# Division of Complex Numbers

## Theorem

Let $z_1 := a_1 + i b_1$ and $z_2 := a_2 + i b_2$ be complex numbers such that $z_2 \ne 0$.

The operation of division is performed on $z_1$ by $z_2$ as follows:

$\dfrac {z_1} {z_2} = \dfrac {a_1 a_2 + b_1 b_2} {a_2^2 + b_2^2} + i \dfrac {a_2 b_1 - a_1 b_2} {a_2^2 + b_2^2}$

## Proof 1

 $\ds \frac {z_1} {z_2}$ $=$ $\ds z_1 \paren {z_2}^{-1}$ Definition of Complex Division $\ds$ $=$ $\ds \paren {a_1 + i b_1} \dfrac {a_2 - i b_2} {a_2^2 + b_2^2}$ Inverse for Complex Multiplication $\ds$ $=$ $\ds \frac {\paren {a_1 a_2 + b_1 b_2} + i \paren {a_2 b_1 - a_1 b_2} } {a_2^2 + b_2^2}$ Definition of Complex Multiplication $\ds$ $=$ $\ds \frac {a_1 a_2 + b_1 b_2} {a_2^2 + b_2^2} + i \frac {a_2 b_1 - a_1 b_2} {a_2^2 + b_2^2}$

$\blacksquare$

## Proof 2

 $\ds \frac {z_1} {z_2}$ $=$ $\ds \frac {z_1 \overline {z_2} } {\cmod {z_2}^2}$ Complex Division as Product with Conjugate over Square of Modulus $\ds$ $=$ $\ds \dfrac {\paren {a_1 + i b_1} \paren {a_2 - i b_2} } {\cmod {z_2}^2}$ Definition of Complex Conjugate $\ds$ $=$ $\ds \dfrac {\paren {a_1 + i b_1} \paren {a_2 - i b_2} } {\paren {\sqrt { {a_2}^2 + {b_2}^2} }^2}$ Definition of Complex Modulus $\ds$ $=$ $\ds \frac {\paren {a_1 a_2 + b_1 b_2} + i \paren {a_2 b_1 - a_1 b_2} } { {a_2}^2 + {b_2}^2}$ Definition of Complex Multiplication and simplification $\ds$ $=$ $\ds \frac {a_1 a_2 + b_1 b_2} { {a_2}^2 + {b_2}^2} + i \frac {a_2 b_1 - a_1 b_2} { {a_2}^2 + {b_2}^2}$

$\blacksquare$

## Also presented as

The operation of complex division on $z_1$ by $z_2$ can also be presented as:

$\dfrac {z_1} {z_2} = \dfrac {a_1 a_2 + b_1 b_2 + i \paren {a_2 b_1 - a_1 b_2} } { {a_2}^2 + {b_2}^2}$