Division of Complex Numbers/Proof 1
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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
\(\ds \frac {z_1} {z_2}\) | \(=\) | \(\ds z_1 \paren {z_2}^{-1}\) | 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 | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {a_1 a_2 + b_1 b_2} {a_2^2 + b_2^2} + i \frac {a_2 b_1 - a_1 b_2} {a_2^2 + b_2^2}\) |
$\blacksquare$
Also presented as
The operation of complex division on $z_1$ by $z_2$ can also be presented as:
- $\dfrac {z_1} {z_2} = \dfrac {a_1 a_2 + b_1 b_2 + i \paren {a_2 b_1 - a_1 b_2} } { {a_2}^2 + {b_2}^2}$
Sources
- 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Fundamental Operations with Complex Numbers: $4$. Division