Geometrical Description of $ \arg\left(\frac{z+1+i}{z-1-i} \right) = \pm \frac{\pi}{2} $ 

The question is in an Argand Diagram, $P$ is a point represented by the complex number. Give a geometrical description of the locus of $P$ as $z$ satisfies the equation:
$$ \arg\left(\frac{z+1+i}{z-1-i} \right) = \pm \frac{\pi}{2} $$


What I have done:
Consider $$\left(\frac{z+1+i}{z-1-i} \right)$$
Let $z=x+iy$
$$\left(\frac{x+iy+1+i}{x+iy-1-i} \right) \Leftrightarrow \left(\frac{(x+1)+i(y+1)}{(x-1)+i(y-1)} \right) $$
$$ \Leftrightarrow \left(\frac{(x+1)+i(y+1)}{(x-1)+i(y-1)} \right) \cdot \left(\frac{(x-1)-i(y-1)}{(x-1)-i(y-1)} \right) $$
$$ \Leftrightarrow \left( \frac{ (x+1)(x-1) -i(y+1)(x+1) + i(y+1)(x-1) - i^2(y+1)(y-1)}{(x-1)^2 -i^2(y-1)^2} \right)$$
$$ \Leftrightarrow \left (\frac{ x^2 -1+y^2-1 +i(xy-y+x-1)-i(xy+y+x+1)}{(x-1)^2 +(y-1)^2} \right) $$
$$ \Leftrightarrow \left (\frac{ x^2 -1+y^2-1 +i(xy-y+x-1)-i(xy+y+x+1)}{(x-1)^2 +(y-1)^2} \right) $$
$$ \Leftrightarrow \left (\frac{ (x^2 +y^2-2) +i(-2-2y)}{(x-1)^2 +(y-1)^2} \right) $$
$$ \Leftrightarrow \left( \frac{x^2+y^2-2}{(x-1)^2 +(y-1)^2} \right) + i \left( \frac{-2-2y}{(x-1)^2 +(y-1)^2} \right) $$
As the $\arg = \pm \frac{\pi}{2}$ , which means locus of $P$ is purely imaginary hence $\Re(z)=0$
$$\Rightarrow \frac{x^2+y^2-2}{(x-1)^2 +(y-1)^2} = 0 $$
$$ \therefore x^2 + y^2 = 2 $$
So the Locus of $P$ should be a circle with centre $(0,0)$ and radius $\sqrt{2}$
However my answer key states that the answer is $|z|=2$ which is a circle with centre $(0,0)$ and radius $2$ so where exactly did I go wrong?

Let's do it by another way and see if the results are the same (which would be good).
What you have inside the $\arg(\cdot)$, $\frac{z+z_0}{z-z_0}$ with $z_0=1+i$ is a particular case of a Möbius transformation
let's break it into pieces.
You can rewrite $\frac{z+z_0}{z-z_0}$ as
$$\frac{z+z_0}{z-z_0}=\frac{z-z_0}{z-z_0}+\frac{2z_0}{z-z_0}=1+2z_0\frac{1}{z-z_0}$$
So we can say that your tranformation is given by the next 4 transformations in order:


*

*translation by $-z_0=-1-i$: $f_1\left(z\right)=z-z_0$

*inversion: $f_2\left(z\right)=\frac{1}{z}$

*multiplication by $2z_0=2+2i$: $f_3 (z)=2z_0z$ 

*translation by $1$: $f_4(z) = 1 + z $


In order to get your transformation yo have to apply them in order:
$$\frac{z+z_0}{z-z_0}=f_4(f_3(f_2(f_1(z))))$$
As we know where we want to get, which is the imaginary axis, $\arg(z)=\pm\frac{\pi}{2}$ we can perform the inverse transformations and in the inverse order from that point, so i'll be calling $P$ the set of points we have in each step.
First of all $P=\{z=x+iy;\,x=0\}$ the imaginary axis:
Step 1) $f^{-1}_4(z) = z - 1$ 
 which is a translation of the points one unit to the left in the Argand plane, so now $P=\{z=x+iy;\,x=-1\}$
Step 2) $f_3 (z)=\frac{1}{2z_0}z=\frac{1-i}{4}\left(-1+yi\right)=\frac{-1+y+(1+y)i}{4}$ so, now $P$ becomes the line that passes through the point $z_A =-\frac{1}{4}+\frac{1}{4}i$ and is at an angle of $\frac{\pi}{4}$.
$P = \{(-\frac{1}{4},\frac{1}{4})+\lambda\left(1,1\right);\,\lambda\in\mathbb{R}\}$
Step 3) $f^{-1}_2(z) = \frac{1}{z}$. An inversion transforms every line which doesn't pass by the origin in a circle that touches the origin. We can find this circle by looking for the oposite point of the one that is transformed into the origin. This point is the closest to the origin, so be have that, the point $z_A$ is transformed into $-2-2i$, so our circle $P$ is the one that has radius $\sqrt2$ and center $\left(-1,-1\right)$
Final Step) $f_1^{-1}(z) = z+z_0$ by performing the translation we get that the whole circle $P$ moves, becoming the circle with center the origin, and the same radius $\sqrt2$
So finally, yes, you are right, $P$ is the circle of radius $\sqrt2$ centered on the origin. 
