Convert this complex to cartesian?

Question

I can convert an ordinary complex function to cartesian but this one has stumped me.

$$\arg\frac{z-8}{z-2}=\frac\pi 2$$

Thanks

Answer 1

An argument of a quotient of two numbers is a difference between arguments of those numbers: $$\arg\frac{z-8}{z-2}=\arg(z-8)-\arg(z-2)$$ (give or take $2\pi$).

Values $(z-8)$ and $(z-2)$ correspond to vectors from points $8$ and $2$ at the real axis, respectively, to $z$. The condition on the arguments' difference being $\frac\pi 2$ means the two vectors are orthogonal.

And you know from the Thales theorem that the right angle between two vectors from fixed points $A$ and $B$ to a common $z$ implies $z$ being on the circle, whose diameter is $AB$.

...and the only part remaining to complete the solution is finding which semicircle corresponds to the positive value given in the problem.

EDIT:

Let's try to do that algebraically only. Argument equal $\frac\pi 2$ means the value is imaginary with imaginary component positive: $$\arg\frac{z-8}{z-2}=\frac\pi2 \implies \frac{z-8}{z-2} = ki,\ k\in\mathbb R^+.$$ Let $z=x+yi$ with $x,y \in \mathbb R$: $$x-8+yi = (x-2+yi)\cdot ki = -ky + k(x-2)i$$ The complex equality implies equality of both real and imaginary parts: $$\begin{cases}x-8 = -ky \\ y = k(x-2)\end{cases}$$ Multiply both sides of 1st equation by $(-1)$ and both sides of second by $y$: $$\begin{cases}8-x = ky \\ y^2 = ky(x-2)\end{cases}$$ and plug $ky$ from 1st to 2nd: $$y^2 = (8-x)(x-2)$$ $$y^2 = -x^2+10x-16 = -(x-5)^2 + 9$$ and here is the circle: $$(x-5)^2+y^2 = 3^2$$

For the ratio $\frac{z-8}{z-2}$ to be defined, $z$ must not be equal $2$.
And the $x$ span of the circle is $[2,8]$. So we have the circle without its leftmost point.

However, when we apply both constraints $(k>0, x>2)$ to the second equation of our system $(y=k(x-2))$ we get $y>0.$

Finally, the answer is: an upper ($y$-positive) part of the circle with radius $3$, centered at $(5,0)$:

$$\left\{(x,y): (x-5)^2+y^2=3^2 \land y>0\right\}$$