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

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.

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