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

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Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.1 | Set 1

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- $\frac{-7i}{(√2)}$

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Last Updated : 05 Apr, 2021