# Show that $arg(z_1) \equiv \arg(z_2) \pmod{2\pi}$ assuming $|a|^2 - 4|b| \leq 0$ and $\arg b \equiv 2\arg a \pmod{2\pi}$

Let $a, b \in \mathbb{C}^{*}$ and $(E)$ the following equation : $z^2 - az + b = 0$.

Assuming : \left\{ \begin{equation} \begin{aligned} & \arg b \equiv 2\arg a \pmod{2\pi} \\ & |a|^2 - 4|b| \geq 0 \end{aligned} \end{equation} \right.

How should I start in order to show that $z_1$ and $z_2$ (roots of $(E)$) would have the same arguments ($\arg z_1 = \arg z_2 \pmod{2\pi}$).

I have proved the first implication by assuming $z_1$ and $z_2$ have the same arguments, hence the set of conditions above is true.

For the reciprocal, I tried to start again from $b = z_1 z_2$ and $a = z_1 + z_2$ and manipulating these equations, but I have difficulty in wrapping my mind around the simultaneous usage of these two conditions.

I wonder also if that's provable through counter example?

I think the condition you want is actually $|a|^2-4|b|\ge 0$ (otherwise, take $a = b = 2$, and the roots of $z^2-2z+2$ are actually $1\pm i$, which have different arguments)
If the roots have the same argument, then they should have the same argument as their sum, i.e. $a$. As such, let $y = z/a$. We then have $$z^2-az+b = (ay)^2-a(ay)+b = a^2y^2 - a^2y + b = a^2\left(y^2 - y + \frac{b}{a^2}\right).$$ Note that if $y_1$ and $y_2$ are the roots of $y^2-y+\frac{b}{a^2}$, then $y_1 = z_1/a$ and $y_2 = z_2/a$. It would suffice to show that $y_1$ and $y_2$ are both positive reals to conclude that $z_1$ and $z_2$ have the same argument.
Since $\arg b\equiv 2\arg a \equiv \arg(a^2)\mod 2\pi$, it follows that $b$ and $a^2$ have the same argument, and hence $\frac{b}{a^2}$ is real. Furthermore, by the quadratic formula, the roots $y_1$ and $y_2$ are given by $$y_{1,2} = \frac{1\pm\sqrt{1-4\left(\frac{b}{a^2}\right)}}{2}.$$ Now use the condition $|a|^2-4|b|\ge 0$ to conclude that both roots are real and positive.