If z=x+iy satisfies the equation z^2+az+a^2=0 | vedclass

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(A) Given the equation $z^2+az+a^2=0$ where $a \in R$. This is a quadratic equation in $z$. The roots are given by the quadratic formula: $z = \frac{-

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$|z|=|a|$

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(A) Given the equation $z^2+az+a^2=0$ where $a \in R$. This is a quadratic equation in $z$. The roots are given by the quadratic formula: $z = \frac{-a \pm \sqrt{a^2 - 4(1)(a^2)}}{2} = \frac{-a \pm \sqrt{-3a^2}}{2} = \frac{-a \pm i a \sqrt{3}}{2}$. Now,we find the modulus of $z$: $|z| = \left| \frac{-a}{2} \pm i \frac{a \sqrt{3}}{2} \right|$. $|z| = \sqrt{\left( \frac{-a}{2} \right)^2 + \left( \frac{a \sqrt{3}}{2} \right)^2}$. $|z| = \sqrt{\frac{a^2}{4} + \frac{3a^2}{4}} = \sqrt{\frac{4a^2}{4}} = \sqrt{a^2} = |a|$. Thus,$|z|=|a|$.

If $z=x+iy$ satisfies the equation $z^2+az+a^2=0$,where $a \in R$,then:

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