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

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

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

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