# Need help with this complex numbers problem.?

if |z1 + z2|>|z1-z2| then prove that -pi/2<arg(z1/z2)<pi/2 where z1 and z2 are complex numbers

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- IndicaLv 73 years ago
z₂=0 ⟹ |z₁| > |z₁| so z₂≠0. Hence can divide by z₂ to get |w+1| > |w−1| where w=z₁/z₂

Squaring and using |u|²=uu’ where u’ is complex conjugate gives

(w+1)(w’+1) > (w−1)(w’−1) which simplifies to w+w’ > 0 or Re(w) > 0

∴ w lies in either Q₁ or Q₄ and so it follows that -π/2 < arg(w) < π/2

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