$

On squaring on both sides and then adding them we will get:

$$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}$$

Therefore, Modulus = $$\sqrt{2}$$ [ Since, r > 0]

Now, $$\sqrt{2}\cos\Theta = -1 \;and\; \sqrt{2}\sin\Theta = 1 \\ \Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;and\; \sin\Theta = \frac{1}{\sqrt{2}}$$

Here, $$\Theta$$ lies in 2nd quadrant.

Therefore, $$\Theta$$ = $$(\pi- \frac{\pi}{4}) = \frac{3\pi}{4}$$

Therefore, the required polar form is:

$$-1 + i = r\cos\Theta + ir\sin\Theta = \sqrt{2}\cos(\frac{3\pi}{4}) + i\sqrt{2}\sin(\frac{3\pi}{4})\\$$

=$$\\ \sqrt{2}[\cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4})]$$

2: Here, z = $$\frac{1 \;+ \;3i}{1\; -\; 2i}$$

$$\frac{1 \;+ \;3i}{1 \;- \;2i} = \frac{1 \;+ \;3i}{1 \;- \;2i}\times\frac{1 \;+\; 2i}{1\; + \;2i}\\$$

=$$\\ \frac{1 \;+ 2i \;+ \;3i\; – 6}{1\; + \;4} = \frac{-5 \;+\; 5i}{5}$$ = -i + 1

Assuming $$r\cos\Theta = -1 \;and\; r\sin\Theta = 1$$

On squaring both sides and then adding them we will get:

$$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1)^{2} + (1)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1 + 1 \\ \Rightarrow r^{2}(1) = 2 \\ \Rightarrow r = \sqrt{2}$$

Therefore, Modulus = $$\sqrt{2}$$ [ Since, r > 0]

Now,$$\sqrt{2}\cos\Theta = -1 \;and\; \sqrt{2}\sin\Theta = 1$$

$$\Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;and\; \sin\Theta = \frac{1}{\sqrt{2}}$$

Here, $$\Theta$$ lies in 2nd quadrant.

Now, $$\Theta$$ = $$(\pi- \frac{\pi}{4}) = \frac{3\pi}{4}$$

Therefore, the required polar form is:

$$-1 + i = r\cos\Theta + ir\sin\Theta = \sqrt{2}\cos(\frac{3\pi}{4}) + i\sqrt{2}\sin(\frac{3\pi}{4})\\$$

=$$\\ \sqrt{2}\;[\cos(\frac{3\pi}{4}) + i\sin(\frac{3\pi}{4})]$$

Q-6: Solve the equation given below:

3y2 – 4y + $$\frac{20}{3}$$ = 0

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Here, 3y2 – 4y + $$\frac{20}{3}$$ = 0

The given equation can be also be written as,

9y2 – 12y + 20 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 9, b = -12 and c = 20

Now, the discriminant is:

$$D = b^{2} – 4ac = (-12)^{2} – 4\times 9\times 20 = 144 – 720 = -576$$

Now, the solution is:

$$y=\frac{-b \;\pm \;\sqrt{D}}{2a}$$ $$= \frac{-(12) \;\pm \;\sqrt{-576}}{2(9)}$$ $$= \frac{12\; \pm \;\sqrt{576i^{2}}}{18}\\$$

=$$\\ \frac{12 \;\pm \;24\;i}{18}= \frac{2 \;\pm \; 4\;i}{3}$$

$$=\frac{2}{3}\; \pm \;\frac{4}{3}\;i$$

Therefore, y = $$\boldsymbol{= \frac{2}{3}\; \pm \;\frac{4}{3}\;i}$$

Q-7: Solve the equation given below:

y2 – 2y + $$\frac{3}{2}$$ = 0

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Here, y2 – 2y + $$\frac{3}{2}$$= 0

The given equation can be also be written as:

2y2 – 4y + 3 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 2, b = -4 and c = 3

Now, the discriminant is:

$$D = b^{2} – 4ac = (-4)^{2} – 4\times (2)\times (3) = 16 – 24 = -8$$

Now, the solution is:

$$y = \frac{-b \;\pm \;\sqrt{D}}{2a}$$ $$= \frac{-(-4) \;\pm\;\sqrt{-8}}{2(2)}$$ $$= \frac{4 \;\pm \;\sqrt{8i^{2}}}{4}\\$$

