$\dfrac{1}{a+ib}$ $\,=\,$ $\dfrac{a-ib}{a^2+b^2}$

Complex numbers are involved in multiplication. If the product of any two non-zero complex numbers is equal to one, then each one of them is called the multiplicative inverse of the other one.

Let $z_1$ and $z_2$ denote two non-zero complex numbers and assume that their product is equal to one.

$z_1 \times z_2$ $\,=\,$ $1$

Let us assume that the value of the complex number $z_1$ is known but the value of the complex number $z_2$ is unknown. However, the complex number $z_2$ can be evaluated from the value of the complex number $z_1$ by the above equation.

$\implies$ $z_2$ $\,=\,$ $\dfrac{1}{z_1}$

Now, the complex number $z_2$ is called the reciprocal of the complex number $z_1$.

The concept of the multiplicative inverse of a complex number can be understood now from an example.

Let’s take $z_1 = a+ib$. Now, let us learn how to find the reciprocal of the complex number $z_1$ and it is represented by $z_2$ in mathematics.

$\implies$ $z_2$ $\,=\,$ $\dfrac{1}{z_1}$ $\,=\,$ $\dfrac{1}{a+ib}$

$\,\,\,\,\,\,=\,\,\,$ $\dfrac{1}{a+ib}$ $\times$ $1$

$\,\,\,\,\,\,=\,\,\,$ $\dfrac{1}{a+ib}$ $\times$ $\dfrac{a-ib}{a-ib}$

$\,\,\,\,\,\,=\,\,\,$ $\dfrac{1 \times (a-ib)}{(a+ib) \times (a-ib)}$

$\,\,\,\,\,\,=\,\,\,$ $\dfrac{a-ib}{a^2-(ib)^2}$

$\,\,\,\,\,\,=\,\,\,$ $\dfrac{a-ib}{a^2-(i^2 \times b^2)}$

$\,\,\,\,\,\,=\,\,\,$ $\dfrac{a-ib}{a^2-\big((-1) \times b^2\big)}$

$\,\,\,\,\,\,=\,\,\,$ $\dfrac{a-ib}{a^2-(-b^2)}$

$\,\,\,\,\,\,=\,\,\,$ $\dfrac{a-ib}{a^2+b^2}$

$\,\,\,\,\,\,\therefore\,\,\,$ $z_2$ $\,=\,$ $\dfrac{a-ib}{a^2+b^2}$

A mathematical operation of finding the multiplicative inverse of a complex number is called the reciprocal of a complex number.

$z_1 \times z_2$ $\,=\,$ $1$

$(1).\,\,$ $\dfrac{1}{3+4i}$

$(2).\,\,$ $\dfrac{1}{2-3i}$

$(3).\,\,$ $\dfrac{1}{3+4i}$

The list of questions on finding the multiplicative inverse of complex numbers with solutions to learn how to find the reciprocal of any complex number.

Latest Math Topics

Dec 13, 2023

Jul 20, 2023

Jun 26, 2023

Latest Math Problems

Jan 30, 2024

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved