NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations

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NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations

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NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations

Exercise 5.3

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Q1.Solve the given equation x² + 3 =0.

Answer.

Using the quadratic equation formula,we have

x² +0.x + 3 =0

Comparing it with ax² +bx +c =0

a = 1, b = 0, c = 3

${\color{DarkBlue} \mathbf{x = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x = \frac{0\pm \sqrt{0^{2}-4\times 1\times 3}}{2\times 1}}}$

${\color{DarkBlue} \mathbf{x =\frac{\pm \sqrt{-12}}{2}}}$

${\color{DarkBlue} \mathbf{x =\frac{\pm \sqrt{12i^{2}}}{2}}}$

${\color{DarkBlue} \mathbf{x =\frac{\pm 2i\sqrt{3}}{2}}}$

x = ±i√3

Q2. Solve the equation 2x² + x + 1 =0.

Answer.

Comparing the equation 2x² + x + 1 =0 with ax² + bx + c = 0,we have

a =2, b =1, c = 1

${\color{DarkBlue} \mathbf{x =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}} }$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm \sqrt{1^{2}-4\times 2\times\times 1 }}{2\times 2}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm \sqrt{-7 }}{4}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm \sqrt{7i^{2} }}{4}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm i\sqrt{7 }}{4}}}$

Q3.Solve the equation x² + 3x + 9 = 0.

Answer.Comparing the equation x² + 3x + 9 = 0 with ax² +bx +c =0

a = 1, b=3, c=9

${\color{DarkBlue} \mathbf{x =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x =\frac{-3\pm \sqrt{3^{2}-4\times 1\times 9}}{2\times 1}}}$

${\color{DarkBlue} \mathbf{x =\frac{-3\pm \sqrt{9-36}}{2}}}$

${\color{DarkBlue} \mathbf{x =\frac{-3\pm \sqrt{-27}}{2}}}$

${\color{DarkBlue} \mathbf{x =\frac{-3\pm \sqrt{27i^{2}}}{2}}}$

${\color{DarkBlue} \mathbf{x =\frac{-3\pm i3\sqrt{3}}{2}}}$

Q4. Solve the equation –x² +x – 2 =0.

Answer. Comparing the equation –x² +x – 2 =0 with ax² + bx + c =0

a = –1, b = 1, c = –2

${\color{DarkBlue} \mathbf{x =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm \sqrt{1^{2}-4\times -1\times -2}}{2\times -1}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm \sqrt{-7}}{-2}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm \sqrt{7i^{2}}}{-2}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm i\sqrt{7}}{-2}}}$

Q5.Solve the equation x² + 3x + 5 = 0.

Answer.Comparing the equation x² + 3x + 5 = 0 with ax² + bx + c =0

a =1, b = 3, c = 5

${\color{DarkBlue} \mathbf{x =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x =\frac{-3\pm \sqrt{3^{2}-4\times 1\times 5}}{2\times 1}}}$

${\color{DarkBlue} \mathbf{x =\frac{-3\pm \sqrt{-11}}{2}}}$

${\color{DarkBlue} \mathbf{x =\frac{-3\pm \sqrt{11i^{2}}}{2}}}$

${\color{DarkBlue} \mathbf{x =\frac{-3\pm i\sqrt{11}}{2}}}$

Q6. Solve the equation x² – x + 2 = 0.

Answer.Comparing the equation x² – x + 2 = 0 with ax² + bx + c =0

a =1, b = –1, c =2

${\color{DarkBlue} \mathbf{x =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm \sqrt{1^{2}-4\times -1\times 2}}{2\times -1}}}$

${\color{DarkBlue} \mathbf{x =\frac{-\left ( -1 \right )\pm \sqrt{\left ( -1 \right )^{2}-4\times 1\times 2}}{2\times -1}}}$

${\color{DarkBlue} \mathbf{x =\frac{1\pm \sqrt{-7}}{-2}}}$

${\color{DarkBlue} \mathbf{x =\frac{1\pm \sqrt{7i^{2}}}{-2}}}$

${\color{DarkBlue} \mathbf{x =\frac{1\pm i\sqrt{7}}{-2}}}$

Q7. Solve the equation √2x² +x + √2 =0.

Answer.Comparing the equation √2x² +x + √2 =0 with ax² + bx + c =0

a =√2, b = 1, c =√²

${\color{DarkBlue} \mathbf{x =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm \sqrt{1-4\times \sqrt{2}\times \sqrt{2}}}{2\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm \sqrt{-7}}{2\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm i\sqrt{7}}{2\sqrt{2}}}}$

Q8.Solve the equation √3x² –√2x + 3√3 =0.

Answer.Comparing the equation √3x² –√2x + 3√3 =0 with ax² + bx + c =0

a =√3, b= –√2 , c = 3√3

${\color{DarkBlue} \mathbf{x =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x =\frac{\sqrt{2}\pm \sqrt{\left ( -\sqrt{2} \right )^{2}-4\times \sqrt{3}\times 3\sqrt{3}}}{2\times \sqrt{3}}}}$

${\color{DarkBlue} \mathbf{x =\frac{\sqrt{2}\pm \sqrt{-34}}{2 \sqrt{3}}}}$

${\color{DarkBlue} \mathbf{x =\frac{\sqrt{2}\pm \sqrt{34i^{2}}}{2 \sqrt{3}}}}$

${\color{DarkBlue} \mathbf{x =\frac{\sqrt{2}\pm i\sqrt{34}}{2 \sqrt{3}}}}$

${\color{DarkGreen} { \mathbf{Q9. Solve\; the\: equation\: x^{2}+x+\frac{1}{\sqrt{2}}=0 }}}$

