Polar Form of Reciprocal of Complex Number

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Theorem

Let $z := r \paren {\cos \theta + i \sin \theta} \in \C$ be a complex number expressed in polar form.


Then:

$\dfrac 1 z = \dfrac {\cos \theta - i \sin \theta} r$


Proof

\(\ds \dfrac 1 z\) \(=\) \(\ds \dfrac {\overline z} {z \overline z}\) Inverse for Complex Multiplication
\(\ds \) \(=\) \(\ds \dfrac {r \paren {\cos \theta - i \sin \theta} } {r \paren {\cos \theta + i \sin \theta} r \paren {\cos \theta - i \sin \theta} }\) Polar Form of Complex Conjugate
\(\ds \) \(=\) \(\ds \dfrac {\paren {\cos \theta - i \sin \theta} } {r \paren {\cos^2 \theta - i^2 \sin^2 \theta} }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac {\paren {\cos \theta - i \sin \theta} } {r \paren {\cos^2 \theta + \sin^2 \theta} }\) Definition of Imaginary Unit
\(\ds \) \(=\) \(\ds \dfrac {\cos \theta - i \sin \theta} r\) Sum of Squares of Sine and Cosine

$\blacksquare$


Sources