Locus of complex numbers $z$ with restricted $(z+1-i)/(z-1-i)$ Problem

Describe the locus of the following points on the Argand diagram:
$$\left|\frac{(z+1-i)}{(z-1-i)}\right| = 1$$ and
$$\mathrm{arg}\left[\frac{(z+1+i)}{(z-1-i)}\right] = \pm \frac{\pi}{2}.$$

Progress
I've tried putting $z = x+iy $ then rationalising the denominator by multiplying by the conjugate which gave me some numbers, but I'm not sure what to do with them.
I would really love some help on these 2 questions, all responses much appreciated!
 A: to answer the second one $$arg\left({z + 1 + i \over z - 1 -i} \right) = \pm {\pi \over 2}$$  is a circle with diameter $1+i, -1-i.$  
here is how you can see this: let $${z + 1 + i \over z - 1 -i} = ki \text{  where $k$  is a real number}$$
you can dove for $z$  and get 
$$z = {-ki +k - 1 -i\over 1 -ki} = {[k-1 -i(k+1)] \over 1 - ik} = {[k-1 -i(k+1)](1 + ik) \over 1 + k^2}
\\= {k-1 + k(k+1) + i[k(k-1)-(k+1)] \over 1 + k^2} = {k^2+ 2k - 1 + i(k^2 - 2k - 1)  \over 1 + k^2}$$ 
if $z = x = iy,$ then $$x = {k^2 - 1 + 2k \over k^2 + 1}, y = {k^2 - 1 - 2k \over k^2 + 1} $$  which gives $$x^2 + y^2 = { 2(k^2 - 1)^2 + 8k^2 \over (k^2 + 1)^2} 
= { 2(k^2 + 1)^2  \over (k^2 + 1)^2}  = 2$$ as claimed at the beginning. 
if you look at the geometry of the problem, it says the angle subtended by the point $z$ by the points $\pm(1+i)$ is $90^\circ$ which is the locus of a circle. to derive it algebraically i had to go through all the trouble.
A: The first one is saying that the distance between $z$ and $1 + i$ is the same as the distance between $z$ and $1 - i$. The set of points equidistant from two points is the line bisecting the line segment joining the two points. Hence, the locus is the line $y = 0$.
The second one is saying if you look at the point $P$ given by translating $z$ 1 up and 1 to the right and the point $Q$ given by $z$ translated 1 down and 1 to the left, the angle between $P$ and $Q$ (with respect to the origin) is $\pi/2$. Intuitively, this seems to me to be the line $y = x$, though I haven't checked this rigorously.
A: $$\left|\frac{(z+1-i)}{(z-1-i)}\right| = 1$$
$$\left|\ (z+1-i)\right| = |(z-1-i)|$$
put z=x+iy
$$\left|\ (x+iy+1-i)\right| = |(x+iy-1-i)|$$
$$\left|\ (x+1)+i(y-1)\right| = |(x-1)+i(y-1)|$$
$$\left|\ (x+1)+i(y-1)\right|^2 = |(x-1)+i(y-1)|^2$$
since $$\left|\ (x+iy)\right|^2 = x^2+y^2$$
then
$$\left|\ (x+1)+i(y-1)\right|^2 = |(x-1)+i(y-1)|^2$$
$$
(x+1)^2+(y-1)^2=(x-1)^2+(y-1)^2
$$
$$(x+1)^2=(x-1)^2$$
$$x=0$$
equation of Y-axis.
A: On the Argand diagram, denote the points $A(-1+i), \;B(1+i),\;M(z).$
Then the vectors $\vec{MA}$ and $\vec{MB}$ are given by the difference $\big((-1+i)-z\big)$ and $\big((1+i)-z\big),\;$ respectively. Absolute value $|z+1-i|=|(-1+i)-z|$ is the lenght of the segment $MA$ (or $AM.$) Analogously holds for $MB.$


*

*$\left|\frac{(z+1-i)}{(z-1-i)}\right| = 1,$ or equivalently $|z+1-i|=|z-1-i|$ says: $M$ is equidistant to $A$ and $B$. So the locus of the points $M(z)$ is the line bisector of the segment $AB.$ Since the points $A,\;B$ are symmetric about $y-$axis, the locus is this axis. A short  complex equation is $Re(z)=0,$ while an $x,y-$equation is $x=0.$

*$\mathrm{arg}\left[\frac{(z+1+i)}{(z-1-i)}\right] = \pm \frac{\pi}{2}$ says: The vectors $\vec{MA}$ and $\vec{MB}$ are orthogonal. So the locus of points $M(z)$ is the circle with diameter $AB,$ except the points $A, B.$
$B$ is excluded because the denominator cannot be $0,$ and $A$ because it makes the fraction $0,$ but $\mathrm{arg}(w)$ is only defined for $w\neq 0.$