Answer.Comparing the equation with ax² + bx + c =0

a =1, b = 1, c = 1/√2

${\color{DarkBlue} \mathbf{x =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \boldsymbol{{\color{DarkBlue} }x=\frac{-1\pm\sqrt{1^{2}-4\times 1\times 1/\sqrt{2}} }{2\times 1}}}$

${\color{DarkBlue} \boldsymbol{{\color{DarkBlue} }x=\frac{-1\pm\sqrt{1-4/\sqrt{2}} }{2\times 1}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm \sqrt{\left ( 1-2\sqrt{2} \right )}}{2}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm \sqrt{\left ( 2\sqrt{2}-1 \right )i^{2}}}{2}}}$

${\color{DarkBlue} \mathbf{x =\frac{-1\pm i\sqrt{\left ( 2\sqrt{2}-1 \right )}}{2}}}$

${\color{DarkGreen} \mathbf{Q10.Solve\; the\; equation\; x^{2}+\frac{x}{\sqrt{2}}+1 =0}}$

Answer.

Writing the given equation in the standard form by multiplying with √2

√2x² +x + √2 = 0

Comparing the equation √2x² +x + √2 = 0 with ax² + bx + c =0

a =√2, b = 1, c = √2

${\color{DarkBlue} \mathbf{x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x=\frac{-1\pm \sqrt{1-4\times \sqrt{2}\times \sqrt{2}}}{2\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{x=\frac{-1\pm \sqrt{-7}}{2\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{x=\frac{-1\pm \sqrt{7i^{2}}}{2\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{x=\frac{-1\pm i\sqrt{7}}{2\sqrt{2}}}}$

Miscellanous Exercise

Q1. Evaluate

${\color{DarkGreen} \mathbf{\left [ i^{18}+\left ( \frac{1}{i} \right )^{25} \right ]^{3}}}$

Answer.

The expression given to us is

${\color{DarkBlue} \mathbf{\left [ i^{18}+ \left ( \frac{1}{i} \right )^{25}\right ]^{3}}}$

${\color{DarkBlue} \mathbf{\because i^{2}=-1}}$

${\color{DarkBlue} \mathbf{\left [ \left ( i^{2} \right )^{9}+\frac{1}{\left ( i^{2} \right )^{12}.i} \right ]^{3}}}$

${\color{DarkBlue}= \mathbf{\left [ \left ( -1 \right )^{9}+\frac{1}{\left ( -1 \right )^{12}.i} \right ]^{3}}}$

${\color{DarkBlue} \mathbf{=\left ( -1+\frac{1}{i} \right )^{3}}}$

${\color{DarkBlue} \mathbf{=\left ( -1+\frac{1}{i}\times \frac{i}{i} \right )^{3}}}$

${\color{DarkBlue} \mathbf{=\left ( -1+\frac{i}{i^{2}} \right )^{3}}}$

${\color{DarkBlue} \mathbf{=\left ( -1-i \right )^{3}}}$

${\color{DarkBlue} \mathbf{=\left ( -1 \right )^{3}\left ( 1+i \right )^{3}}}$

${\color{DarkBlue} \mathbf{\because \left ( a +b \right )^{3}=a^{3}+b^{3}+3a^{2}b+3b^{2}a}}$

${\color{DarkBlue} \mathbf{=-\left ( 1^{3}+i^{3}+3i^{2}.1+3.1^{2}.i \right )}}$

${\color{DarkBlue} \mathbf{=-\left ( 1+i^{2}.i-3+3i \right )}}$

= –(1 – i –3 +3i)

= –(–2 + 2i)

= 2 – 2i

Q2.For any two complex number z₁and z₂, prove that

Re(z₁ z₂) = Re z₁Re z₂ – Im z₁Im z₂

Answer.

Let z₁ = a + ib and z₂ = c + id

z₁ z₂ = (a + ib)(c + id)

= ac + iad +ibc +i²bd

= (ac-bd) + i(ad +bc)

Re(z₁z₂) = ac – bd

= Rez₁Rez₂ – Imz₁Imz₂

Hence proved

${\color{DarkGreen} \mathbf{Q3.Reduce\left ( \frac{1}{1-4i}-\frac{2}{1+i} \right )\left ( \frac{3-4i}{5+i} \right )\; to\; standard \; form}}$

Answer.

${\color{DarkBlue} \mathbf{\left [ \frac{1}{1-4i}-\frac{2}{1+i} \right ]\left [ \frac{3-4i}{5+i} \right ]}}$

${\color{DarkBlue}= \mathbf{\left [ \frac{1+i-2+8i}{\left ( 1-4i \right )\left ( 1+i \right )} \right ]\left [ \frac{3-4i}{5+i} \right ]}}$

${\color{DarkBlue}= \mathbf{\left [ \frac{-1+9i}{\left ( 1+i-4i-4i^{2} \right )} \right ]\left [ \frac{3-4i}{5+i} \right ]}}$

${\color{DarkBlue}= \mathbf{\left [ \frac{-1+9i}{\left ( 1+i-4i+4 \right )} \right ]\left [ \frac{3-4i}{5+i} \right ]}}$

${\color{DarkBlue}= \mathbf{\left [ \frac{-1+9i}{\left ( 5-3i \right )} \right ]\left [ \frac{3-4i}{5+i} \right ]}}$

${\color{DarkBlue} \mathbf{=\left [ \frac{-3+4i+27i-36i^{2}}{25+5i-15i-3i^{2}} \right ]}}$

${\color{DarkBlue} \mathbf{=\left [ \frac{-3+4i+27i+36}{25+5i-15i+3} \right ]}}$ ${\color{DarkBlue} \mathbf{=\left [ \frac{33+31i}{28-10i} \right ]}}$

Multiplying denominator and numerator by the conjugate of the denominator:28 +10i

${\color{DarkBlue} \mathbf{=\left [ \frac{33+31i}{28-10i} \right ]\times \frac{28+10i}{28+10i}}}$

${\color{DarkBlue} \mathbf{=\frac{307}{442}+i\frac{599}{442}}}$

${\color{DarkGreen} \mathbf{Q4.}\mathbf{If \: x -iy =\sqrt{\frac{a-ib}{c-id}}\; prove\; that\; \left ( x^{2}+y^{2} \right )^{2}=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}}}$

Answer. We are given that

${\color{DarkBlue} \mathbf{x-iy =\sqrt{\frac{a-ib}{c-id}}}}$

As we know (x +iy)(x –iy) = x² – (iy)² = x² + y²…..(i)

${\color{DarkBlue} \because \mathbf{x-iy=\sqrt{\frac{a-ib}{c-id}}\left [ Given\; to\; us\; \right ]}}$ ………(ii)

${\color{DarkBlue} \therefore \mathbf{x+iy=\sqrt{\frac{a+ib}{c+id}}}}$ …………..(iii)

Multiplying(ii) and (iii),we have

${\color{DarkBlue} \mathbf{\left ( x-iy \right )\left ( x+iy \right )=\sqrt{\frac{\left ( a-ib \right )\left ( a+ib \right )}{\left ( c-id \right )\left ( c+id \right )}}}}$

From equation number (i)

${\color{DarkBlue} \mathbf{x^{2}+y^{2}=\sqrt{\frac{a^{2}+b^{2}}{c^{2}+d^{2}}}}}$

Squaring both sides

${\color{DarkBlue} \mathbf{\left ( x^{2}+y^{2} \right )^{2}=\frac{a^{2}+b^{2}}{c^{2}+d^{2}}}}$

Hence proved

Q5.Convert the following into the polar form

${\color{DarkGreen} \mathbf{\left ( i \right )\frac{1+7i}{\left ( 2-i \right )^{2}}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (ii)\frac{1+3i}{1-2i}}}$

${\color{DarkBlue} \mathbf{\frac{1+7i}{\left ( 2-i \right )^{2}}}}$

${\color{DarkBlue} \mathbf{\Rightarrow \frac{1+7i}{2^{2}+i^{2}-4i}=\frac{1+7i}{4-1-4i}=\frac{1+7i}{3-4i}}}$

Multiplying denominator and numerator by the conjugate of the denominator;

(3 + 4i)

${\color{DarkBlue} \mathbf{=\frac{1+7i}{3-4i}\times \frac{3+4i}{3+4i}}}$

${\color{DarkBlue} \mathbf{\frac{3+4i+21i+28i^{2}}{9-\left ( 4i \right )^{2}}}}$

${\color{DarkBlue}= \mathbf{\frac{3+25i-28}{9+16}}}$

${\color{DarkBlue}= \mathbf{\frac{-25+25i}{25}=-1+i}}$

So, let z = –1 + i

Comparing the terms of this complex number with the polar form of the complex number:

z=rcosθ + irsinθ

rcosθ = –1……..(i)

rsinθ = 1………..(ii)

Squaring and adding both the equations

r²(cos²θ + sin²θ) = (–1)² + 1² =2

r = ±√2 [cos²θ + sin²θ = 1]

r is the modulus, so it is always +

∴r = √2

Placing the value of r in eq.(i) and (ii)

√2cosθ = –1, √2sinθ = 1

${\color{DarkBlue} \mathbf{cos\Theta =-\frac{1}{\sqrt{2}},sin\Theta =\frac{1}{\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{cos\Theta =-cos\frac{\Pi }{4},sin\Theta =sin\frac{\Pi }{4}}}$

cosθ is – and sinθ is + in second quadrant but sinθ and cosθ are + in l quadrant, therefore taking the value of cosθ, in second quadrant the value of argument

=(π – π/4)= 3π/4

Therefore required polar form of the given complex number will be a follows

${\color{DarkBlue} \mathbf{=\sqrt{2}cos\frac{3\Pi }{4}+i\sqrt{2}sin\frac{3\Pi }{4}}}$

(ii) The given complex number is

${\color{DarkBlue} \mathbf{\frac{1+3i}{1-2i}}}$

Multiplying denominator and numerator by the conjugate of the denominator:

(1 + 2i)

${\color{DarkBlue} =\mathbf{\frac{1+3i}{1-2i}\times \frac{1+2i}{1+2i}}}$

${\color{DarkBlue} \mathbf{=\frac{1+2i+3i+6i^{2}}{1-\left ( 2i \right )^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{1+5i-6}{1-4i^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{-5+5i}{1+4}}}$

${\color{DarkBlue} \mathbf{=\frac{-5+5i}{5}=-1+i}}$

Let z = –1 + i,

Comparing the terms of this complex number with the polar form of the complex number:

z=rcosθ + irsinθ

rcosθ =–1……(i) rsinθ = 1…….(ii)

Squaring and adding both the equations

r²(cos²θ + sin²θ) = (–1)² + 1² =2

r = ±√2 [cos²θ + sin²θ = 1]

r is the modulus, so it is always +

∴r = √2

Placing the value of r in eq.(i) and (ii)

√2cosθ = –1, √2sinθ = 1

${\color{DarkBlue} \mathbf{cos\Theta =-\frac{1}{\sqrt{2}},sin\Theta =\frac{1}{\sqrt{2}}}}$

${\color{DarkBlue} \mathbf{cos\Theta =-cos\frac{\Pi }{4},sin\Theta =sin\frac{\Pi }{4}}}$

cosθ is – and sinθ is + in second quadrant but sinθ and cosθ are + in l quadrant, therefore taking the value of cosθ, in second quadrant the value of argument

