A) \[x+y-1=0\]
B) \[x-y-1=0\]
C) \[x+y+1=0\]
D) \[x-y+1=0\]
Correct Answer: D
Solution :
\[z-2-3i=x+iy-2-3i=(x-2)+i(y-3)\] \[{{\tan }^{-1}}\left( \frac{y-3}{x-2} \right)=\frac{\pi }{4}\Rightarrow \,\frac{y-3}{x-2}=\tan \frac{\pi }{4}=1\] Þ \[x-y+1=0.\]
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