[JEE Advanced 1997] $z_1,z_2$ are roots of $z^2+pz+q=0$ ($p,q$ complex). $A,B$ represent them; $\angle AOB=\alpha\ne0$ and $OA=OB$ ($O$ the origin). Prove $p^2=4q\cos^2\frac\alpha2$.
$z_1,z_2$ are roots of $z^2+pz+q=0$ ($p,q$ complex). $A,B$ represent them; $\angle AOB=\alpha\ne0$ and $OA=OB$ ($O$ the origin). Prove $p^2=4q\cos^2\frac\alpha2$.
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Answer: Proved.
$OA=OB$ gives $|z_1|=|z_2|$, so $z_2=z_1e^{i\alpha}$. Then $\frac{p^2}{q}=\frac{(z_1+z_2)^2}{z_1z_2}=2+e^{i\alpha}+e^{-i\alpha}=4\cos^2\frac\alpha2$, i.e. $p^2=4q\cos^2\frac\alpha2$.
JEE Advanced 1997 · Complex Numbers — verified solution by the Vidaara Team.
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