Reciprocal of a Complex Number in Trigonometric Form

The reciprocal of a complex number $ z = r (\cos \alpha + i \sin \alpha) $ in trigonometric form is given by: $$ \frac{1}{z} = \frac{1}{r} \cdot (\cos \alpha - i \sin \alpha) $$ More generally, for any positive integer $ n $: $$ z^{-n} = \frac{1}{r^n} \cdot (\cos (n\alpha) - i \sin (n\alpha)) $$

In shorthand notation, the reciprocal of a complex number is simply expressed as:

$$ z^{-1} = \frac{1}{z} $$

If the exponent is a negative integer:

$$ z^{-n} = \frac{1}{z^n} $$

Note: More generally, when raising a complex number to a negative integer exponent, we can apply the following identity: $$ z^{-n} = [r \cdot (\cos \alpha + i \sin \alpha)]^{-n} $$ $$ z^{-n} = \frac{1}{[r \cdot (\cos \alpha + i \sin \alpha)]^n} $$ $$ z^{-n} = \frac{1}{r^n \cdot (\cos (n\alpha) + i \sin (n\alpha))} $$ $$ z^{-n} = \frac{1}{r^n} \cdot (\cos (n\alpha) - i \sin (n\alpha)) $$

A Practical Example
Proof

A Practical Example

Consider the complex number given in trigonometric form:

$$ z = 5 \cdot [ \cos(30^\circ) + i \sin(30^\circ) ] $$

Applying the reciprocal formula, we get:

$$ \frac{1}{z} = \frac{1}{5} \cdot [ \cos(30^\circ) - i \sin(30^\circ)] $$

Proof

Let’s derive the formula for the reciprocal of a complex number. We start with a general complex number expressed in polar form:

$$ z = r \cdot (\cos \alpha + i \sin \alpha )$$

The reciprocal is simply:

$$ \frac{1}{z} = \frac{1}{r \cdot (\cos \alpha + i \sin \alpha) } $$

To rewrite this expression in a more useful form, we use the fact that $\cos(0) = 1$ and $\sin(0) = 0$, allowing us to multiply by 1 in the form of $ \cos 0 + i \sin 0 $:

$$ \frac{1}{z} = \frac{1 \cdot (\cos 0 + i \sin 0)}{r \cdot (\cos \alpha + i \sin \alpha) } $$

Now the numerator and denominator are both complex numbers in trigonometric form, so we can apply the standard rule for division of complex numbers in polar form:

$$ \frac{1}{z} = \frac{1}{r} \cdot [ \cos (0 - \alpha) + i \sin (0 - \alpha) ] $$

Since cosine is an even function, meaning that $ \cos(-\alpha) = \cos(\alpha) $, we get:

$$ \frac{1}{z} = \frac{1}{r} \cdot [ \cos (\alpha) + i \sin (-\alpha) ] $$

And since sine is an odd function, meaning that $ \sin(-\alpha) = -\sin(\alpha) $, we conclude:

$$ \frac{1}{z} = \frac{1}{r} \cdot (\cos \alpha - i \sin \alpha) $$

Thus, the formula for the reciprocal of a complex number in trigonometric form is fully established.

And that’s the proof!