# How to Find the Reciprocal of a Complex Number

Here you will learn how to find the reciprocal of a complex number with examples.

Let’s begin –

## How to Find the Reciprocal of a Complex Number

The reciprocal is also called multiplicative 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.

Also, $$1\over z$$ = $$a – ib\over a^2 + b^2$$ = $$\bar{z}\over | z |^2$$

Thus, the multiplicative inverse of a non-zero complex number is same as its reciprocal and is given by

$$Re (z)\over | z |^2$$ + $$i{(-Im (z))\over | z |^2}$$ = $$\bar{z}\over | z |^2$$

Example : Find the reciprocal or multiplicative inverse of the following complex numbers.

(i) 3 + 2i

(ii) $$(2 + \sqrt{3}i)^2$$

Solution :

(i) Let z = 3 + 2i. Then,

$$1\over z$$ = $$1\over 3 + 2i$$

= $$3 – 2i\over (3 + 2i)(3 – 2i)$$ = $$3 – 2i\over 9 – 4i^2$$

= $${3\over 13} – {2\over 13}i$$

(ii) Let z = $$(2 + \sqrt{3}i)^2$$. Then,

z = $$4 + 3i^2 + 4\sqrt{3}i$$

= $$4 – 3 + 4\sqrt{3}i$$ = $$1 + 4\sqrt{3}i$$

$$\therefore$$ $$1\over z$$ = $$1\over 1 + 4\sqrt{3}i$$

Multiply numerator and denominator by conjugate of denominator

$$\implies$$ $$1\over z$$ = $$1 – 4\sqrt{3}i\over (1 + 4\sqrt{3}i)(1 + 4\sqrt{3}i)$$

= $$1 – 4\sqrt{3}i\over 1 + 48$$ = $$1\over 49$$ – $$4\sqrt{3}i\over 49$$