Geometry of complex numbers

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$Let z_{1}, z_{2} be the roots of the equation z^{2} + a z + 12 = 0 and z_{1}, z_{2} form an equilateral$ $triangle with origin. Then, the value of | a | is$

Solution

$\begin{array}{l} The given equation is z^{2} + a z + 12 = 0 \\ suppose that z_{1}, z_{2} are the roots of the above equation \\ Hence, z_{1} + z_{2} = - a, and z_{1} z_{2} = 12 \\ {Given that O, z}_{1}, z_{2} forms an equilateral triangle \\ {Hence, O}^{2} + {z_{1}}^{2} + {z_{2}}^{2} = z_{1} z_{2} + z_{2} O + O z_{1} \\ \Rightarrow {z_{1}}^{2} + {z_{2}}^{2} = z_{1} z_{2} \\ \Rightarrow {(z_{1} + z_{2})}^{2} = 3 z_{1} z_{2} \\ \Rightarrow a^{2} = 3 (12) = 36 \\ \Rightarrow |a| = 6 \end{array}$

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Geometry of complex numbers

Remember concepts with our Masterclasses.

Let z1,z2 be the roots of the equation z2+az+12=0 and z1,z2 form an equilateral triangle with origin. Then, the value of |a| is

The given equation is z2+az+12=0suppose that z1,z2 are the roots of the above equationHence, z1+z2=−a, and z1z2=12Given that O, z1,z2 forms an equilateral triangleHence, O2+z12+z22=z1z2+z2O+Oz1⇒z12+z22=z1z2⇒z1+z22=3z1z2⇒a2=312=36⇒a=6

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$Let z_{1}, z_{2} be the roots of the equation z^{2} + a z + 12 = 0 and z_{1}, z_{2} form an equilateral$ $triangle with origin. Then, the value of | a | is$