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
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 2$. Geometrical Representations: $(2.14)$