=(π – π/4)= 3π/4

Therefore required polar form of the given complex number will be a follows

${\color{DarkBlue} \mathbf{rcos\frac{3\Pi }{4}+irsin\frac{3\Pi }{4}}}$

${\color{DarkGreen} { \mathbf{Q6.Solve\; the\; equation\; 3x^{2}-4x +\frac{20}{3}=0}}}$

Answer. The given eq. can be written as 9x² –12x +20= 0

Comparing the equation 9x² –12x +20= 0 with ax² + bx + c =0

a =9, b= –12, c = 20

${\color{DarkBlue} \mathbf{x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x=\frac{12\pm \sqrt{\left ( -12 \right )^{2}-4\times 9\times 20}}{2\times 9}}}$

${\color{DarkBlue} \mathbf{=\frac{12\pm \sqrt{144-720}}{18}}}$

${\color{DarkBlue} \mathbf{=\frac{12\pm \sqrt{-576}}{18}}}$

${\color{DarkBlue} \mathbf{=\frac{12\pm \sqrt{576i^{2}}}{18}}}$

${\color{DarkBlue} \mathbf{=\frac{12\pm 24i}{18}}}$

${\color{DarkBlue} \mathbf{=\frac{12(1\pm 2i)}{18}}}$

${\color{DarkBlue} \mathbf{Hence\; the\; value\: of\; x=\frac{2(1\pm 2i)}{3}}}$

${\color{DarkBlue} \boldsymbol{Q7.Solve\: the \: equation\: x^{2}-2x+\frac{3}{2}=0}}$

${\color{DarkBlue} { { \mathbf{Answer.Comparin\; the\; equation\; x^{2}-2x +\frac{3}{2}=0\; with\; ax^{2}+bx +c=0.}}}}$

a = 1, b= –2

${\color{DarkBlue} \mathbf{c =\frac{3}{2}}}$

${\color{DarkBlue} \mathbf{x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x=\frac{-\left ( -2 \right )\pm \sqrt{\left ( -2 \right )^{2}-4\times 1\times \frac{3}{2}}}{2a}}}$

${\color{DarkBlue} \mathbf{x=\frac{2\pm \sqrt{-2}}{2}}}$

${\color{DarkBlue} \mathbf{x=\frac{2\pm \sqrt{2i^{2}}}{2}}}$

${\color{DarkBlue} \mathbf{x=\frac{2\pm i\sqrt{2}}{2}}}$

${\color{DarkBlue} \mathbf{x =1\pm \frac{i\sqrt{2}}{2}}}$

Q8.Solve the equation 27x² – 10x + 1 = 0.

Answer.Comparing the equation 27x² –10 x + 1 = 0 with ax² + bx + c =0

a =27,b = –10, c = 1

${\color{DarkBlue} \mathbf{x =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x =\frac{-\left ( -10 \right )\pm \sqrt{\left ( -10 \right )^{2}-4\times 27\times 1}}{2a}}}$

${\color{DarkBlue} \mathbf{x =\frac{ 10 \pm \sqrt{100-108 }}{2\times 27}}}$

${\color{DarkBlue} \mathbf{x =\frac{10\pm \sqrt{-8}}{54}}}$

${\color{DarkBlue} \mathbf{x =\frac{10\pm \sqrt{8i^{2}}}{54}}}$

${\color{DarkBlue} \mathbf{x =\frac{10\pm i2\sqrt{2}}{54}}}$

${\color{DarkBlue} \mathbf{x =\frac{5\pm i\sqrt{2}}{27}}}$

Q9.Solve the equation 21x² – 28x + 10 = 0.

Answer.Comparing the equation 21x² – 28x + 10 = 0 with ax² + bx + c =0

a = 21, b = –28, c = 10

${\color{DarkBlue} \mathbf{x =\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}}$

${\color{DarkBlue} \mathbf{x =\frac{-\left ( -28 \right )\pm \sqrt{28^{2}-4\times 21\times 10}}{2\times 21}}}$

${\color{DarkBlue} \mathbf{x =\frac{28\pm \sqrt{784-840}}{42}}}$

${\color{DarkBlue} \mathbf{x =\frac{28\pm \sqrt{-56}}{42}}}$

${\color{DarkBlue} \mathbf{x =\frac{28\pm \sqrt{56i^{2}}}{42}}}$

${\color{DarkBlue} \mathbf{x =\frac{14\pm i\sqrt{14}}{21}}}$

${\color{DarkBlue} \mathbf{x =\frac{2}{3}\pm \frac{i\sqrt{14}}{21}}}$

${\color{DarkGreen} { \mathbf{Q10.If\; z_{1}=2-i,z_{2}=1+i,Find\; \left | \frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1} \right |}}}$

Answer. The given expression is as follows.

