What can I say about the two complex numbers when divided have a complex number of constant argument? Suppose there are two complex numbers $Z_1$ and $Z_2$ We are given that :
$$\arg\left(\frac{Z_1}{Z_2}\right) = k$$ where $k$ is an arbitrary constant.
Is there any way to visualise the complex number $\frac{Z_1}{Z_2}$ on argand plane. I know the traditional method of using
$$\arg\left(\frac{a}{b}\right) = \arg(a) - \arg(b)$$
Actual question was :

*

*$Z_1$ and $Z_2$ were given in terms of $Z$

*I was asked to find the locus of $Z$
I am trying to generalise the question
$Z_1$ and $Z_2$ are $f(Z)$ and we are supposed to find locus of $Z$.
For example :
$$\arg\left(\frac{3Z-6-3i}{2Z-8-6i}\right)=\frac{\pi}{4}$$
 A: If $Z_1$ and $Z_2$  satisfy $arg\left(\frac{Z_1}{Z_2}\right)=k,$ then $\frac{Z_1}{Z_2}$ lies on the line through the origin that makes an angle $k$ with the positive real axis.
A: In response to substantial edits of the question, rather than edit my previous answer, I have decided to leave it and add a new answer because the original question has been so thoroughly changed that it has become a new question.
One way to find all $z$ that satisfy
$$
arg\left(\frac{3z-6-3i}{2z-8-6i}\right)=\frac{\pi}{4}
$$
is to let $z=x+iy$ for arbitrary $x,y\in\mathbb{R}.$ We observe that the above equation says$^*$ that there exists some $r>0$ such that
$$
\left(\frac{3}{2}\right)\frac{x-2+i(y-1)}{x-4+i(y-3)}=\left(\frac{3}{2}\right)\frac{z-2-i}{z-4-3i}=\frac{3z-6-3i}{2z-8-6i}=re^{i\frac{\pi}{4}}.
$$
(In passing we note that $z\ne 4+3i$. Otherwise, the denominator is zero.)
Let $R=\frac{2}{3}r$, then $R>0$ is still arbitrary and we have
$$
\frac{x-2+i(y-1)}{x-4+i(y-3)}=Re^{i\frac{\pi}{4}}=R+iR.
$$
Then
$$
x-2+i(y-1)=(R+iR)(x-4+i(y-3)).
$$
Equating real and imaginary parts, and solving for $R$ in each case we find that (as long as $x+y\ne 7$, and $x-y\ne1$)
$$
R=\frac{y-1}{x+y-7}=\frac{x-2}{x-y-1}.
$$
(We note that the fact that $R>0$ implies some restrictions: if $y> 1$, we need $x+y> 7$; if $y< 1$, we need $x+y< 7$; if $x> 2,$ we need $x-y>1$; if $x< 2,$ we need $x-y<1.$)$^{**}$
Multiplying both sides by $(x+y-7)(x-y-1),$ rearranging, and completing the square (for both $x$ and $y$), we obtain
$$
(x-4)^2+(y-1)^2=4.
$$
This describes a circle of radius $2$ centered at $4+i$. The set of points on this circle can be written as
$$
\{z=4+i+2e^{it}\ :\ t\in\mathbb{R} \text{ and } -\pi\leq t<\pi\}.
$$
Recall, however, the restrictions on $z$, including those marked $^{**}$. Taken together (after checking all cases) these tell us that the set
$$
L=\{z=4+i+2e^{it}\ :\ t\in\mathbb{R} \text{ and } -\pi< t<\frac{\pi}{2}\}
$$
is the locus of points that satisfy the given equation. Geometrically this, is $\frac{3}{4}$ of the circle of radius $2$ about $4+i$.
*Here we consider $arg(0)$ to be undefined, but we could say that $z=2+i$ satisfies the equation, if we took another interpretation, in which case we would have
$$
L=\{z=4+i+2e^{it}\ :\ t\in\mathbb{R} \text{ and } -\pi\leq t<\frac{\pi}{2}\}.
$$
