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)