When the arguments of two roots of a quadratic equation are equal? Let $az^2+bz+c=0$ be a quadratic equation with complex coefficients $a,b,c$ and roots $z_1, z_2.$  
How can I obtain the condition for $$\arg z_1=\arg z_2$$ containing $a,b,c?$
At present I have,   Since $$z_1z_2=\dfrac{c}{a}$$ If $\arg z_1=\arg z_2,$
 then $$\arg z_1=\arg z_2=\dfrac{1}{2}(\arg c-\arg a)$$
Is there any reference discuss about roots of quadratic equations with complex coefficients?
Here, I have ask a similar question about $|z_1|=|z_2|,$ which has a nice solution. 
 A: $arg(c/a) = 2.arg(b/a)$ and $|b|^2 \ge 4|a|.|c|$ seems to work.
Assume the equation has roots of the form required, i.e. $r_1 e^{i\theta}$ and $r_2 e^{i\theta}$, then $(z-r_1 e^{i\theta} )(z-r_2 e^{i\theta} ) = 0 = z^2 - (r_1 + r_2).e^{i\theta}  .z + r_1.r_2.e^{2i\theta}$ Equating the coefficients gives 


*

*$ b/a =- (r_1 + r_2).e^{i\theta}$ 

*$c/a = r_1.r_2.e^{2i\theta}$


From this it seems (with $a$ non-zero) we can find $r_1  $ and $r_2$ as the real roots of $r^2 + |b/a|.r + |c/a| = 0$, provided $|b|^2 \ge 4|a|.|c|$, and clearly $arg(c/a) = 2.arg(b/a)$.
(Note: if $c = 0$ the solution is degenerate: $z_1 = 0, z_2 = -b/a$ You might still regard $z_1 = 0.arg (z_2)$)
(Note 2: in line with the second answer, the condition that $arg(c/a) = 2.arg(b/a)$ is equivalent to  to $b/ac $ is real as $b/ac = (b/a)^2. a/c = (r_1 + r_2)^2.e^{2i\theta}. 1/(r_1.r_2). e^{-2i\theta} = (r_1 + r_2)^2/(r_1.r_2)$ and so together with the condition $|b|^2 \ge 4|a|.|c|$ this is equivalent to $b/ac$ is real and $\ge 4$)
A: If $a = 0$ then the equation has (at most) one solution, and if $c = 0$
then $z_1 = 0$ is one solution for which $\arg z_1$ is not defined.
Therefore in the following I am assuming that both $a$ and $c$ are non-zero.
Assume that $a z^2 + b z + c$ has the two solutions $z_1, z_2$
with $\arg z_1 = \arg z_2$. Then $z_2  = \lambda z_1$ with a real number
$\lambda > 0$ and
$$
   a z^2 + b z + c = a (z - z_1)(z-\lambda z_1) 
$$
gives the equations
$$
   a \lambda z_1^2 = c \, , \quad - a(1+\lambda)z_1 = b \quad .
$$
Substituting $z_1$ from the second equation into the first gives
$$
    \frac {b^2}{a c} = \frac {(1+\lambda)^2}{\lambda} = 
   \frac {(1-\lambda)^2}{\lambda} + 4
    \ge 4 \quad.
$$
So a necessary condition is that $b^2/(ac)$ is positive real and $\ge 4$.
This condition is also sufficient. If $b^2/(ac) \ge 4$ then it is easy to
see that 
$$
   \frac {(1+\lambda)^2}{\lambda} = \frac {b^2}{a c}
$$
has a real solution $\lambda > 0$ and then
$$
   z_1 = -\frac{b}{a(1+\lambda)} \, , \quad z_2 = -\frac{\lambda b}{a(1+\lambda)}
$$
are solutions of  $a z^2 + b z + c$ with $z_2  = \lambda z_1$ .
A: The roots $(-b\pm\sqrt{b^2-4ac})/2a$ are a real positive multiples of each other.
For some positive $\mu$,
$$-b+\sqrt{b^2-4ac}=\mu(-b-\sqrt{b^2-4ac})$$
$$(\mu-1)^2b^2=(\mu+1)^2(b^2-4ac)$$
$$b^2=\frac{(\mu+1)^2}{\mu}ac$$
The latter function of $\mu$ can take any value not smaller than $4$, and the condition is
$$b^2=\lambda ac,$$with $\lambda\ge4$, which can be expressed as $2\arg b=\arg a+\arg c\land |b|^2\ge4|ac|$.
