A cool complex number proof that uses cosine identity!

Proof:

$$\ z + \frac{1}{z} = 2\cos(\theta) $$

Step 1: Multiply both sides by $\ z $

$$ z \left( z + \frac{1}{z} \right) = z \cdot 2 \cos(\theta) $$ $$ z^2 + 1 = 2 \cdot z \cos(\theta) $$

Step 2: Completing the square

Rearrange the terms to complete the square:

$$ z^2 - 2\cos(\theta)z + \cos^2(\theta) = -1 + \cos^2(\theta) $$

$$ (z - \cos(\theta))^2 = -(1 - \cos^2(\theta)) $$

Since $$\ 1 - \cos^2(\theta) = \sin^2(\theta) $$ , we have:

$$ (z - \cos(\theta))^2 = -\sin^2(\theta) $$

Step 3: Take the square root

Taking the square root of both sides, we get:

$$ z - \cos(\theta) = \pm i \sin(\theta) $$

Thus:

$$ z = \cos(\theta) + i \sin(\theta) \quad \text{or} \quad z = \cos(\theta) - i \sin(\theta) $$

Step 4: Add $$\ z $$ and $$\ \frac{1}{z} $$

We know that $$\ z = \cos(\theta) + i \sin(\theta) $$ and its reciprocal is $$\ \frac{1}{z} = \cos(\theta) - i \sin(\theta) $$ because:

$$\ \frac{1}{a + bi} = \frac{a - bi}{a^2 + b^2} = {a - bi} \quad \text{if} \quad a^2 + b^2 = 1 $$

Therefore:

$$ z + \frac{1}{z} = \left( \cos(\theta) + i \sin(\theta) \right) + \left( \cos(\theta) - i \sin(\theta) \right) $$

$$ z + \frac{1}{z} = 2\cos(\theta) $$

Thus, we have proven:

$$ z + \frac{1}{z} = 2\cos(\theta) $$

*This proof uses the case where $$\ z = \cos(\theta) + i \sin(\theta) $$, and it only holds true when the modulus(length) of the complex number is one.

Visualization Proof:

No longer true when modulus isn't one: