Question

Find the multiplicative inverse of the complex number \[z = a + ib\] when \[a = 2\] and \[b = - 3\].

Answer 1

Hint: In this question, we will proceed by substituting the given values in the complex number. Then we will find the multiplicative inverse and simplify it by rationalising with its denominator to obtain the required solution.

Complete step-by-step answer:
Given a complex number is \[z = a + ib\].
Put \[a = 2\] and \[b = - 3\] which is given in the question itself then, we have
\[ \Rightarrow z = 2 - 3i\]
We know that if \[z\] is a complex number then its multiplicative inverse is \[\dfrac{1}{z}\].
So, the multiplicative inverse of the given complex number is
\[ \Rightarrow \dfrac{1}{z} = \dfrac{1}{{2 - 3i}}\]
On rationalising the denominator, we have
By using the formula \[\left( {x - iy} \right)\left( {x + iy} \right) = {x^2} + {y^2}\], we have
\[
   \Rightarrow \dfrac{1}{z} = \dfrac{{2 + 3i}}{{{2^2} + {3^2}}} \\
   \Rightarrow \dfrac{1}{z} = \dfrac{{2 + 3i}}{{4 + 9}} \\
   \Rightarrow \dfrac{1}{z} = \dfrac{{2 + 3i}}{{13}} \\
\]
Separating the terms, we get
\[
   \Rightarrow \dfrac{1}{z} = \dfrac{2}{{13}} + \dfrac{{3i}}{{13}} \\
  \therefore \dfrac{1}{z} = \dfrac{2}{{13}} + i\dfrac{3}{{13}} \\
\]
Thus the multiplicative inverse of \[z = a + ib\] when \[a = 2\] and \[b = - 3\] is \[\dfrac{2}{{13}} + i\dfrac{3}{{13}}\].

Note:In a complex number \[z = a + ib\], ‘\[a\]’ is the real part of the complex number and ‘\[b\]’ is the imaginary part of the complex number. Rationalization, as the name suggests, is the process of making fractional rational. The need for rationalization arises when there are irrational numbers, surds or roots and complex numbers in the denominator of a fraction.