Question:

<p><strong>Step 1:</strong> Let w = kz/(k+1). Rewrite as w(k+1) = kz, so wk + w = kz, giving w = k(z-w), thus k = w/(z-w).</p><p><strong>Step 2:</strong> Eliminate k: From w = kz/(k+1), we get w(k+1) = kz, so w = k(z-w). This means as k varies over ℝ, the point w traces the line through origin and z (excluding z itself). Actually, rearranging: w = kz/(k+1) implies the locus is a line in the complex plane.</p><p><strong>Step 3:</strong> The parametric form kz/(k+1) can be written as z/(1+1/k). As k varies, this traces a line. Specifically, let z = x+iy. Then kz/(k+1) = k(x+iy)/(k+1) traces a line passing through origin with direction z, excluding the point at infinity.</p><p><strong>Step 4:</strong> More precisely: the locus of w = kz/(k+1) for k∈ℝ, k≠-1 is the line through the origin in direction of z. The condition |w+2i| > √2 for all such w means: distance from the line {tw : t∈[0,1)} to point -2i must exceed √2.</p><p><strong>Step 5:</strong> The distance from point -2i to the line through origin with direction z = x+iy is |(-2i)·(perpendicular direction)|/|z|. This distance = |2y|/√(x²+y²) > √2/|z|... </p><p><strong>Step 6:</strong> More directly: distance from -2i to line through origin in direction z equals |Im(z·(-2i)*)/|z|| = 2|Re(z)|/|z|. For this to exceed √2: 2|x|/√(x²+y²) > √2, so 4x² > 2(x²+y²), giving 2x² > 2y², thus |x| > |y|.</p><p><strong>Step 7:</strong> Since z is in fourth quadrant: x > 0, y < 0, with |x| > |y| meaning x > -y. In fourth quadrant, arg(z)∈(-π/2, 0). The condition |x| > |y| gives -π/4 > arg(z) > -π/2.</p><p>∴ <strong>Answer: arg(z) ∈ (-π/2, -π/4)</strong></p>