Reciprocal of Complex Number

Theorem

Let $z = a + i b$ be a complex number.

The reciprocal of $z$ is:

$\dfrac 1 z = \dfrac {a - i b} {a^2 + b^2}$


Proof

\(\ds \dfrac 1 z\) \(=\) \(\ds \dfrac 1 {a + i b}\) Definition of $z$
\(\ds \) \(=\) \(\ds \dfrac {a - i b} {\paren {a + i b} \paren {a - i b} }\) multiplying top and bottom by $a - i b$
\(\ds \) \(=\) \(\ds \dfrac {a - i b} {a^2 + b^2}\) Product of Complex Number with Conjugate

$\blacksquare$


Also see


Sources

  • 1964: Milton Abramowitz and Irene A. Stegun: Handbook of Mathematical Functions ... (previous) ... (next): $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.24$