NCERT Solutions For Class 11 Maths Chapter 5

NCERT Solutions For Class 11 Maths Chapter 5 PDF Free Download

NCERT solutions for class 11 maths chapter 5 with heading as Complex Numbers and Quadratic Equations, are provided here so that students can learn the chapter with solutions. Solving NCERT solutions will be an added advantage for students who are appearing this year for Class 11 exams and competitive exams. Students who want to score good marks in Maths subject can click here, to get the solutions for all the chapters of 11th standard.

Class 11 Maths NCERT Solutions – Complex Numbers and Quadratic Equations

NCERT maths class 11 solutions PDF format of chapter 5, Complex Numbers and Quadratic Equations, is provided here so that students can refer to it if they face any difficulties while solving the problems. Let us see what topics are covered in chapter 5 of 11th class Maths subject.

  1. Complex Numbers
  2. Algebra of Complex Numbers
  3. Addition and Difference of two Complex Numbers
  4. Multiplication and Division of two Complex Numbers
  5. Power of i
  6. The square roots of a negative real number
  7. Identities
  8. The Modulus and Conjugate of a Complex number
  9. Argand Plane and Polar Representation
  10. And Quadratic equations

Along with solving NCERT solutions for class 11 maths chapter 5, students should also solve sample papers and previous year question papers to get an idea of the paper pattern and type of questions asked in the main exam. Students can get NCERT Question papers for class 11 here and practice them accordingly.

BYJU’S is also providing exemplar problems with solutions designed by experts to help students practice more for a given concept and have a deep understanding of it. Click here to get the class 11 exemplar problems based on NCERT books for Maths subject.

Complex Numbers Class 11 Extra Questions


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

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