# Division of Complex Numbers

Here you will learn what is the division of complex numbers with examples.

Let’s begin –

## Division of Complex Numbers

The division of a complex number $$z_1$$ by a non-zero complex number $$z_2$$ is defined as the multiplication of $$z_1$$ by the multiplicative inverse of $$z_2$$ and is denoted by $$z_1\over z_2$$.

Thus, $$z_1\over z_2$$ = $$z_1.{z_2}^{-1}$$ = $$z_1$$.($$1\over z_2$$)

How to Find Muliplicative Inverse :

Let z = a + ib be a non-zero complex number. Then,

$$1\over z$$ = $$1\over a + ib$$

Multiply numerator and denominator by conjugate of denominator,

$$1\over z$$ = $$1\over a + ib$$ $$\times$$ $$a – ib\over a – ib$$

$$\implies$$ $$1\over z$$ = $$a – ib\over a^2 – i^2b^2$$ = $$a – ib\over a^2 + b^2$$

$$\implies$$ $$1\over z$$ = $$a\over a^2 + b^2$$ + $$i(-b)\over a^2 + b^2$$

Clearly, $$1\over z$$ is equal to the multiplicatve inverse of z.

How to Divide Two Complex Numbers :

Let $$z_1$$ = $$a_1 + ib_1$$ and $$z_2$$ = $$a_2 + ib_2$$. Then

$$z_1\over z_2$$ = ($$a_1 + ib_1$$){$$a_2\over {a_2}^2 + {b_2}^2$$ + $$i{(-b_2)\over {a_2}^2 + {b_2}^2}$$}

[ $$\because$$ z = a + ib $$\implies$$ $$1\over z$$ = $$a\over a^2 + b^2$$ + $$i(-b)\over a^2 + b^2$$

By definition of multiplication,

$$z_1\over z_2$$ = ($$a_1a_2 + b_1b_2\over {a_2}^2 + {b_2}^2$$) + i($$a_2b_1 – a_1b_2\over {a_2}^2 + {b_2}^2$$)

Example : If $$z_1$$ = 2 + 3i and $$z_2$$ = 1 + 2i, then find $$z_1\over z_2$$.

Solution : We have $$z_1$$ = 2 + 3i and $$z_2$$ = 1 + 2i,

$$\implies$$ $$1\over z_2$$ = $$1\over 1 + 2i$$ = $${1\over 5} – {2\over 5}i$$

Now,

$$z_1\over z_2$$ = $$z_1$$.$$1\over z_2$$ = (2 + 3i)( $${1\over 5} – {2\over 5}i$$)

= ($${2\over 5} + {6\over 5}$$) + i($${-4\over 5} + {3\over 5}$$) = $$8\over 5$$ – $$1\over 5$$i