2
$\begingroup$

$d$ and $g$ are complex numbers and $g$ is not eqaul to $0$. Prove that if the roots of the equation $$x^2 + dx + g^2 = 0$$ have the same absolute value, then $d/g$ is a real number.

I tried to solve the problem by finding the roots and then transforming the results into the form of $d/g$. But it seems that I am going in the wrong direction.

Could somebody tell as to how the problem could be solved?

$\endgroup$
  • $\begingroup$ math.stackexchange.com/questions/1091646/… $\endgroup$ – Bumblebee Mar 29 '15 at 19:50
  • $\begingroup$ @randomgirl: $x_1 = 1$ and $x_2 = i$ is a valid example which the roots have the same absolute value but they are not of the form $a\pm bi$. Note that $d$ and $g^2$ are complex. $\endgroup$ – kennytm Mar 29 '15 at 20:08
0
$\begingroup$

(I assume we already knew $e^{i\phi}=\cos\phi+i\sin\phi$ here.)

Let the two roots be $x_1 = Re^{i\theta_1}$ and $x_2 = Re^{i\theta_2}$. Then we get:

\begin{align} x_1 + x_2 &= R(e^{i\theta_1} + e^{i\theta_2}) = -d \\ x_1x_2 &= R^2 e^{i(\theta_1 + \theta_2)} = g^2 \end{align}

Thus $g = \pm R e^{i(\theta_1 + \theta_2)/2}$. The sign will not affect whether $d/g$ is real so we just choose "−" here.

Then,

\begin{align} \frac dg &= \frac{e^{i\theta_1} + e^{i\theta_2}}{e^{i(\theta_1+\theta_2)/2}} \\ &= (e^{i\theta_1} + e^{i\theta_2})(e^{-i(\theta_1+\theta_2)/2)}) \\ &= e^{i(\theta_1 - \theta_2)/2} + e^{i(\theta_2 - \theta_1)/2} \\ &= 2\cos\frac{\theta_1-\theta_2}2 \in \mathbb R. \end{align}

$\endgroup$

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.