${\color{DarkBlue} \mathbf{\left | \frac{z_{1}+z_{2}+1}{z_{1}-z_{2}+1} \right |}}$

We are given that

z1 = 2–i, z2 = 1+i

${\color{DarkBlue} \mathbf{=\left | \frac{2-i+1+i+1}{2-i-1-i+1} \right |}}$

${\color{DarkBlue} \mathbf{=\left | \frac{4}{2-2i} \right |}}$

${\color{DarkBlue} \mathbf{=\left | \frac{4}{2\left ( 1-i \right )} \right |}}$

${\color{DarkBlue} \mathbf{=\left | \frac{2}{1-i } \right |}}$

${\color{DarkBlue} \mathbf{=\left | \frac{2\left ( 1+i \right )}{\left ( 1-i \right )\left ( 1+i \right ) } \right |}}$

${\color{DarkBlue} \mathbf{=\left | \frac{2\left ( 1+i \right )}{1-i^{2} } \right |}}$

i² =–1

${\color{DarkBlue} \mathbf{=\left | \frac{2\left ( 1+i \right )}{1+1 } \right |}}$

${\color{DarkBlue} \mathbf{=\left | 1 +i \right |}}$

As we know the modulus of complex number z=a + ib = √(a² +b²)

$\mathbf{{\color{DarkBlue} \mathbf{=\left | 1 +i \right |}}=\sqrt{1^{2}+1^{2}}=\sqrt{2}}$

${\color{DarkGreen} \mathbf{Q11.If\; a +ib=\frac{\left ( x+i \right )^{2}}{2x^{2}+1}, prove\; that\; a^{2}+b^{2}=\frac{\left ( x^{2}+1 \right )^{2}}{\left ( 2x^{2}+1 \right )^{2}}}}$

Answer. Evaluating the right hand part

${\color{DarkBlue} \mathbf{a+ib =\frac{\left ( x^{2}+i \right )^{2}}{2x^{2}+1}}}$

${\color{DarkBlue} \mathbf{a +ib=\frac{x^{4}+i^{2}+2x^{2}i}{2x^{2}+1}}}$

${\color{DarkBlue} \mathbf{a +ib=\frac{x^{2}-1+2xi}{2x^{2}+1}}}$

${\color{DarkBlue} \mathbf{a+ib=\frac{x^{2}-1}{2x^{2}+1}+\frac{i2x}{2x^{2}+1}}}$

Comparing L.H.S and R.H.S, we have

${\color{DarkBlue} \mathbf{a=\frac{x^{2}-1}{2x^{2}+1}, b=\frac{2x}{2x^{2}+1}}}$

${\color{DarkBlue} \mathbf{\therefore a^{2}+b^{2}=\frac{\left ( x^{2}-1 \right )^{2}}{\left ( 2x^{2}+1 \right )^{2}}+\frac{\left ( 2x \right )^{2}}{\left ( 2x^{2}+1 \right )^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{\left ( x^{2}-1 \right )^{2}+4x^{2}}{\left ( 2x^{2}+1 \right )^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{ x^{4}+1 -2x^{2}+4x^{2}}{\left ( 2x^{2}+1 \right )^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{ x^{4}+1 +2x^{2}}{\left ( 2x^{2}+1 \right )^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{\left ( x^{2}+1 \right )^{2}}{\left ( 2x^{2}+1 \right )^{2}}}}$

So,it has been proved that

${\color{DarkBlue}\mathbf{a^{2}+b^{2}} \mathbf{=\frac{\left ( x^{2}+1 \right )^{2}}{\left ( 2x^{2}+1 \right )^{2}}}}$

Q12. Let z₁=2+i,z₂= -2 + i, Find

$\boldsymbol{{\color{DarkBlue} {\left ( i \right )\:Re \left ( \frac{z_{1}z_{2}}{\bar{z}_{1}} \right )}}}$

${\color{DarkBlue} \boldsymbol{\left ( ii \right )Im\left ( \frac{1}{z_{1}\bar{z}_{1}} \right )}}$

Answer. (i) We have to find out

$\boldsymbol{{\color{DarkBlue} {\left ( i \right )\:Re \left ( \frac{z_{1}z_{2}}{\bar{z}_{1}} \right )}}}$

z1.z2 =(2 –i)(–2 +i)

= –4 + 2i + 2i –i²

= –4 +4i –(–1)

=–4 +4i + 1

= –3 +4i

${\color{DarkBlue} \mathbf{z_{1}=2-i\Rightarrow \bar{z_{1}}=2+i}}$

Therefore

${\color{DarkBlue} \mathbf{\frac{z_{1}z_{2}}{\bar{z_{1}}}=\frac{-3+4i}{2+i}}}$

Multiplying dinominator and numerator by the conjugate of denominator in R.H.S

${\color{DarkBlue} \mathbf{=\frac{\left ( -3+4i \right )\left ( 2-i \right )}{\left ( 2+i \right )\left ( 2-i \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{-6+3i+8i-4i^{2} }{4-i^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{-6+3i+8i-4\left ( -1 \right ) }{4-\left ( -1 \right )}}}$

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

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

${\color{DarkBlue} \mathbf{Re\left [ \frac{z_{1}z_{2}}{\bar{z_{1}}} \right ]=\frac{-2}{5}}}$

${\color{DarkBlue} \boldsymbol{\left ( ii \right )Im\left ( \frac{1}{z_{1}\bar{z}_{1}} \right )}}$

${\color{DarkBlue} \mathbf{z_{1}=2+i, \bar{z_{1}}= 2 -i}}$

${\color{DarkBlue}\therefore \mathbf{z_{1}.\bar{z_{1}}=\left ( 2-i \right )\left ( 2+i \right )}}$

${\color{DarkBlue} \mathbf{z_{1}.\bar{z_{1}}= 2^{2}-i^{2}}}$

${\color{DarkBlue} \mathbf{z_{1}.\bar{z_{1}}= 2^{2}-\left ( -1 \right )=5}}$

${\color{DarkBlue} \mathbf{\therefore \frac{1}{z_{1}\bar{z_{1}}}=\frac{1}{5}}}$

So, writing it in the standard form of the complex number a + ib

${\color{DarkBlue} \mathbf{ \frac{1}{z_{1}\bar{z_{1}}}=\frac{1}{5}+0.i}}$

Comparing L.H.S and R.H.S, we have

${\color{DarkBlue} \mathbf{\therefore Im\left ( \frac{1}{z_{1}\bar{z_{1}}} \right )=0}}$

