<p>Let \(z\) be a non-real complex number with \(|z|=1\). Then:</p>
\(z = e^{i\theta}\) for some \(\theta\)
\(z + \dfrac{1}{z} = 2\cos\theta\)
\(z - \dfrac{1}{z} = 2i\sin\theta\)
All of the above
Step-by-Step Solution
Key Concept: For |z|=1: z = e^(i\theta) = cos\theta+isin\theta, 1/z = e^(-i\theta). Sum = 2cos\theta, difference = 2isin\theta.
<p>All three identities follow from \(z=e^{i\theta}\) and \(1/z = \bar{z} = e^{-i\theta}\). \(z+\bar{z}=2\cos\theta\), \(z-\bar{z}=2i\sin\theta\). ✓ All options A, B, C, D are correct.</p>
Correct Answer: ABCD