Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.1 | Set 1

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

Question 1. (5i) $(\frac{-3i}{5})$

Solution:

Let the given number be a,
a= (5i)* $(\frac{-3i}{5})$
a= $\frac{-15i^2}{5}$
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. ( $\frac{1}{5}+\frac{2}{5}i$ )-(4+ $\frac{5i}{2}$ )

Solution:

Let the given number be a,
a = $(\frac{1}{5}+\frac{2i}{5})-(4+\frac{5i}{2})$
a = $(\frac{1}{5}-4)+(\frac{2i}{5}-\frac{5i}{2})$
a = $(\frac{-19}{5})+(\frac{2}{5}-\frac{5}{2})i$
a = ( $\frac{-19}{5}$ )+( $\frac{-21i}{10}$ )
a = $\frac{-38-21i}{10}$

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

Solution:

Let the given number be a,
a = ( $\frac{1}{3}$ + $\frac{7i}{3}$ )+(4+ $\frac{i}{3}$ )-( $\frac{-4}{3}$ +i)
a = ( $\frac{1}{3}$ +4+ $\frac{4}{3}$ )+( $\frac{7i}{3}+\frac{i}{3}$ -i)
a = ( $\frac{5}{3}$ +4)+( $\frac{8i}{3}$ -i)
a = $\frac{17}{3}+ \frac{5i}{3}$
a = $\frac{17+5i}{3}$

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. ( $\frac{1}{3}$ +3i)³

Solution:

Let the given number be a,
a = ( $\frac{1}{3}$ +3i)³
As we know, (a+b)³= (a³+b³+3ab(a+b))
a = (( $\frac{1}{27}$ )+(3i)³+3( $\frac{1}{3}$ )*(3i)( $\frac{1}{3}$ +3i))
a = ( $\frac{1}{27}$ +(-27i)+ 3i*( $\frac{1}{3}$ +3i))
a = ( $\frac{1}{27}$ +(-27i)+i+9i²)
a = (( $\frac{1}{27}$ )-9+(-27)i+i)
a = (( $\frac{-242}{27}$ )-26i)

Question 10. (-2-( $\frac{i}{3}$ ))³

Solution:

Let the given number be a,
a = (-2- $\frac{i}{3}$ )³
a = -((2+ $\frac{i}{3}$ )³)
As we know, (a+b)³= (a³+b³+3ab(a+b))
a = -((8)+( $\frac{i}{3}$ )³ +3(2)*( $\frac{i}{3}$ )(2+ $\frac{i}{3}$ ))
a = -(8+( $\frac{-i}{27}$ )+ 2i*(2+ $\frac{i}{3}$ ))
a = -(8- $\frac{i}{27}$ +4i+ $\frac{2i^2}{3}$ )
a = -(8- $\frac{2}{3}$ +( $\frac{-i}{27}$ )+4i)
a = -( $\frac{22}{3}$ +( $\frac{107i}{27}$ ))
a = $\frac{-22}{3}-\frac{107i}{27}$