Q13.Find the modulus and argument of the complex number

${\color{DarkGreen} \mathbf{\frac{1+2i}{1-3i}}}$

Answer. The given complex number is

${\color{DarkBlue} { \mathbf{\frac{1+2i}{1-3i}}}}$

Multiplying its denominator and numerator by the conjugate of the denominator:

1 +3i

${\color{DarkBlue} \mathbf{=\frac{\left ( 1+2i \right )\left ( 1+3i \right )}{\left ( 1-3i \right )\left ( 1+3i \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{1+3i+2i+6i^{2}}{1-9i^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{1+5i-6}{1+9}}}$

${\color{DarkBlue} \mathbf{=\frac{-5+5i}{10}}}$

${\color{DarkBlue} \mathbf{\frac{-5}{10}+\frac{5i}{10}=\frac{-1}{2}+\frac{1}{2}i}}$

Comparing it with the polar form of the complex number,we have

${\color{DarkBlue} \mathbf{rcos\Theta =\frac{-1}{2},rsin\Theta =\frac{1}{2}}}$

Squaring adding both the terms

r²(cos²θ + sin²θ) =(–1/2)² +(1/2)²

r² = 1/4 + 1/4

${\color{DarkBlue} \mathbf{r^{2}= \frac{2}{4}=\frac{1}{2}}}$

${\color{DarkBlue} \mathbf{r=\pm \frac{1}{\sqrt{2}}}}$

r is the modulus,so neglecting minus sign

${\color{DarkBlue} \mathbf{r= \frac{1}{\sqrt{2}}}}$

We have

${\color{DarkBlue} \mathbf{\frac{1}{\sqrt{2}}cos\Theta =-\frac{1}{2}}}$ ${\color{DarkBlue} ,\mathbf{\frac{1}{\sqrt{2}}sin\Theta =\frac{1}{2}}}$

${\color{DarkBlue} \mathbf{cos\Theta =-\frac{1}{\sqrt{2}}}}$ ${\color{DarkBlue} ,\mathbf{sin\Theta =\frac{1}{\sqrt{2}}}}$

cosθ is – and sinθ is + in ll quadrant but sinθ is + in l quadrant also,therefore taking the value of cosθ

${\color{DarkBlue} \mathbf{cos\Theta =-\frac{1}{\sqrt{2}}}}$

Value of argument will lie in ll quadrant and its value will be = π – θ

${\color{DarkBlue} \mathbf{cos\Theta =-cos\frac{\Pi }{4}=cos\left ( \Pi -\frac{\Pi }{4} \right )}}$

${\color{DarkBlue} \mathbf{cos\Theta =cos\frac{3\Pi }{4}}}$

${\color{DarkBlue} \mathbf{\Theta =\frac{3\Pi }{4}}}$

Therefore the modulus and argument of the given complex number is 1/√2 and 3π/4.

Q14. Find the real numbers x and y if (x –iy)(3 + 5i) is the conjugate of (–6 –24i).

Answer.We are given that (x –iy)(3 + 5i) is the conjugate of

(–6 –24i)

(x –iy)(3 + 5i) = 3x + 5xi –i3y –5i²y

= 3x + (5x –3y)i +5y [i² = –1]

=(3x + 5y) + (5x –3y)i…..(i)

conjugate of (–6 –24i) = –6 +24i…..(ii)

(i) is conjugate of (ii)

3x + 5y = –6……(iii)

5x –3y = 24……..(iv)

Multiplying eq.(iii) by 5 and eq.(iv) by 3 and adding both of the equation.

We get y = -3

Putting value of y=–3, in equation (iii)

3x + 5×–3 = –6

3x = –6 + 15

x = 3

Therefore the value of real number x and y will be x =3 and y =-3

${\color{DarkGreen} \mathbf{Q15.Find\; the\: modulus\; of\; \frac{1+i}{1-i}-\frac{1-i}{1+i}}}$

Answer.

The complex number we are given that

${\color{DarkBlue} \mathbf{\frac{1+i}{1-i}-\frac{1-i}{1+i}}}$

Multiplying denominators and numerators of both the fractions by the conjugates of the denominators of both the fractions (1 +i) and (1 –i) respectively.

${\color{DarkBlue} \mathbf{=\frac{\left ( 1+i \right )\left ( 1+i \right )}{\left ( 1-i \right )\left ( 1+i \right )}-\frac{\left ( 1-i \right )\left ( 1-i \right )}{\left ( 1+i \right )\left ( 1-i \right )}}}$

${\color{DarkBlue} \mathbf{=\frac{1+i^{2}+2i}{1-i^{2}}-\frac{1+i^{2}-2i}{1-i^{2}}}}$

${\color{DarkBlue} \mathbf{=\frac{1-1+2i}{1+1}-\frac{1-1-2i}{1+1}}}$

${\color{DarkBlue} \mathbf{=\frac{2i}{2}-\frac{-2i}{2}}}$

= i + i = 2i

Therefore

${\color{DarkBlue} \mathbf{\frac{1+i}{1-i}-\frac{1-i}{1+i}=2i}}$

2i = 0 + 2i

So,

${\color{DarkBlue} \mathbf{\left | \frac{1+i}{1-i}-\frac{1-i}{1+i} \right |=\sqrt{0^{2}+2^{2}}=2}}$

${\color{DarkBlue} \mathbf{Q16.\; If \left ( x+iy \right )^{3}= u+iv, then\; show\; that\; \frac{u}{x}+\frac{u}{y}=4\left ( x^{2}-y^{2} \right )}}$

Answer. We are given that

${\color{DarkBlue} \mathbf{\left ( x+iy \right )^{3}=u+iv}}$

${\color{DarkBlue} \mathbf{\because \left ( a+b \right )^{3}= a^{3}+b^{3}+3ab\left ( a+b \right )}}$

