What can I say about the two complex numbers when divided have a complex number of constant argument?

Question

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 : $$\arg\left(\frac{3Z-6-3i}{2Z-8-6i}\right)=\frac{\pi}{4}$$

Answer 1

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.

Answer 2

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 $$ L=\{z=4+i+2e^{it}\ :\ t\in\mathbb{R} \text{ and } -\pi\leq t<\frac{\pi}{2}\}. $$