<p>Let \(z, w\) be non-zero complex numbers with \(|z|=|w|\) and \(\arg(z)+\arg(w)=\pi\). Then \(z\) equals:</p>
Step-by-Step Solution
Key Concept: If arg(z) + arg(w) = \pi and |z|=|w|=r, then z = r \cdot e^(i\alpha) and w = r \cdot e^(i(\pi-\alpha)). So z \cdot w = r^2e^(i\pi) = -r^2 < 0. This gives z = -w̄ (reflecting w across origin and taking conjugate).
<p>Let \(w = r e^{i\phi}\). Then \(\arg(z) = \pi-\phi\), so \(z = re^{i(\pi-\phi)} = re^{i\pi}e^{-i\phi} = -re^{-i\phi} = -\bar{w}\). So \(z = -\bar{w}\). Checking option C: \(z = w \Rightarrow \arg(z)=\arg(w)\Rightarrow 2\arg(w)=\pi\)... C holds only if \(\arg(w)=\pi/2\).</p>
Correct Answer: BC