Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.1 | Set 2

For Q.11 to Q.13 find the multiplicative inverse of the given number

Question 11. 4-3i

Solution:

Let’s denote given number as a,
the complement of a = $\overline{a}$ = $\overline{4-3i}$
$\overline{a}$ = (4+3i)
Modulus of a = (|a|) = √((4)²+(3)²)
|a|= √(16+9)=√(25)
|a| = 5
$\frac{1}{a}$ = $\frac{\overline{a}}{|a|^2}$
$\frac{1}{a}$ = $\frac{(4+3i)}{25}$

Question 12. √5+3i

Solution:

Let’s denote given number as a,
the complement of a = $\overline{a} = \overline{√5+3i}$
$\overline{a}$ = √5-3i
Modulus of a (|a|) = √((√5)²+(-3)²)
|a|= √(5+9)=√(14)
|a|=√(14)
$\frac{1}{a} = \frac{\overline{a}}{|a|^2}$
$\frac{1}{a} = \frac{(5-3i)}{14}$

Question 13. -i

Solution:

Let’s denote given number as a,
the complement of a = $\overline{a} = \overline{-i}$
$\overline{a}$ = i
Modulus of a (|a|) = √((0)²+(-1)²)
|a|= √(1)
|a|=1
$\frac{1}{a} = \frac{\overline{a}}{|a|^2}$
$\frac{1}{a} = \frac{(i)}{1}$ = i

Express the following expression in form of a+ib

Question 14. $\frac{(3+√5i)(3-√5i)} {(√3+√2i)-(√3-√2i)}$

Solution:

Let’s denote the given expression as z,
z = $\frac{(3+√5i)(3-√5i)} {(√3+√2i)-(√3-√2i)}$
z = $\frac{(3+√5i)(3-√5i)} {(√3-√3+√2i+√2i)}$
z = $\frac{(3+√5i)(3-√5i)} {(2√2i)}$
As we know that (a+b)(a-b) = a² – b²
z = $\frac{(3)^2-(√5i)^2}{2√2i}$
z = $\frac{(9-5i^2)}{(2√2i)}$
z = $\frac{(9+5)}{(2√2i)}$
z = $\frac{14}{(2√2i)}$
z = $\frac{7}{(√2)} . \frac{1}{(i)}$
As we can write $\frac{1}{(i)} = \frac{1}{(i)}. \frac{i}{(i)} = \frac{i}{(i^2)}$ = -i
z = $\frac{-7i}{(√2)}$
z = 0- $\frac{-7i}{(√2)}$