Let $λ,z_1,z_2,z_3 ∈ \mathbb{C}$ (Complex Numbers) are such that
$$\frac{z_3-z_1}{z_2-z_1} = \lambda.$$
Now if $$λ=e^{it}$$ where $t ∈ \mathbb{R}$ and $z_2, z_3$ are fixed, then what will be the locus of $z_1$.
NOTE: I am considering complex numbers $z_1, z_2, z_3$ as points on the Argand's Plane.
Since, $\dfrac{z_3-z_1}{z_2-z_1}$ is a complex number then it can be expressed as \begin{align*} \dfrac{z_3-z_1}{z_2-z_1}=\left|\dfrac{z_3-z_1}{z_2-z_1}\right|e^{i\phi}=\dfrac{|\vec{AC}|}{|\vec{AB}|}e^{i\phi} \hspace{1cm}...... (1) \end{align*} where we have assumed that, $|z_3-z_1|$ is the distance between point $z_3$ and $z_1$ similarly $|z_2-z_1|$ is defined and named as $AC$ & $AB$ respectively. $\phi$ is the argument of the complex number $\dfrac{z_3-z_1}{z_2-z_1}$. But it will be easy if we consider $\phi$ as the angle between the vector $\vec{AC}$ and $\vec{AB}$. It is given that $$\dfrac{z_3-z_1}{z_2-z_1}=\lambda=e^{it}$$ hence we can say $|\vec{AC}|=|\vec{AB}|$. It means $z_1$ moves in such a way that $t$ is the angle between $z_3-z_1$ and $z_2-z_1$ and the distance between $z_3-z_1$ and $z_2-z_1$ is equal. Geometrically it is the perpendicular bisector of the line joining $z_3$ and $z_2$. Hence locus of the equation $\dfrac{z_3-z_1}{z_2-z_1}=e^{it}$ is a straight line.