So, ${\color{DarkBlue} \therefore \mathbf{x^{3 }+i^{2}y^{3}i+3xyi\left ( x+iy \right )=u+iy}}$

${\color{DarkBlue} \mathbf{\left ( x+iy \right )^{3}= x^{3}+\left ( iy \right )^{3}+3xyi\left ( x+iy \right )}}$

${\color{DarkBlue}= \mathbf{x^{3 }-y^{3}i+3x^{2}yi+3i^{2}y^{2}x=u+iy}}$

${\color{DarkBlue} \mathbf{\Rightarrow x^{3}-y^{3}i+3x^{2}yi-3y^{2}x=u+iy}}$

${\color{DarkBlue} \mathbf{\Rightarrow \left ( x^{3}-3y^{2}x \right )+\left ( 3x^{2}y-y^{3} \right )i=u+iv}}$

Comparing terms of both sides,we have

${\color{DarkBlue} \mathbf{u=x^{3}-3y^{2}x,\; v =3x^{2}y-y^{3}}}$

We have to prove

${\color{DarkBlue} \mathbf{\frac{u}{x}+\frac{v}{y}=4\left ( x^{2}-y^{2} \right )}}$

L.H.S

${\color{DarkBlue} \mathbf{\frac{u}{x}+\frac{v}{y}}}$

Substituting the values of x and y in

${\color{DarkBlue} =\mathbf{\frac{x^{3}-3y^{2}x}{x}+\frac{3x^{2}y-y^{3}}{y}}}$

${\color{DarkBlue} \mathbf{=\frac{x^{3}y-3y^{3}x+3x^{3}y-y^{3}x}{xy}}}$

${\color{DarkBlue} \mathbf{=\frac{4x^{3}y-4y^{3}x}{xy}}}$

${\color{DarkBlue} =\mathbf{\frac{4xy\left ( x^{2} -y^{2}\right )}{xy}}}$

= 4(x² – y²)

Hence it has been proved that

${\color{DarkBlue} \mathbf{\frac{u}{x}+\frac{v}{y}=4\left ( x^{2}-y^{2} \right )}}$

${\color{DarkGreen} { \mathbf{Q17.If\; \alpha\; and\: \beta\; are\; different\; complex\; numbers\; with\left | \beta \right |=1,then find \left | \frac{\beta -\alpha }{1-\bar{\alpha }\beta } \right |}}}$

Answer.

Let α = a + ib and β = c + id

${\color{DarkBlue} \mathbf{\because \left | \beta \right |=1}}$

${\color{DarkBlue} \mathbf{\therefore \left | c+id \right |=\sqrt{c^{2}+d^{2}}}}$

${\color{DarkBlue} \mathbf{\mathbf{\Rightarrow \sqrt{c^{2}+d^{2}}=1}}}$

⇒ c² + d² = 1….(i)

${\color{DarkBlue} \mathbf{\bar{\alpha }=a -ib}}$ ………(ii)

${\color{DarkBlue} \mathbf{\left | \frac{\beta -\alpha }{1-\bar{\alpha \beta }} \right |=\left | \frac{c+id-a-ib}{1-\left ( a-ib \right )\left ( c+id \right )} \right |}}$

${\color{DarkBlue} \mathbf{=\left | \frac{\left ( c-a \right )+i\left ( d-b \right )}{1-ac-iad+ibc+i^{2}bd} \right |}}$

${\color{DarkBlue} \mathbf{=\left | \frac{\left ( c-a \right )+i\left ( d-b \right )}{(1-ac-bd)-i(ad-bc)} \right |}}$

${\color{DarkBlue} \mathbf{\because \left |\frac{x}{y} \right |=\frac{\left | x \right |}{\left | y \right |}}}$

${\color{DarkBlue} \therefore \mathbf{\frac{\left | \left (c-a \right )+i\left (d-b \right ) \right |}{\left | \left ( 1-ac-bd \right )-i\left ( ad-bc \right ) \right |}}}$

${\color{DarkBlue} =\mathbf{\frac{\sqrt{\left ( c-a \right )^{2}+\left ( d-b \right )^{2}}}{\sqrt{\left ( 1-ac-bd \right )^{2}+\left [ -\left ( ad-bc \right ) \right ]^{2}}}}}$ ${\color{DarkBlue}= \mathbf{\frac{\sqrt{c^{2}+a^{2}-2ac+d^{2}+b^{2}-2bd}}{\sqrt{1+a^{2}c^{2}+b^{2}d^{2}-2ac+2abcd-2bd+a^{2}d^{2}+b^{2}c^{2}-2abcd}}}}$

${\color{DarkBlue}= \mathbf{\frac{\sqrt{c^{2}+d^{2}+a^{2}+b^{2}-2ac-2bd}}{\sqrt{1+a^{2}(c^{2}+d^{2})+b^{2}(c^{2}+d^{2})-2ac+2abcd-2bd-2abcd}}}}$

${\color{DarkBlue} \mathbf{\because c^{2}+d^{2}=1}}$

${\color{DarkBlue} \mathbf{=\frac{\sqrt{1+b^{2}+d^{2}-2ac-2bd}}{\sqrt{1+b^{2}+d^{2}-2ac-2bd}}}}$

${\color{DarkBlue} \mathbf{\therefore \left | \frac{\beta -\alpha }{1-\bar{\alpha \beta }} \right |=1}}$

