Find locus of $z$, where $z$ satisfies $\arg \frac{(z- z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} = \pi $ How to find the locus of the point $z$, satisfying
$$\arg \frac{(z- z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)} = \pi $$
Can anyone please extensively describe how one should tackle these kind of problems?
 A: Note, that $$\arg \frac{b}{a} = \arg b - \arg a = \angle (Ox, b) - \angle (Ox, a) = \angle (a,b)$$ is the directed angle between vectors $a$ and $b$. From this we have $$\arg \frac{z-z_1}{z-z_3} + \arg \frac{z_2-z_3}{z_2-z_1} = \angle (\vec {Z_3Z},\vec{Z_1Z}) + \angle (\vec {Z_1Z_2},\vec{Z_3Z_2}) = \pi$$
which means, that point $Z$ lies on the arc $Z_1Z_3$ of the circle $Z_1Z_2Z_3$, which contains point $Z_2$.

A: Let $w= \frac{(z- z_1)(z_2 - z_3)}{(z - z_3)(z_2 - z_1)}$. Then, the given condition $\text{arg}(w) = \pi$ implies $w = \bar w =r e^{i\pi}$, or
$$\frac{(z- z_1)(z_2 - z_3)}{( z -z_3)(z_2 - z_1)} 
=\frac{(\bar z- \bar z_1)(\bar z_2 - \bar z_3)}{(\bar z - \bar z_3)(\bar z_2 - \bar z_1)}$$
Use the shorthand 
$$a = (z_2 - z_3)(\bar z_2 - \bar z_1)$$
to rearrange above equation as follows,
$$a(z- z_1)(\bar z - \bar z_3)=\bar a( z -z_3)(\bar z- \bar z_1)$$
$$a[|z|^2- (z_1\bar z + \bar z_3z)+z_1\bar z_3]
-\bar a[|z|^2- (z_3\bar z + \bar z_1z)+\bar z_1 z_3]=0$$
$$|z|^2- \frac{az_1-\bar az_3}{a-\bar a}\bar z -\frac{a \bar z_3-\bar az_3}{a-\bar a}z
=\frac{\bar a\bar z_1 z_3 - a z_1\bar z_3}{a-\bar a}$$
Recognize that above $z$ satisfies the equation of a circle, i.e.
$$\left| z - c\right|^2
=r^2$$
with
$$c=\frac{az_1-\bar az_3}{a-\bar a},\>\>\>\>\>\>\>r^2=\left|\frac{az_1-\bar az_3}{a-\bar a}\right|^2+\frac{\bar a\bar z_1 z_3 - a z_1\bar z_3}{a-\bar a}\tag 1$$
Thus, the locus of the point $z$ lies on the circle with its center at $c$ and its radius $r$ given by (1). (See the graph below.)

A: $$\arg\Bigl( \frac{z_1-z}{z_3-z} \Bigr)+\arg\Bigl( \frac{z_3-z_2}{z_1-z_2}\Bigr)=π$$
Sum of angle about $z_2$ and $z$ is $π$.
Hence locus of $z$ is arc of circle opposite to $z_2$ with respect to chord joining $z_1$ and $z_3$.
