Suppose we have relation $$z^2 + az + b=0 $$ where $a$ and $b$ are real and roots of this equation $z_1$ and $z_2$ form equilateral triangle with origin then what could be relation between $a$ and $b$ ? I simply applied quadratic formula in equation $$ z = \frac{-a \pm \sqrt{a^2 -4b}}{2} $$ now since it forms equilateral triangle with origin so $$|z_1| =|z_2|$$ applying which $$( \frac{-a + \sqrt{a^2 -4b}}{2})^2= (\frac{-a - \sqrt{a^2 -4b}}{2})^2$$ I arrived at $$a^2 = 4b$$ but my answer is incorrect , why?

  • $\begingroup$ Are you given that $a$ and $b$ are real? $\endgroup$ – MJD Dec 5 '14 at 14:23
  • $\begingroup$ Yes they are real $\endgroup$ – Tesla Dec 5 '14 at 14:24
  • $\begingroup$ It's hard to say what you did wrong, because you didn't show us your mistake. You're right that $z=\frac{-a\pm\sqrt{a^2-4b}}2$. Butyou don't say how you got from there to $a^2=4b$. $\endgroup$ – MJD Dec 5 '14 at 14:27
  • $\begingroup$ Okay i have now edited my steps $\endgroup$ – Tesla Dec 5 '14 at 14:32
  • $\begingroup$ The problem here is that $|z_1| = |z_2|$ is true whether or not the roots form an equilateral triangle; it is true whenever $z_1$ and $z_2$ are roots of the same quadratic, because the two roots are always conjugate. But the equilateral triangle condition is stricter. $\endgroup$ – MJD Dec 5 '14 at 14:34

Hint :

$z_1,z_2$ are complex numbers $$(z-z_1)(z-z_2)=z^2+az+b$$ $$z_1+z_2=-a$$


Because of equal triangle, We know that $$z_1=M e^{i \alpha }$$ $$z_2=M e^{i (\alpha+ \pi /3) }$$ Thus you can use a relation between two complex numbers.

$$z_2/z_1=e^{i \pi /3}= \frac{ 1}{2}+i \frac{ \sqrt3}{2} $$

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  • 2
    $\begingroup$ That picture is very misleading, because $z_1$ and $z_2$ are conjugate. $\endgroup$ – MJD Dec 5 '14 at 14:49

Hint; The triangle in the complex plane whose verticies are the origin and the points $z_{1}$ and $z_{2}$ is equalateral if and only if \begin{equation} z_{1}+z_{2} = z_{1}z_{2} \end{equation} (Further hint: The reasoning behind this is that the distance from the origin to $z_{1}$ is $\sqrt{z_{1}\bar{z_{1}}}$ etc...)

  • $\begingroup$ Suppose $z_1=\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $z_2=1$. Then the right hand side is $z_1z_2=z_1=\frac{1}{2}+\frac{\sqrt{3}}{2}i$. That's not equal to the left hand side $z_1+z_2=\frac{3}{2}+\frac{\sqrt{3}}{2}i$. $\endgroup$ – khalatnikov Oct 19 '18 at 6:12

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