[JEE Advanced 1995] If $|Z|\le1,\ |W|\le1$, show that $|Z-W|^2\le(|Z|-|W|)^2+(\arg Z-\arg W)^2$.
If $|Z|\le1,\ |W|\le1$, show that $|Z-W|^2\le(|Z|-|W|)^2+(\arg Z-\arg W)^2$.
1 Answer
VAVidaara Admin
✓ Vidaara Team
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· 1mo ago
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Answer: Proved.
Write $|Z-W|^2=(|Z|-|W|)^2+4|Z||W|\sin^2\frac{\theta}2$ where $\theta=\arg Z-\arg W$. Since $|Z||W|\le1$ and $4\sin^2\frac\theta2\le\theta^2$, the bound follows.
JEE Advanced 1995 · Complex Numbers — verified solution by the Vidaara Team.
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