Complex number and quadratic equation

Question: Let $z_1$ and $z_2$ be two roots of the equation $z^2 + az + b= 0$, where $z$ is complex. Further, assume the origin, $z_1$ and $z_2$ to form an equilateral triangle. Then, find the relation between $a$ and $b$.

I cannot think of a way to attempt this question. I tried drawing a diagram, but that didn't help. How should I begin the question?

• @MrYouMath From any of the information given, it is possible to get the relation $z_1^2 + z_2^2 -z_1z_2 = 0$? – Gummy bears Sep 20 '15 at 10:22

By Vieta's formulas, we have $z_1+z_2=-a$ and $z_1z_2=b$.

From the given condition, we have $$z_2=z_1\left(\cos\left(\pm\frac{\pi}{3}\right)+i\sin\left(\pm\frac{\pi}{3}\right)\right)=\frac{1\pm i\sqrt 3}{2}z_1,$$ i.e. $$2z_2-z_1=\pm i\sqrt 3z_1.$$ Squaring the both sides gives $$4z_2^2-4z_1z_2+z_1^2=-3z_1^2$$ Can you take it from here?

• Yes. Definitely. Thanks for the tip! Is this a condition for an equilateral triangle? Is there any relation between $z_1$ and $z_2$ if they, with the origin, form an equilateral triangle? – Gummy bears Sep 20 '15 at 10:24
• @Gummybears: $\alpha,\beta,\gamma$ form an equilateral triangle if and only if $\frac{\gamma-\alpha}{\beta-\alpha}=\cos(\pm\frac{\pi}{3})+i\sin (\pm\frac{\pi}{3})$. In our case, take $\alpha=0,\beta=z_1,\gamma=z_2$. – mathlove Sep 20 '15 at 10:27
• Aha..... That makes sense. Thanks for the help! – Gummy bears Sep 20 '15 at 10:29

From equilateral triangle we get two informations. a) Angle between $z_1$ and $z_2$ needs to be $\pi/3$ and we know that both roots have the same distance to the Origin ($r$). Now we can rewrite the roots in polar form: $$z_1=re^{i\phi}$$ $$z_2=re^{i(\phi+\pi/3)}=re^{i\phi}e^{i\pi/3}=e^{i\pi/3}z_1$$

Apply the fundamental theorem of algebra

$$z^2+az+b=(z-z_1)(z-z_2)$$

Compare the coefficients of the polynomials and see what relationship exists between $a,b$ and $z_1,z_2$