A) \[\{(x,y)\in {{R}^{2}}\left| {{x}^{2}}+{{y}^{2}}-3y+2=0\} \right.\]
B) \[\{(x,y)\in {{R}^{2}}\left| {{x}^{2}}+{{y}^{2}}-3x+2=0\} \right.\]
C) \[\{(x,y)\in {{R}^{2}}\left| {{x}^{2}}+{{y}^{2}}+3x+2=0\} \right.\]
D) none of above
Correct Answer: D
Solution :
Consider, \[\frac{z-2}{z-1}\,=\frac{(x-2)\,+iy}{(x-1)\,+iy}\,\times \,\frac{(x-1)-iy}{(x-1)\,-iy}\] So, \[\operatorname{Re}\left( \frac{z-2}{z-1} \right)=0\Rightarrow \,\,\frac{(x-2)\,(x-1)\,+{{y}^{2}}}{{{(x-1)}^{2}}+{{y}^{2}}}\,=0\] \[\Rightarrow \,\,{{x}^{2}}+{{y}^{2}}\,-3x+2=0\] But z= 1 (reject). So locus of P(z)is \[{{x}^{2}}\,+{{y}^{2}}\,-3x+2=0,\] where (1, 0) is excluded.
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