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 the points $z_A = -1-i$ and $z_B=1+i$. Then we look for the locus of all points Pwith $z_P=z$ such that  $\vert\arg \frac{z-z_A}{A-z_B} \vert = \pi$ (in other words, the angle $\angle APB=\pi$). This is the circle in the complex plane with diameter $AB$, since we know that the angle under which the segment $AB$ is seen from a point $P$ on the circle with the same diameter is $\pi$.
