Find complex numbers $\alpha$,$\beta$,$\gamma$ such that... Is it possible to find three complex numbers $\alpha$, $\beta$, $\gamma$ such that all the ratios $\frac{\beta-\alpha}{\gamma-\alpha}$,$\frac{\gamma-\beta}{\alpha-\beta}$,$\frac{\alpha-\gamma}{\beta-\gamma}$ have negative real part?
Of course one can start writing $\alpha=\alpha_1+\alpha_2 i$, $\beta=\beta_1+\beta_2 i$ and $\gamma=\gamma_1+\gamma_2 i$ and start doing a huge amount of awful computations... There should be a smarter way to tackle this.
Any suggestions?
 A: Note that a fraction of the form $\frac{a+bi}{c+di}$ has a negative real part iff $ac+bd<0$. So yes, I think the only way to do this is by writing it in full. However, you can do this efficiently. Assuming we can find such a triplet, we derive the following set of inequalities
$$(\beta_1-\alpha_1)(\gamma_1-\alpha_1)+(\beta_2-\alpha_2)(\gamma_2-\alpha_2) < 0,$$
$$(\gamma_1-\beta_1)(\alpha_1-\beta_1)+(\gamma_2-\beta_2)(\alpha_2-\beta_2) < 0,$$
$$(\alpha_1-\gamma_1)(\beta_1-\gamma_1)+(\alpha_2-\gamma_2)(\beta_2-\gamma_2) < 0.$$
Adding those together we derive:
$$\alpha_1^2 + \beta_1^2 + \gamma_1^2 - \alpha_1\beta_1 - \beta_1\gamma_1 - \gamma_1\alpha_1 +\alpha_2^2 + \beta_2^2 + \gamma_2^2 - \alpha_2\beta_2 - \beta_2\gamma_2 - \gamma_2\alpha_2<0.$$
Now for $a,b,c\in\mathbb{R}$ we note that
$$(a-b)^2 + (b-c)^2 + (c-a)^2 = 2\big(a^2+b^2+c^2 - ab -bc-ca\big).$$
A: For distinct $a,b,c \in \mathbb{C}$, let
\begin{align*}
x &= \frac{b-a}{c-a}\\[4pt]
y &= \frac{c-b}{a-b}\\[4pt]
z &= \frac{a-c}{b-c}\\[4pt]
\end{align*}
The question is whether it is possible for $x,y,z$ to all have negative real part.

In fact, at least two of $x,y,z$ must have positive real part.

For example, consider $x$ and $y$.
\begin{align*}
\text{Then}\;\;y + \frac{1}{x} &=  \frac{c-b}{a-b} +  \frac{c-a}{b-a}\\[4pt]
&=  \frac{c-b}{a-b} -  \frac{c-a}{a-b}\\[4pt]
&=  \frac{(c-b)-(c-a)}{a-b}\\[4pt]
&=  \frac{a-b}{a-b}\\[4pt]
&= 1\\[8pt]
\text{But}\;\;y + \frac{1}{x}&=1\\[4pt]
\implies\;y + \frac{\bar{x}}{|x|^2}&=1\\[4pt]
\end{align*}
It follows that at least one of $\bar{x},y$ has positive real part.

But $x$ and $\bar{x}$ have the same real part, hence at least one of $x,y$ has positive real part.

Analogously, one can show that at least one of $y,z$ has positive real part, and at least one of $z,x$ has positive real part.
