Division of Complex Numbers/Proof 2
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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
\(\ds \frac {z_1} {z_2}\) | \(=\) | \(\ds \frac {z_1 \overline {z_2} } {\cmod {z_2}^2}\) | 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:
- $\dfrac {z_1} {z_2} = \dfrac {a_1 a_2 + b_1 b_2 + i \paren {a_2 b_1 - a_1 b_2} } { {a_2}^2 + {b_2}^2}$