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Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.1 | Set 2
  • Last Updated : 13 Jan, 2021

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)

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