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If the amplitude of $z-2-3 i$ is $\pi / 4$, then the locus of $z=x+i y$ is
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Verified Answer
The correct answer is:
$x-y+1=0$
$\begin{aligned}
& \text { Given, } \arg (z-2-3 i)=\frac{\pi}{4} \Rightarrow z=x+i y \\
& z-2-3 i=x+i y-2-3 i=(x-2)+(y-3) i \\
& \arg (z-2-3 i)=\tan ^{-1}\left(\frac{y-3}{x-2}\right)=\frac{\pi}{4} \Rightarrow \frac{y-3}{x-2}=\tan \frac{\pi}{4} \\
& \quad y-3=x-2 \Rightarrow x-y+1=0
\end{aligned}$
$\therefore$ Locus of $z$ is $x-y+1=0$.
& \text { Given, } \arg (z-2-3 i)=\frac{\pi}{4} \Rightarrow z=x+i y \\
& z-2-3 i=x+i y-2-3 i=(x-2)+(y-3) i \\
& \arg (z-2-3 i)=\tan ^{-1}\left(\frac{y-3}{x-2}\right)=\frac{\pi}{4} \Rightarrow \frac{y-3}{x-2}=\tan \frac{\pi}{4} \\
& \quad y-3=x-2 \Rightarrow x-y+1=0
\end{aligned}$
$\therefore$ Locus of $z$ is $x-y+1=0$.
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