Let A and B represent z_1 and z_2 in the Arga | vedclass

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(C) Given that $z_1$ and $z_2$ are roots of $Z^2+pZ+q=0$,we have $z_1+z_2 = -p$ and $z_1z_2 = q$.Since $OA=OB$,we have $|z_1| = |z_2|$.Let $z_1 = re^{

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$4q \cos^2 \left(\frac{\alpha}{2}\right)$

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(C) Given that $z_1$ and $z_2$ are roots of $Z^2+pZ+q=0$,we have $z_1+z_2 = -p$ and $z_1z_2 = q$.Since $OA=OB$,we have $|z_1| = |z_2|$.Let $z_1 = re^{i	heta_1}$ and $z_2 = re^{i	heta_2}$.Given $\angle AOB = \alpha$,we have $|	heta_1 - 	heta_2| = \alpha$.Then $z_1/z_2 = e^{i(	heta_1-	heta_2)} = e^{\pm i\alpha}$.From $z_1+z_2 = -p$,we have $p^2 = (z_1+z_2)^2 = z_1^2 + z_2^2 + 2z_1z_2$.Also $p^2 - 4q = (z_1+z_2)^2 - 4z_1z_2 = (z_1-z_2)^2$.Thus $p^2 = 4q + (z_1-z_2)^2 = 4q + z_2^2(z_1/z_2 - 1)^2$.Using $z_1/z_2 = e^{i\alpha}$,we get $p^2 = 4q + z_2^2(e^{i\alpha}-1)^2 = 4q + z_2^2 e^{i\alpha}(e^{i\alpha/2} - e^{-i\alpha/2})^2$.Since $z_1z_2 = q$,$z_2^2 e^{i\alpha} = z_2^2 (z_1/z_2) = z_1z_2 = q$.So $p^2 = 4q + q(2i \sin(\alpha/2))^2 = 4q - 4q \sin^2(\alpha/2) = 4q(1-\sin^2(\alpha/2)) = 4q \cos^2(\alpha/2)$.

Let $A$ and $B$ represent $z_1$ and $z_2$ in the Argand plane and $z_1, z_2$ be the roots of the equation $Z^2+pZ+q=0$,where $p, q$ are complex numbers. If $O$ is the origin,$OA=OB$ and $\angle AOB=\alpha$,then $p^2=$

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