A) a circle with centre \[(0,\,\,0)\] and radius 1
B) a circle with centre and radius 1
C) a circle with centre (0, 1) and radius 1
D) a straight line
Correct Answer: A
Solution :
Given, \[i{{z}^{3}}-{{z}^{2}}+1z+i=0\] \[\therefore \] \[i{{z}^{3}}+{{i}^{2}}{{z}^{2}}+z+i=0\] \[\Rightarrow \] \[i{{z}^{2}}(z+i)+(z+i)=0\] \[\Rightarrow \] \[(i{{z}^{2}}+1)\,(z+i)=0\] \[\Rightarrow \] \[z=-i\] or \[{{z}^{2}}=-\frac{1}{i}=\frac{{{i}^{2}}}{i}=i\] \[\Rightarrow \] Now, \[{{z}^{2}}=i\] On taking mod both sides, we get \[|z{{|}^{2}}=|i|\] \[\therefore \] \[{{x}^{2}}+{{y}^{2}}=1\] Hence, centre and radius of circle are \[(0,\,0)\] and 1.
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