=$$\\\frac{4 \;\pm \;2\sqrt{2}\;i}{4}= \frac{2 \;\pm \;\sqrt{2}\;i}{2}$$

$$= 1 \;\pm \;\frac{\sqrt{2}}{2}i$$

Therefore, y $$\boldsymbol{= 1 \;\pm \;\frac{\sqrt{2}}{2}i}$$

Q-8: Solve the equation given below:

27y2 – 10y + 1 = 0

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Here, 27y2 – 10y + 1 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 27, b = -10 and c = 1

Now, the discriminant is:

$$D = b^{2} – 4ac = (-10)^{2} – 4\times (27)\times (1) = 100 – 108 = -8$$

Now, the solution is:

$$y = \frac{-b \;\pm \;\sqrt{D}}{2a}$$ $$= \frac{-(-10)\; \pm \;\sqrt{-8}}{2(27)}$$ $$= \frac{10 \;\pm \;\sqrt{8i^{2}}}{54}$$

=$$\\\frac{10 \;\pm \;2\sqrt{2}\;i}{54} = \frac{5 \pm \sqrt{2}\;i}{27}$$

$$= \frac{5}{7}\; \pm \;\frac{\sqrt{2}}{27}\;i\\$$

Therefore, y $$\boldsymbol{= \frac{5}{7} \;\pm \;\frac{\sqrt{2}}{27}\;i}$$

Q-9: Solve the equation given below:

21y2 – 28y + 10 = 0

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Here, 21y2 – 28y + 10 = 0

On comparing the given equation with ax2 + bx + c = 0, we will get:

a = 21, b = -28 and c = 10

Now, the discriminant is:

$$D = b^{2} – 4ac = (-28)^{2} – 4\times (21)\times (10) = 784 – 840 = -56$$

Now, the solution is:

Since, $$y = \frac{-b \;\pm \;\sqrt{D}}{2a}$$

$$= \frac{-(-28) \;\pm \;\sqrt{-56}}{2(21)}$$ $$= \frac{28 \;\pm \;\sqrt{56i^{2}}}{42}\\$$

=$$\\\frac{28 \;\pm \;2\sqrt{14}\;i}{42} = \frac{14 \;\pm \;\sqrt{14}\;i}{21}$$$$= \frac{3}{2} \;\pm \;\frac{\sqrt{14}}{21}\;i$$

Therefore, y $$\boldsymbol{= \frac{3}{2} \;\pm \;\frac{\sqrt{14}}{21}\;i}$$

Q-10: z1 = 2 – i, z2 = 1 + i, find $$\left | \frac{z_{1} + z_{2} + 1}{z_{1} – z_{2} + i} \right |$$

Sol:

Here, z1 = 2 – i, z2 = 1 + i

$$\left | \frac{z_{1} + z_{2} + 1}{z_{1} – z_{2} + i} \right | = \left | \frac{(2 – i) + ( 1 + i) + 1}{(2 – i) – ( 1 + i) + 1} \right |\\$$

=$$\\ \left | \frac{4}{2 – 2i} \right | = \left | \frac{2}{1 – i} \right |= \left | \frac{2}{1 – i}\times\frac{1 + i}{1 + i} \right | = \left | \frac{2(1 + i)}{1 – i^{2}} \right |\\$$

=$$\\ \left | \frac{2(1 + i)}{1 + 1} \right | = \left | \frac{2(1 + i)}{2} \right |= \left | 1 + i \right | = \sqrt{1^{2} + 1^{2} } = \sqrt{2}$$

Q-11: If x + iy = $$\frac{(a + i)^{2}}{2a^{2} + 1}$$ then prove that

x2 + y2 = $$\frac{(a^{2} + 1)^{2}}{(2a + 1)^{2}}$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Here, x + iy = $$\frac{(a + i)^{2}}{2a^{2} + 1}\\$$

=$$\\\frac{a^{2} + i^{2} + 2ai}{2a^{2} + 1} = \frac{a^{2} – 1 + 2ai}{2a^{2} + 1} = \frac{a^{2} – 1}{2a^{2} + 1} + i\frac{2a}{2a^{2} + 1}\\$$

