If amplitude of $(z-2-3 i)$ is $\frac{3 \pi}{4}$, then locus of $z$ is (where $\mathrm{z}=\mathrm{x}+$ iy)
If amplitude of $(z-2-3 i)$ is $\frac{3 \pi}{4}$, then locus of $z$ is (where $\mathrm{z}=\mathrm{x}+$ iy)
$x+y=1$
$x+y=5$
$x-y=-5$
$x-y=1$
Solution
Amplitude of $(z-2-3 i)$ is $\frac{3 \pi}{4}$ and we have $z=x+$ iy
$\therefore \operatorname{Amp}[(\mathrm{x}-2)+\mathrm{i}(\mathrm{y}-3)]$ is $\frac{3 \pi}{4}$
$\therefore \tan ^{-1}\left(\frac{\mathrm{y}-3}{\mathrm{x}-2}\right)=\frac{3 \pi}{4} \Rightarrow \tan \left(\frac{3 \pi}{4}\right)=\frac{\mathrm{y}-3}{\mathrm{x}-2}$
$\therefore-1=\frac{y-3}{x-2} \Rightarrow-x+2=y-3 \Rightarrow x+y=5$