<p>A square is drawn in the complex plane with one vertex at the origin, one side along a ray at 45° and the opposite vertex at distance \(2\sqrt{2}\). The sum of the \(x\)-coordinates of all four vertices of the square is:</p>
Step-by-Step Solution
Key Concept: Represent the square's vertices using complex numbers by rotating from the origin. If one side is along a 45° ray from origin, the vertices are at 0, r·e^(iπ/4), r·e^(iπ/4)·(1+i), and r·e^(iπ/4)·i, where the diagonal constraint gives the side length.
<p><strong>Step 1:</strong> Set up vertices. With one vertex at origin O=0 and one side along 45°, let the side length be s. The side direction is e^(iπ/4) = (1+i)/√2.</p><p><strong>Step 2:</strong> The four vertices are: V₁ = 0, V₂ = s·e^(iπ/4) = s(1+i)/√2, V₃ = s·e^(iπ/4)·(1+i) = s(1+i)²/2 = si, V₄ = s·i·e^(iπ/4) = s(i-1)/√2.</p><p><strong>Step 3:</strong> The opposite vertex to origin is V₃ = si. Its distance from origin is s. Given this equals 2√2, we have s = 2√2.</p><p><strong>Step 4:</strong> Find x-coordinates (real parts):<br/>V₁: Re(0) = 0<br/>V₂: Re(2√2·(1+i)/√2) = Re(2(1+i)) = 2<br/>V₃: Re(2√2·i) = 0<br/>V₄: Re(2√2·(i-1)/√2) = Re(2(i-1)) = -2</p><p><strong>Step 5:</strong> Sum of x-coordinates = 0 + 2 + 0 + (-2) = <strong>0</strong></p><p>∴ Answer: A</p>
Correct Answer: A