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