NCERT Solutions For Class 11 Maths Chapter 5

NCERT Solutions Class 11 Maths Complex Numbers and Quadratic Equations

Ncert Solutions For Class 11 Maths Chapter 5 PDF Free Download

Exercise: 5.1

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

Sol:

\(\\(4i)\;(-\frac{7}{4}i)\)  = -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:

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

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

= i + (-i)

= 0

 

 

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

 

Sol:

\(i^{-39} = i^{-4\times 9 – 3}\\\)

\(\\= (i^{4})^{-9}\cdot i^{-3}\)

\(\\= (1)^{-9}\cdot i^{-3}\\\)   [ 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:

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:

(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:

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

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

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

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

 

 

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:

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

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

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

=\(\\\boldsymbol{ \frac{17}{3} + i\frac{5}{3}}\)

 

 

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

Sol:

 

(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:

 

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

=\(\\-[8 + \frac{i^{3}}{27} + 4i + \frac{2i^{2}}{3})]\) = \(-[8 – \frac{i}{27} + 4i – \frac{2}{3}]\\\)      [ 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:

 

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:

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:

 

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:

 

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

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:

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:

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:

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:

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:

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:

 

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:

\(z = \sqrt{3} + i\)

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:

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:

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

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:

\(\left [ \frac{1}{1 – 4i} – \frac{2}{1 + i} \right ]\;\left [ \frac{3 – 4i}{5 + i} \right ] = \left [ \frac{(1 + i) -2(1 – 4i)}{(1- 4i)(1 + i)} \right ]\left [ \frac{3 – 4i}{5 + i} \right ]\)

\(= \left [ \frac{1 + i – 2 + 8i}{1 + i – 4i -4i^{2}} \right ]\left [ \frac{3 – 4i}{5 + i} \right ] = \left [ \frac{-1 + 9i}{5 – 3i} \right ]\left [ \frac{3 – 4i}{5 + i} \right ]\)

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

\(= \frac{33 + 31i}{2(14 – 5i)}\times\frac{14 + 5i}{14 + 5i}\)

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:

\(a – ib =\sqrt{\frac{x \;-\; iy}{u \;- \;iv}}\)

\(= \sqrt{\frac{x – iy}{u – iv}\times\frac{x + iy}{u + iv}}\\\)

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:

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:

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:

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:

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:

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:

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:

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:

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:

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:

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

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

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

 

 

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

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

 

Sol:

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:

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:

|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:

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:

\((\frac{1 \;+ \;i}{1\; -\; i})^{x} = 1\\\)

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

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

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

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

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