Q18. Find the number of non-zero integral solutions of the equation

${\color{DarkGreen} \mathbf{\left | 1-i \right |^{x}=2^{x}}}$

Answer.The given eq. is

${\color{DarkBlue} \mathbf{\left | 1-i \right |^{x}= 2^{x}}}$

${\color{DarkBlue} \mathbf{\left | 1-i \right |=\sqrt{1^{2}+\left ( -1 \right )^{2}}=\sqrt{2}}}$

${\color{DarkBlue} \mathbf{\Rightarrow (\sqrt{2})^{x}=2^{x}}}$

${\color{DarkBlue} \Rightarrow \mathbf{2^{\frac{x}{2}}= 2^{x}}}$

${\color{DarkBlue} \mathbf{\Rightarrow \frac{x}{2}=x}}$

${\color{DarkBlue} \mathbf{\Rightarrow \frac{x}{2}-x=0}}$

${\color{DarkBlue} \mathbf{\Rightarrow -\frac{x}{2}=0}}$

x =0

Number of non-zero integral solutions of the equation are=0

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

NCERT Solutions for class 9 maths

Chapter 1- Number System	Chapter 9-Areas of parallelogram and triangles
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3- Coordinate Geometry	Chapter 11-Construction
Chapter 4- Linear equations in two variables	Chapter 12-Heron’s Formula
Chapter 5- Introduction to Euclid’s Geometry	Chapter 13-Surface Areas and Volumes
Chapter 6-Lines and Angles	Chapter 14-Statistics
Chapter 7-Triangles	Chapter 15-Probability
Chapter 8- Quadrilateral

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Chapter 1-Matter in our surroundings	Chapter 9- Force and laws of motion
Chapter 2-Is matter around us pure?	Chapter 10- Gravitation
Chapter3- Atoms and Molecules	Chapter 11- Work and Energy
Chapter 4-Structure of the Atom	Chapter 12- Sound
Chapter 5-Fundamental unit of life	Chapter 13-Why do we fall ill ?
Chapter 6- Tissues	Chapter 14- Natural Resources
Chapter 7- Diversity in living organism	Chapter 15-Improvement in food resources
Chapter 8- Motion	Last years question papers & sample papers

NCERT Solutions for class 10 maths

Chapter 1-Real number	Chapter 9-Some application of Trigonometry
Chapter 2-Polynomial	Chapter 10-Circles
Chapter 3-Linear equations	Chapter 11- Construction
Chapter 4- Quadratic equations	Chapter 12-Area related to circle
Chapter 5-Arithmetic Progression	Chapter 13-Surface areas and Volume
Chapter 6-Triangle	Chapter 14-Statistics
Chapter 7- Co-ordinate geometry	Chapter 15-Probability
Chapter 8-Trigonometry

CBSE Class 10-Question paper of maths 2021 with solutions

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NCERT Solutions for Class 10 Science

Chapter 1- Chemical reactions and equations	Chapter 9- Heredity and Evolution
Chapter 2- Acid, Base and Salt	Chapter 10- Light reflection and refraction
Chapter 3- Metals and Non-Metals	Chapter 11- Human eye and colorful world
Chapter 4- Carbon and its Compounds	Chapter 12- Electricity
Chapter 5-Periodic classification of elements	Chapter 13-Magnetic effect of electric current
Chapter 6- Life Process	Chapter 14-Sources of Energy
Chapter 7-Control and Coordination	Chapter 15-Environment
Chapter 8- How do organisms reproduce?	Chapter 16-Management of Natural Resources

NCERT Solutions for class 11 maths

Chapter 1-Sets	Chapter 9-Sequences and Series
Chapter 2- Relations and functions	Chapter 10- Straight Lines
Chapter 3- Trigonometry	Chapter 11-Conic Sections
Chapter 4-Principle of mathematical induction	Chapter 12-Introduction to three Dimensional Geometry
Chapter 5-Complex numbers	Chapter 13- Limits and Derivatives
Chapter 6- Linear Inequalities	Chapter 14-Mathematical Reasoning
Chapter 7- Permutations and Combinations	Chapter 15- Statistics
Chapter 8- Binomial Theorem	Chapter 16- Probability

CBSE Class 11-Question paper of maths 2015

CBSE Class 11 – Second unit test of maths 2021 with solutions

NCERT solutions for class 12 maths

Chapter 1-Relations and Functions	Chapter 9-Differential Equations
Chapter 2-Inverse Trigonometric Functions	Chapter 10-Vector Algebra
Chapter 3-Matrices	Chapter 11 – Three Dimensional Geometry
Chapter 4-Determinants	Chapter 12-Linear Programming
Chapter 5- Continuity and Differentiability	Chapter 13-Probability
Chapter 6- Application of Derivation	CBSE Class 12- Question paper of maths 2021 with solutions
Chapter 7- Integrals
Chapter 8-Application of Integrals

Class 12 Solutions of Maths Latest Sample Paper Published by CBSE for 2021-22 Term 2

Class 12 Maths Important Questions-Application of Integrals

Class 12 Maths Important questions on Chapter 7 Integral with Solutions for term 2 CBSE Board 2021-22

Solutions of Class 12 Maths Question Paper of Preboard -2 Exam Term-2 CBSE Board 2021-22

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NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations

NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations

NCERT Solutions for Class 11 Maths Chapter 5 Complex number and Quadratic equations

NCERT solutions of class 11 maths

NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations

Exercise 5.3

Miscellanous Exercise

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NCERT Solutions for class 9 maths

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NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations

NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations

NCERT Solutions for Class 11 Maths Chapter 5 Complex number and Quadratic equations

NCERT solutions of class 11 maths

NCERT Solutions for Class 11 Maths Exercise 5.3 Chapter 5 Complex number and Quadratic equations

Exercise 5.3

Miscellanous Exercise

NCERT Solutions of Science and Maths for Class 9,10,11 and 12

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