<s>[INST] 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. 

 [/INST] This is indeed impossible: Using Cauchy-Schwarz, we have
$$ x-y = (1)(x+1) + (-1)(y-1) -2 \le \sqrt{2}\sqrt{(x+1)^2+(y-1)^2} - 2 $$
and hence $\frac{x-y}{\sqrt{2}} < \sqrt{(x+1)^2+(y-1)^2}$ for all $x$ and $y$.
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