Inverse for Complex Multiplication

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Theorem

Each element $z = x + i y$ of the set of non-zero complex numbers $\C_{\ne 0}$ has an inverse element $z^{-1}$ under the operation of complex multiplication:

$\forall z \in \C_{\ne 0}: \exists z^{-1} \in \C_{\ne 0}: z \times z^{-1} = 1 + 0 i = z^{-1} \times z$

This inverse can be expressed as:

$\dfrac 1 z := \dfrac {x - i y} {x^2 + y^2} = \dfrac {\overline z} {z \overline z}$

where $\overline z$ is the complex conjugate of $z$.


Proof

\(\ds \paren {x + i y} \frac {x - i y} {x^2 + y^2}\) \(=\) \(\ds \frac {\paren {x \cdot x - y \cdot \paren {-y} } + i \paren {x \cdot \paren {-y} + x \cdot y} } {x^2 + y^2}\)
\(\ds \) \(=\) \(\ds \frac {\paren {x^2 + y^2} + 0 i} {x^2 + y^2}\)
\(\ds \) \(=\) \(\ds 1 + 0 i\)

Similarly for $\dfrac {x - i y} {x^2 + y^2} \paren {x + i y}$.

So the inverse of $x + i y \in \struct {\C_{\ne 0}, \times}$ is $\dfrac {x - i y} {x^2 + y^2}$.

As $x^2 + y^2 > 0 \iff x, y \ne 0$ the inverse is defined for all $z \in \C: z \ne 0 + 0 i$.

$\Box$


From the definition, the complex conjugate $\overline z$ of $z = x + i y$ is $x - i y$.

From the definition of the modulus of a complex number, we have:

$\cmod z = \sqrt {a^2 + b^2}$

From Modulus in Terms of Conjugate, we have that:

$\cmod z^2 = z \overline z$

Hence the result:

$\dfrac 1 z = \dfrac {\overline z} {z \overline z}$

$\blacksquare$


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Examples

Example: $\dfrac 1 {1 + i}$

$\dfrac 1 {1 + i} = \dfrac {1 - i} 2$


Example: $\dfrac 1 {3 + 2 i}$

$\dfrac 1 {3 + 2 i} = \dfrac 3 {13} + \dfrac {2 i} {13}$


Sources

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