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Let $z_{1}\: and \: z_{2}$ be two roots of the equation $z^{2}+az+b=0,z$ being complex further, assume that the origin , $z_{1}\: and \: z_{2}$ form an equilateral triangle then

Option 1)
$a^{2}= 2b$

Option 2)
$a^{2}= 3b$

Option 3)
$a^{2}= 4b$

Option 4)
$a^{2}= b$

Answers (1)

best_answer

As we have learned

Rotation -

$\frac{z_{3}-z_{1}}{z_{2}-z_{1}}=\frac{\left |z_{3}-z_{1} \right |}{\left |z_{2}-z_{1} \right |}.e^{i\Theta }$

- wherein

$e^{i \theta}=\cos \theta+i\sin \theta$

$z_1+z_2=-a$

$z_1z_2=b$

we have ,

$\frac{z_1}{z_2}= \frac{|z_1|}{|z_2|}e^{i\pi /3}$

$= \frac{1}{1}e^{i\pi /3}$

$z_1= z_2}e^{i\pi /3}$

$\Rightarrow z_2(e^{i\pi /3}+1) = -a$

and $\Rightarrow z_2^2(e^{i\pi /3}) = b$

$\therefore be^{-i\pi /3}=\frac{a^2}{(1+e^{i\pi/3})^2}$

$\Rightarrow a^2=b(1+e^{i2\pi/3}+2e^{i\pi/3})e^{-i\pi/3}=b(e^{-i\pi/3+e^{i\pi/3}+2})= b(2\cos \pi/3+2)= 3b$

Option 1)

$a^{2}= 2b$

Option 2)

$a^{2}= 3b$

Option 3)

$a^{2}= 4b$

Option 4)

$a^{2}= b$

Posted by

Himanshu

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