On comparing both sides, we will get:

$$x = \frac{a^{2} – 1}{2a^{2} + 1}\;and\; y = \frac{2a}{2a^{2} + 1}\\$$

Therefore, $$x^{2} + y^{2} = (\frac{a^{2} – 1}{2a^{2} + 1})^{2} + (\frac{2a}{2a^{2} + 1})^{2}\\$$

$$= \frac{a^{4} + 1 – 2a^{2} + 4a^{2}}{(2a + 1)^{2}} = \frac{a^{4} + 1 + 2a^{2}}{(2a^{2} + 1)^{2}}$$ $$= \frac{(a^{2} + 1)^{2}}{(2a^{2} + 1)^{2}}\\$$

Hence, proved

Q-12: If z1 = 2 – i and z2 = -2 + i is given find out,

1: $$Im(\frac{1}{z_{1}\overline{z_{1}}})$$

2:$$Re(\frac{z_{1}z_{2}}{z_{1}})$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Here, z1 = 2 – i and z2 = -2 + i

1. $$Im(\frac{1}{z_{1}\overline{z_{1}}})$$

$$= \frac{1}{(2 + i)(-2 + i)} = \frac{1}{2^{2} – i^{2}} \\ = \frac{1}{4 + 1} = \frac{1}{5}$$

On comparing both sides, we will get:

$$Im(\frac{1}{z_{1}\overline{z_{1}}}) = 0\\$$

2. $$Re(\frac{z_{1}z_{2}}{z_{1}})$$

z1z2 = (2 – i)(- 2 + i) = -4 + 2i + 2i – i2 = -4 + 4i – (-1) = -3 + 4i

$$\overline{z_{1}} = 2 + i\\$$ $$\frac{z_{1}z_{2}}{\overline{z_{1}}} = \frac{-3 + 4i}{2 + i}$$

Now multiplying and dividing it by (2 – i), we will get:

=$$\frac{-3 + 4i}{2 + i}\times\frac{2 – i}{2 – i}$$

=$$\frac{-6 + 3i + 8i – 4i^{2}}{2^{2} – i^{2}}$$

=$$\frac{-6 + 11i – 4(-1)}{4 + 1}$$

=$$\frac{2 + 11i}{5} = \frac{-2}{5} + \frac{11}{5}i$$

On comparing both the sides, we will get:

$$\boldsymbol{Re(\frac{z_{1}\;z_{2}}{\overline{z_{1}}}) = \frac{-2}{5}}$$

Q-13: Find out the modulus and argument of the given complex number

$$\frac{1 \;+\;2i}{1\; -\; 3i}$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Here, z = $$\frac{1 + 2i}{1 – 3i}$$

$$\frac{1\; +\; 2i}{1\; -\; 3i}\times\frac{1\; +\; 3i}{1 \;+ \;3i} = \frac{1 \;+\; 3i \;+\; 2i \;+ \;6i^{2}}{1^{2} \;- \;9i^{2}}\\$$

=$$\\\frac{1\; +\; 3i \;+\; 2i \;+ \;6i^{2}}{1^{2} +\; 9} = \frac{-5\; +\; 5i}{10}$$

=$$\frac{-5}{10}\; +\; \frac{5}{10}i = \frac{-1}{2}\; +\; \frac{1}{2}i$$

Assuming $$r\cos\Theta = -1/2 \;and\; r\sin\Theta = 1/2$$

On squaring both sides and then adding them we will get:

$$(r\cos\Theta)^{2} + (r\sin\Theta)^{2} = (-1/2)^{2} + (1/2)^{2} \\ \Rightarrow r^{2}(\cos^{2}\Theta + \sin^{2}\Theta ) = 1/4 + 1/4 \\ \Rightarrow r^{2}(1) = 1/2 \\ \Rightarrow r = \frac{1}{\sqrt{2}}\\$$

Therefore, Modulus = $$\frac{1}{\sqrt{2}}$$ [ Since, r > 0]

Now, $$\frac{1}{\sqrt{2}}\cos\Theta = \frac{-1}{2} \;and\; \frac{1}{\sqrt{2}}\sin\Theta = \frac{1}{2}\\$$

$$\Rightarrow \cos\Theta = \frac{-1}{\sqrt{2}} \;and\; \sin\Theta = \frac{1}{\sqrt{2}}$$

Here, $$\Theta$$ lies in 2nd quadrant.

