0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ @MrYouMath From any of the information given, it is possible to get the relation $z_1^2 + z_2^2 -z_1z_2 = 0$? $\endgroup$ – Gummy bears Sep 20 '15 at 10:22
1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Gummy bears Sep 20 '15 at 10:24
  • $\begingroup$ @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 $\alpha=0,\beta=z_1,\gamma=z_2$. $\endgroup$ – mathlove Sep 20 '15 at 10:27
  • $\begingroup$ Aha..... That makes sense. Thanks for the help! $\endgroup$ – Gummy bears Sep 20 '15 at 10:29
0
$\begingroup$

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 $z_1,z_2$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.