Find the multiplicative inverse of the following complex numbers : √

1.	Find the multiplicative inverse of the following complex numbers : √5 + 3i
Answer» Given complex number is Z= √5+3i We know that the multiplicative inverse of a complex number Z is Z^-1(or) \(\frac{1}{Z}\). ⇒ Z^-1 = \(\frac{1}{√{5} + 3i}\) Multiplying and dividing with √5+3i ⇒ Z^-1 =\(\frac{1}{√5+3i}\times\frac{√5- 3i}{√5- 3i}\) ⇒ Z^-1 = \(\frac{√5 - 3i}{(√5 )^2-(3i)^2}\) ⇒ Z^-1 = \(\frac{√5+3i}{5-9i^2}\) We know that i²=-1 ⇒ Z^-1 = \(\frac{√5+3i}{5-9(-1)}\) ⇒ Z^-1 = \(\frac{√5+3i}{14}\) ∴ The Multiplicative inverse of √5+3i is \(\frac{√5+3i}{14}\)

Find the multiplicative inverse of the following complex numbers : √5 + 3i

Answer»

Given complex number is Z= √5+3i

We know that the multiplicative inverse of a complex number Z is Z^-1(or) \(\frac{1}{Z}\).

⇒ Z^-1 = \(\frac{1}{√{5} + 3i}\)

Multiplying and dividing with √5+3i

⇒ Z^-1 =\(\frac{1}{√5+3i}\times\frac{√5- 3i}{√5- 3i}\)

⇒ Z^-1 = \(\frac{√5 - 3i}{(√5 )^2-(3i)^2}\)

⇒ Z^-1 = \(\frac{√5+3i}{5-9i^2}\)

We know that i²=-1

⇒ Z^-1 = \(\frac{√5+3i}{5-9(-1)}\)

⇒ Z^-1 = \(\frac{√5+3i}{14}\)

∴ The Multiplicative inverse of √5+3i is \(\frac{√5+3i}{14}\)