Locus of $z$ satisfying $\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$

Question

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?

Answer 1

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

Answer 2

Here is perhaps a better approach. Note $\arg\left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$ implies

$$\frac{z-8}{z-2}=re^{i \frac{\pi}{2}}\implies \frac{z-8}{z-2}= - \frac{\bar z-8}{\bar z-2}\implies |z|^2 -5(z+\bar z )+16=0 $$ which is $|z-5|^2=9$, or $|z-5|=3$, i.e. a circle of center $(5,0)$ and radius $3$.

Note that $z=\frac{8-i 2r }{1-i r} =\frac{(8+2r^2)+i 6r}{1+r^2}$, indicating a positive imaginary and the upper half circle.

Answer 3

If $A(z_A),B(z_B),M(z_M)$ are distinct points in Argand plane, then $\arg \left(\frac{z_M-z_B}{z_M-z_A}\right)$ is an oriented angle between vectors $(\vec{AM},\vec{BM}).$

In the present case $z_M=z,z_A=2,z_B=8.$

$\arg \left(\frac{z-8}{z-2}\right)=\frac{\pi}{2}$ says that the vectors are orthogonal, and $M$ lies on an open semi-circle with diameter $AB.$
(It is the upper half of the circle with diameter $AB,$ without $A,B.$)