A) centre at \[(-3/2,1/2)\] and radius \[\frac{1}{\sqrt{2}}\]
B) centre at \[(3/2,-1/2)\] and radius \[\frac{1}{\sqrt{2}}\]
C) centre at \[(3/2,-1/2)\] and radius \[\sqrt{2}\]
D) centre at \[(-3/2,1/2)\]and radius \[\sqrt{2}\]
Correct Answer: A
Solution :
[a] We have \[|z-(-1+i)|=1\] \[\Rightarrow \,\,\,\,\,|z+1-i|=1\] Now, \[\omega =\frac{z+i}{1-i}\] \[\Rightarrow \,\,\,\,\,\,(1-i)\omega =z+i\] \[\Rightarrow \,\,\,\,\,\,(1-i)\omega -2i+1=z+1-i\] \[\Rightarrow \,\,\,\,\,\,|(1-i)\omega -2i+1|\,\,=\,\,|z+1-i|\] \[\Rightarrow \,\,\,|1-i|\,\,\left| \omega +\frac{1-2i}{1-i} \right|=1\Rightarrow \left| \omega +\frac{(1-2i)(1+i)}{(1+i)(1-i)} \right|=\frac{1}{\sqrt{2}}\]\[\Rightarrow \,\,\,\,\left| \omega -\frac{-3+i}{2} \right|=\frac{1}{\sqrt{2}}\] Therefore, locus of \[\omega \]is a circle having centre at \[(-3/2,\,1/2)\] and radius \[\frac{1}{\sqrt{2}}\].
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