Therefore, $$\Theta = (\pi- \frac{\pi}{4}) = \frac{3\pi}{4}$$

Q-14: Find the real number a and b if (3 + 5i) (a – ib) is the conjugate if -6 – 24i.

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming z = (3 + 5i) (a – ib)

z = 3a + 5ai – 3bi – 5bi2 = 3a + 5ai – 3bi + 5b = (3a + 5b) + i(5a – 3b)

Therefore, $$\overline{z} = (3a + 5b) – i(5a – 3b)$$

Here, given that $$\overline{z} = -6 – 24i$$

Therefore, (3a + 5b) – i(5a – 3b) = -6 – 24i

On comparing both sides, we will get:

3a + 5b = -6 . . . . . . . . . . . . (a)

5a – 3b = 24 . . . . . . . . . . . . (b)

On solving these two equations, we will get:

9a + 15b = -18

25a – 15b = 120

34a = 102

Therefore, a = 3

Now, 3(3) + 5b = -6 [From equation (a)]

5b = -15

Therefore, b = -3

Hence, a = 3 and b = -3

Q-15: Find the modulus of $$\frac{1\; +\; i}{1\; – \;i}\; -\; \frac{1\; -\; i}{1 \;+\; i}$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

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

=$$\\ \frac{1 \;+ \;i^{2} \;+\; 2i \;-\; 1\; -\; i^{2}\; +\;2i}{1 \;+\; 1} = \frac{4i}{2} = 2i\\$$

$$\\\Rightarrow$$ $$\left | \frac{1\; +\; i}{1\; -\; i} \;- \;\frac{1\; -\; i}{1 \;+\; i} \right | = \left | 2i \right | = \sqrt{2^{2}} = 2$$

Q-16: If (a + ib)3 = u + iv, then prove that:

$$\frac{u}{a} + \frac{v}{b} = 4(a^{2} – b^{2})$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Here, (a + ib)3 = u + iv

a3 + (ib)3 + 3a(ib)(a + ib) = u + iv

a3 + i3b3 + 3a2bi + 3ab2i2 = u + iv

a3 – ib3 + 3a2bi – 3ab2 = u + iv

(a3 – 3ab2) + i(3a2b – b3) = u + iv

On comparing both the sides, we will get:

u = a3 – 3ab2 and v = 3a2b – b3

Hence, Proved

Q-17: If A and B are two different complex numbers with |B| = 1, then find $$\left | \frac{B – A}{1 – \overline{A}B} \right |$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Assuming, A = x + iy and B = u + iv

Here, |B| = 1,

Therefore, $$\sqrt{u^{2} + v^{2}} = 1\\$$

$$\\\Rightarrow$$ u2 + v2 = 1 . . . . . . . . . . . . (a)

$$\\\left | \frac{B – A}{1\; – \;\overline{A}B} \right | \;= \;\left | \frac{(x \;+\; iy) \;-\; (u \;+\; iv)}{1 – (u\; – \;iv)(x \;+\; iy)} \right |\\$$

=$$\\ \left | \frac{(x \;- \;u) \;+ \;i(y\; -\; v)}{1 \;- \;(xu \;+\; uiy \;+ \;ivx \;+ \;yv)} \right |\\$$

=$$\\ \left | \frac{(x\; – \;u) + i(y \;- \;v)}{(1 \;- \;xu\; – \;yv) \;+\; i(xv \;-\; yu)} \right | = \frac{\left | (x \;-\; v) \;+\; i(y\; -\; v) \right |}{\left | (1 \;- \;xu \;-\; yv) \;+\; i(xv\; – \;yu) \right |}\\$$

Since, $$\\\left | \frac{z_{1}}{z_{2}} \right | = \frac{\left | z_{1} \right |}{\left | z_{2} \right |}\\$$

