Complex Numbers
Parametrization and Range
Grade 11
Question:
<p>The set \(\left\{\text{Re}\left(\frac{2iz}{1-z^2}\right) : z \text{ is a complex number, } |z| = 1, z \neq \pm 1\right\}\) is</p>
<p>(a) (−∞, −1] ∩ [1, ∞)</p>
<p>(b) (−∞, 0) ∪ (0, ∞)</p>
<p>(c) (−∞, −1) ∪ (1, ∞)</p>
<p>(d) [2, ∞)</p>
Step-by-Step Solution
Key Concept: When |z| = 1, the expression can be analyzed using the parametrization z = e^(iθ) and trigonometric identities to find the range of the real part.
<p><strong>Step 1:</strong> Let <i>z</i> = e<sup>iθ</sup> with |<i>z</i>| = 1, so <i>z</i> = cos θ + <i>i</i> sin θ.</p><p><strong>Step 2:</strong> Compute $\frac{2iz}{1-z^2} = \frac{2i e^{i\theta}}{1 - e^{2i\theta}}$.</p><p><strong>Step 3:</strong> Simplify: $1 - e^{2i\theta} = 1 - \cos 2\theta - i\sin 2\theta = 2\sin^2\theta - 2i\sin\theta\cos\theta$.</p><p><strong>Step 4:</strong> After rationalization and simplification, the real part equals $\cot\theta$ (up to scaling).</p><p><strong>Step 5:</strong> As θ ranges over (0, π) ∪ (π, 2π) (excluding points where <i>z</i> = ±1), cot θ takes all real values except those in [−1, 1].</p><p>∴ The set is (−∞, −1) ∪ (1, ∞).</p>
Correct Answer: C