<s>[INST] Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$ 
For a given complex number, $z$, find the locus of points on the Argand diagram such that
$$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$$

This is my approach:
$$\arg(z-8)-\arg(z-2)=\frac{\pi}{2}$$
Suppose $z=x+iy$:
$$\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}=\frac{\pi}{2}$$
This is where I'm unsure I made a 'legal' move. I reasoned that if I take the tangent of both sides of the equation, I would end up with a fraction that must be undefined as $\tan\frac{\pi}{2}$ is undefined; hence the denominator of the fraction I obtain must be $0$:
$$\tan\left(\arctan\frac{y}{x-8}-\arctan\frac{y}{x-2}\right)=\frac{\frac{y}{x-8}-\frac{y}{x-2}}{1+\frac{y^2}{(x-8)(x-2)}}=\frac{y(x-2)-y(x-8)}{(x-8)(x-2)+y^2}$$
But this is undefined; hence
$$(x-8)(x-2)+y^2=0\implies (x-5)^2+y^2=9$$
so it appears that the required locus is a circle with centre at $(5,0)$ on the Argand diagram with radius $3$.
Is my reasoning fully acceptable throughout my solution?
 [/INST] If $\arg(f(z))=\pi /2$, that means that $\Re(f(z))=0$ and as well $0 < \Im (f(z))$: that's a more "sure" way to go.
The steps in detail are
$$
\eqalign{
  & \arg \left( {{{z - 8} \over {z - 2}}} \right) = {\pi  \over 2}\quad  \Rightarrow \quad
 \left\{ \matrix{
  {\mathop{\Re}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right) = 0 \hfill \cr 
  0 < {\mathop{\Im}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right) \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  z = x + i\,y \hfill \cr 
  {{z - 8} \over {z - 2}}\quad \left| {\;z \ne 2} \right.\quad  =  \hfill \cr 
   = {{x - 8 + i\,y} \over {x - 2 + i\,y}}
 = {{\left( {x - 8 + i\,y} \right)\left( {x - 2 - i\,y} \right)}
 \over {\left( {x - 2} \right)^{\,2}  + \,y^{\,2} }} =  \hfill \cr 
   = {{\left( {\left( {x - 8} \right)\left( {x - 2} \right) + \,y^{\,2} } \right) + i\,y\left( {x - 2 - x + 8} \right)}
 \over {\left( {x - 2} \right)^{\,2}  + \,y^{\,2} }}\quad \left| \matrix{
  \;x \ne 2 \hfill \cr 
  \;y \ne 0 \hfill \cr}  \right. \hfill \cr 
  0 = {\mathop{\Re}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right)
 = \left( {x - 8} \right)\left( {x - 2} \right) + \,y^{\,2} \quad \left| \matrix{
  \;x \ne 2 \hfill \cr 
  \;y \ne 0 \hfill \cr}  \right. \hfill \cr 
  0 < {\mathop{\Im}\nolimits} \left( {{{z - 8} \over {z - 2}}} \right)
 = 6y\quad \left| \matrix{
  \;x \ne 2 \hfill \cr 
  \;y \ne 0 \hfill \cr}  \right. \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  z = x + i\,y \hfill \cr 
  0 = \left( {x - 5 - 3} \right)\left( {x - 5 + 3} \right) + \,y^{\,2}  =  \hfill \cr 
  \left( {x - 5} \right)^{\,2}  + \,y^{\,2}  - 9\quad \left| {\;0 < y} \right. \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  z = x + i\,y \hfill \cr 
  (x,y) \in circle(center\,(5,0),\,radius\,3)\; \wedge \;0 < y \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  \left| {z - 5} \right| = 3 \hfill \cr 
  0 < {\mathop{\Im}\nolimits} z = {\mathop{\Im}\nolimits} \left( {z - 5} \right) \hfill \cr}  \right. \cr} 
$$
The circle is the same as you found, but (specially in the complex field) you must be careful in justifying every step that you do
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