Therefore, $$\\= \frac{\sqrt{(x \;-\; u)^{2} \;+\; (y\; – \;v)^{2} }}{\sqrt{(1 \;- \;xu\; -\; yv)+ i(xv\; -\; uy)}}\\$$

=$$\\ \frac{\sqrt{x^{2} \;+\; u^{2}\; -\; 2ux\; + \;y^{2}\; +\; v^{2} \;- \;2vy}}{\sqrt{1 \;+\; u^{2}x^{2} \;+\; v^{2}y^{2} \;-\; 2ux \;+ \;2uvxy\; – \;2vy \;+ \;v^{2}x^{2} \;+\; u^{2}y^{2}\; -\; 2uvxy}}\\$$

=$$\\ \frac{\sqrt{(x^{2} \;+\; y^{2})\; +\; u^{2}\; +\; v^{2} \;- \;2ux\; – \;2vy}}{\sqrt{1\; +\; u^{2}(x^{2}\; +\; y^{2})\; +\; v^{2}\;(x^{2} \;+\; y^{2}) \;-\;2xu\; -\;2yv }}\\$$

=$$\\ \frac{\sqrt{1 \;+\; u^{2}\; +\; v^{2}\; – \;2xu\; – \;2yv}}{\sqrt{1 \;+ \;u^{2} \;+ \;v^{2} \;-\; 2xu\;- \;2yv}}=1\\$$ [ From equation (a) ]

Therefore, $$\\\left | \frac{B \;- \;A}{1 \;- \;\overline{A}B} \right | = 1$$

Q-18: Find the number of non-zero solutions of the equation |1 – i|y = 2y

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

|1 – i|y = 2y

$$\Rightarrow \left ( \sqrt{1^{2} + (-1)^{2}} \right )^{y} = 2^{y}\\$$

$$\\\Rightarrow$$ $$(\sqrt{2})^{y} = 2^{y}$$ $$\Rightarrow 2^{\frac{y}{2}} = y$$

$$\Rightarrow$$ $$y = 2y \Rightarrow 2y \;- \;y = 0 \Rightarrow y = 0$$

Therefore, there is only one integral solution 0.

Q.19: If (s + it) (u + iv) (w + ix) (y + iz) = X + iY, then prove that:

(s2 + t2) (u2 + v2) (w2 + x2) (y2 + z2) = X2 + Y2.

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

Here, (s + it) (u + iv) (w + ix) (y + iz) = X + iY

Therefore, |(s + it)(u + iv)(w + ix)(y + iz)| = |X + iY|

|(s + it)| × |(u + iv)| × |(w + ix)| × |(y + iz)| = |X + iY| [because |z1z2| = |z1||z2|]

$$\\\Rightarrow \sqrt{s^{2} + t^{2}}\times \sqrt{u^{2} + v^{2}}\times \sqrt{w^{2} + x^{2}}\times \sqrt{y^{2} + z^{2}} = \sqrt{X^{2} + Y^{2}}\\$$

Now, on squaring both the sides, we will get:

(s2 + t2) (u2 + v2) (w2 + x2) (y2 + z2) = X2 + Y2

Hence, Proved

Q.20: Find the least positive value of x for the following:

$$(\frac{1 + i}{1 – i})^{x} = 1$$

Sol: Based on formulae given in Complex Numbers and Quadratic Equations

$$(\frac{1 \;+ \;i}{1\; -\; i})^{x} = 1\\$$ $$\\\Rightarrow \left ( \frac{1\; +\; i}{1 \;-\; i}\times\frac{1 \;+\; i}{1 \;+\; i} \right )^{x} = 1\\$$ $$\\\Rightarrow \left ( \frac{(1\; +\; i)^{2}}{1\; +\; 1} \right )^{x} = 1\\$$

$$\\\Rightarrow$$ $$\left ( \frac{1\; +\; i^{2}\;+ \;2i}{2} \right )^{x} = 1\\$$

$$\\\Rightarrow \left ( \frac{2i}{2} \right )^{x} = 1 \Rightarrow i^{x} = 1\\$$

Hence, x = 4m, where m belongs to Z.

Thus, the least positive integer = 1

Therefore, the least positive integral for given problem = x = 4m = 4 × 1 = 4