<p>The maximum value of arg\(\left(\frac{1}{1-z}\right)\) for |<i>z</i>| = 1, <i>z</i> ≠ 1, is</p>
Step-by-Step Solution
Key Concept: For |z| = 1, the argument of 1/(1−z) can be related to the angle θ in z = e^(iθ), with the maximum occurring at a geometric extremum.
<p><strong>Step 1:</strong> Let <i>z</i> = e<sup>iθ</sup> with |<i>z</i>| = 1.</p><p><strong>Step 2:</strong> Then $1 - z = 1 - e^{i\theta} = 1 - \cos\theta - i\sin\theta = 2\sin^2\frac{\theta}{2} - 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}$.</p><p><strong>Step 3:</strong> So $\frac{1}{1-z} = \frac{1}{2\sin\frac{\theta}{2}(\sin\frac{\theta}{2} - i\cos\frac{\theta}{2})}$.</p><p><strong>Step 4:</strong> The argument of (sin(θ/2) − i cos(θ/2)) is −(π/2 − θ/2) = θ/2 − π/2 for appropriate range.</p><p><strong>Step 5:</strong> As θ varies in (0, 2π), the maximum argument is π/2, achieved as θ → π.</p>
Correct Answer: C