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

• Reciprocal of Complex Number in terms of Conjugate and Modulus

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: $3.7.24$