What line is determined by the following complex equation? $$\left | z+1-i \right |= \frac{\Re z-\Im z}{\sqrt{2}}$$
This leads to the following equation, if we write $z=x+iy$ :
$$\sqrt{\left ( x+1 \right )^{2}+\left( y-1 \right)^2}=\frac{x-y}{\sqrt{2}}$$
which according to desmos, defines a set in $\mathbb{C}$ which is empty. 


Suppose that there exist $(x,y)$ such that
$$\sqrt{(x+1)^2+(y-1)^2}=\frac{x-y}{\sqrt 2}$$
(note here that $x-y\ge 0$)
Then we have
$$\begin{align}(x+1)^2+(y-1)^2=\frac{(x-y)^2}{2}&\Rightarrow x^2+y^2+2x-2y+2=\frac{x^2}{2}+\frac{y^2}{2}-xy\\&\Rightarrow \frac{x^2}{2}+\frac{y^2}{2}+xy+2x-2y+2=0\\&\Rightarrow x^2+y^2+2xy+4x-4y+4=0\\&\Rightarrow (x+y)^2+4=4(y-x)\end{align}$$
The LHS is positive and the RHS is non-positive. This is a contradiction.
