NCERT Solutions Class 11 – Chapter 4 Complex Numbers And Quadratic Equations – Exercise 4.1

For Q.1 to Q.10 express each complex number in form of a+ib

Question 1. (5i)[Tex](\frac{-3i}{5})[/Tex]

Solution:

Let the given number be a,

a= (5i)*[Tex](\frac{-3i}{5})[/Tex]

a= [Tex]\frac{-15i^2}{5}[/Tex]

a= (-3)*i²

a= (-3)*(-1)

a= 3+0i

Question 2. i⁹+i¹⁹

Solution:

Let the given number be a,

a = i⁹ * (1+i¹⁰)

a = ((i⁴)²*i )(1 + (i⁴)² (i²))

a = (1*i)(1+i²)

a = (i)*(0)

a = 0+0i

Question 3. i^-39

Solution:

Let the given number be a and let z = i³⁹ ,

z = (i)*(i²)¹⁹

z = (i)*(-1)¹⁹

z = -i

a = i^-39 

a = 1/i³⁹

a = 1/z

a = 1/-i

a = (i⁴)/-i

a = -i³ = -(i²*i)

a = -1*-i

a = 0+i

Question 4. 3(7+7i) + i(7+7i)

Solution:

Let the given number be a,

a = 3*(7+7i)+i*(7+7i)

a = 21+21i+7i+7i²

a = 21+7i²+28i

a = 21-7+28i

a = 14+28i

Question 5. (1-i)-(-1+i6)

Solution:

Let the given number be a,

a = (1-i)-(-1+6i)

a = 1-i+1-6i

a = 2-7i         

Question 6. ([Tex]\frac{1}{5}+\frac{2}{5}i  [/Tex])-(4+[Tex]\frac{5i}{2}  [/Tex])

Solution:

Let the given number be a,

a = [Tex](\frac{1}{5}+\frac{2i}{5})-(4+\frac{5i}{2})[/Tex]

a = [Tex](\frac{1}{5}-4)+(\frac{2i}{5}-\frac{5i}{2})[/Tex]

a = [Tex](\frac{-19}{5})+(\frac{2}{5}-\frac{5}{2})i[/Tex]

a = ([Tex]\frac{-19}{5}  [/Tex])+([Tex]\frac{-21i}{10}  [/Tex])

a =[Tex]\frac{-38-21i}{10}[/Tex]

Question 7. [([Tex]\frac{1}{3}+\frac{7i}{3}  [/Tex])+(4+[Tex]\frac{i}{3}  [/Tex]]-([Tex]\frac{-4}{3}  [/Tex]+i)

Solution:

Let the given number be a,

a = ([Tex]\frac{1}{3}  [/Tex]+[Tex]\frac{7i}{3}  [/Tex])+(4+[Tex]\frac{i}{3}  [/Tex])-([Tex]\frac{-4}{3}  [/Tex]+i)

a = ([Tex]\frac{1}{3}  [/Tex]+4+[Tex]\frac{4}{3}  [/Tex])+([Tex]\frac{7i}{3}+\frac{i}{3}  [/Tex]-i)

a = ([Tex]\frac{5}{3}  [/Tex]+4)+([Tex]\frac{8i}{3}  [/Tex]-i)

a = [Tex]\frac{17}{3}+ \frac{5i}{3}[/Tex]

a = [Tex]\frac{17+5i}{3}[/Tex]

Question 8. (1-i)⁴

Solution:

Let the given number be a,

a = ((1-i)²)²                               

As we know , (a-b)²= (a²+b²-2ab)

a = (1+i²-2i)²

a = (1-1-2i)²

a = (-2i)²

a = 4i²

a = -4+0i

Question 9. ([Tex]\frac{1}{3}  [/Tex]+3i)³

Solution:

Let the given number be a,

a = ([Tex]\frac{1}{3}  [/Tex]+3i)³

As we know, (a+b)³= (a³+b³+3ab(a+b))     

a = (([Tex]\frac{1}{27}  [/Tex])+(3i)³ +3([Tex]\frac{1}{3}  [/Tex])*(3i)([Tex]\frac{1}{3}  [/Tex]+3i))

a = ([Tex]\frac{1}{27}  [/Tex] +(-27i)+ 3i*([Tex]\frac{1}{3}  [/Tex]+3i))

a = ([Tex]\frac{1}{27}  [/Tex]+(-27i)+i+9i²)

a = (([Tex]\frac{1}{27}  [/Tex])-9+(-27)i+i)

a = (([Tex]\frac{-242}{27}  [/Tex])-26i)

