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 (a’) = (4-3i)’

a’ = 4+3i 

Modulus of a (|a|) = √((4)²+(3)²)

|a|= √(16+9)=√(25)

|a|=5

a^-1 = (a)’/ |a|²

a^-1 = (4+3i)/25

Question 12. √5+3i

Solution:

Let’s denote given number as a,

the complement of a (a’) = (√5+3i)’

a’ = √5-3i

Modulus of a (|a|) = √((√5)²+(-3)²)

|a|= √(5+9)=√(14)

|a|=√(14)

a^-1 = (a)’/ |a|²

a^-1 = (√5-3i)/14

Question 13. -i

Solution:

Let’s denote given number as a,

the complement of a (a’) = (-i)’

a’ = +i

Modulus of a (|a|) = √((0)²+(-1)²)

|a|= √(1)

|a|=1

a^-1 = (a)’/ |a|²

a^-1 = (i)/1 

a^-1 = i

Express the following expression in form of a+ib

Question 14. ((3+√5i)(3-√5i)) / ((√3+√2i)-(√3-√2i))

Solution:

Let’s denote the given expression as z,

z = ((3+√5i)(3-√5i)) / ((√3+√2i)-(√3-√2i))

z = ((3+√5i)(3-√5i)) / (√3-√3+√2i+√2i)

z = ((3+√5i)(3-√5i)) / (2√2i)

As we know that (a+b)(a-b) = a² – b²

z = ((3)²-(√5i)²) / (2√2i)

z = (9-5i²) / (2√2i)

z = (9+5) / (2√2i)

z = (14) / (2√2i)

z = (7 / (√2)) * (1 / i)

As we can write 1/i = i³ = -i

z = (-7i) / (√2)

z = 0-(7i/√2)