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

  • $\begingroup$ Are you given $Z = \frac{{{Z_1}}}{{{Z_2}}}$? $\endgroup$ – user328032 Apr 5 '16 at 8:34
  • $\begingroup$ The ratio of two complex number will always have a constant argument, because they are just numbers. Do you mean that $Z_1$ and $Z_2$ are curves or solutions to some equations and thereby change? $\endgroup$ – EHH Apr 5 '16 at 8:35
  • $\begingroup$ @Benedict yes , I want to know what can we comment about Z .. can we locate it on argand plane accurately? $\endgroup$ – brainst Apr 5 '16 at 8:35
  • $\begingroup$ @EHH actually $Z$ is something like $(x + k )+ \iota (y + m)$ where x and y are variables and k and m are constants $\endgroup$ – brainst Apr 5 '16 at 8:36
  • $\begingroup$ In future you need all of this detail in the question otherwise it's very hard for people to see exactly what you are asking. But the answer is as given by spinoza since $\frac{Z_1}{Z_2}$ is just another complex number and you are restricting it's argument to be $k$. $\endgroup$ – EHH Apr 5 '16 at 8:41

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.

  • $\begingroup$ I think the question was not really clear , I've made an edit , please have a look and change the answer accordingly . Thanks! $\endgroup$ – brainst Apr 5 '16 at 8:51

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


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