Question 10. (-2-([Tex]\frac{i}{3}  [/Tex]))³

Solution:

Let the given number be a,

a = (-2-[Tex]\frac{i}{3}  [/Tex])³

a = -((2+[Tex]\frac{i}{3}  [/Tex])³)                      

As we know, (a+b)³= (a³+b³+3ab(a+b))  

a = -((8)+([Tex]\frac{i}{3}  [/Tex])³ +3(2)*([Tex]\frac{i}{3}  [/Tex])(2+[Tex]\frac{i}{3}  [/Tex]))

a = -(8+([Tex]\frac{-i}{27}  [/Tex])+ 2i*(2+[Tex]\frac{i}{3}  [/Tex]))

a = -(8-[Tex]\frac{i}{27}  [/Tex]+4i+[Tex]\frac{2i^2}{3}  [/Tex]) 

a = -(8-[Tex]\frac{2}{3}  [/Tex]+([Tex]\frac{-i}{27}  [/Tex])+4i)

a = -([Tex]\frac{22}{3}  [/Tex] +([Tex]\frac{107i}{27}  [/Tex]))

a = [Tex]\frac{-22}{3}-\frac{107i}{27}[/Tex]

Question 11. 4-3i

Solution:

Let’s denote given number as a,

the complement of a = [Tex]\overline{a}  [/Tex] = [Tex]\overline{4-3i}[/Tex]

[Tex]\overline{a} [/Tex] = (4+3i)

Modulus of a = (|a|) = √((4)²+(3)²)

|a|= √(16+9)=√(25)

|a| = 5

[Tex]\frac{1}{a}  [/Tex] = [Tex]\frac{\overline{a}}{|a|^2}[/Tex]

[Tex]\frac{1}{a}  [/Tex] = [Tex]\frac{(4+3i)}{25}[/Tex]

Question 12. √5+3i

Solution:

Let’s denote given number as a,

the complement of a = [Tex]\overline{a} = \overline{√5+3i}[/Tex]

[Tex]\overline{a}  [/Tex] = √5-3i

Modulus of a (|a|) = √((√5)²+(-3)²)

|a|= √(5+9)=√(14)

|a|=√(14)

[Tex]\frac{1}{a} = \frac{\overline{a}}{|a|^2}[/Tex]

[Tex]\frac{1}{a} = \frac{(5-3i)}{14}[/Tex]

Question 13. -i

Solution:

Let’s denote given number as a,

the complement of a = [Tex]\overline{a} = \overline{-i}[/Tex]

[Tex]\overline{a}  [/Tex] = i

Modulus of a (|a|) = √((0)²+(-1)²)

|a|= √(1)

|a|=1

[Tex]\frac{1}{a} = \frac{\overline{a}}{|a|^2}[/Tex]

[Tex]\frac{1}{a} = \frac{(i)}{1}  [/Tex] = i

Express the following expression in form of a+ib

Question 14.[Tex] \frac{(3+√5i)(3-√5i)} {(√3+√2i)-(√3-√2i)}[/Tex]

Solution:

Let’s denote the given expression as z,

z = [Tex]\frac{(3+√5i)(3-√5i)} {(√3+√2i)-(√3-√2i)}[/Tex]

z = [Tex]\frac{(3+√5i)(3-√5i)} {(√3-√3+√2i+√2i)}[/Tex]

z = [Tex]\frac{(3+√5i)(3-√5i)} {(2√2i)}[/Tex]

As we know that (a+b)(a-b) = a² – b²

z = [Tex]\frac{(3)^2-(√5i)^2}{2√2i}[/Tex]

z = [Tex]\frac{(9-5i^2)}{(2√2i)}[/Tex]

z = [Tex]\frac{(9+5)}{(2√2i)}[/Tex]

z = [Tex]\frac{14}{(2√2i)}[/Tex]

z = [Tex]\frac{7}{(√2)} . \frac{1}{(i)}[/Tex]

As we can write [Tex]\frac{1}{(i)} = \frac{1}{(i)}. \frac{i}{(i)} = \frac{i}{(i^2)}  [/Tex] = -i

z = [Tex]\frac{-7i}{(√2)}[/Tex]

z = 0-[Tex]\frac{-7i}{(√2)}[/Tex]