If z is a complex number satisfying the condition $|\mathrm{z}-2-\mathrm{i}|=\left|\mathrm{z}\right|\sin \left(\frac{{\displaystyle \mathrm{\pi }}}{{\displaystyle 4}}-\arg \mathrm{z}\right)$ , then locus of z is

Let $\mathrm{Z}=\mathrm{x}+\mathrm{iy}$
Polar form of a complex number $\mathrm{Z}=\mathrm{r}(\cos \mathrm{\theta }+\mathrm{isin}\mathrm{\theta })$
where $\mathrm{r}=\left|\mathrm{Z}\right|\text{ and }\mathrm{\theta }=\arg \left(\mathrm{z}\right)$
$\Rightarrow \mathrm{x}+\mathrm{iy}=\mathrm{r}(\cos \mathrm{\theta }+\mathrm{isin}\mathrm{\theta })\Rightarrow \mathrm{x}=\mathrm{rcos}\mathrm{\theta },\mathrm{y}=\mathrm{rsin}\mathrm{\theta }$
Given $|\mathrm{z}-2-\mathrm{i}|=\left|\mathrm{z}\right|\sin \left(\frac{\mathrm{\pi }}{4}-\arg \mathrm{z}\right)$
$\begin{array}{l}\Rightarrow |\mathrm{x}+\mathrm{iy}-2-\mathrm{i}|=\mathrm{rsin}\left(\frac{\mathrm{\pi }}{4}-\mathrm{\theta }\right)\\ \Rightarrow \left|\right(\mathrm{x}-2)+\mathrm{i}(\mathrm{y}-1\left)\right|=\mathrm{r}\left(\frac{1}{\sqrt{2}}\cos \mathrm{\theta }-\frac{1}{\sqrt{2}}\sin \mathrm{\theta }\right)\\ \Rightarrow \sqrt{(\mathrm{x}-2{)}^2+(\mathrm{y}-1{)}^2}=\frac{1}{\sqrt{2}}\mathrm{rcos}\mathrm{\theta }-\frac{1}{\sqrt{2}}\mathrm{rsin}\mathrm{\theta }\\ \Rightarrow \sqrt{(\mathrm{x}-2{)}^2+(\mathrm{y}-1{)}^2}=\frac{1}{\sqrt{2}}(\mathrm{x}-\mathrm{y})\end{array}$
$\Rightarrow$parabola with focus (2,1)  directrix  x - y = 0
( Locus of a moving point ‘P’ in a plain which such that the distance from the point ‘P’ to fixed point is equal to distance from the point ‘P’ to fixed line is parabola.)