A) Line \[y=x\]
B) Line \[y=-x\]
C) Circle \[{{x}^{2}}+{{y}^{2}}=1\]
D) Line \[x=0\] or on a line segment joining \[(-1,0)\] to \[(1,0)\]
Correct Answer: D
Solution :
[d] \[|z-|z+1|{{|}^{2}}=\,\,|z+|z-1|{{|}^{2}}\] \[\therefore \,\,\,\,(z-|z+1|)(\bar{z}-|z+1|)=(z+|z-1|)(\bar{z}+|z-1|)\]\[z\bar{z}-z|z+1|-\bar{z}|z+1|+|z+1{{|}^{2}}\] \[=z\bar{z}+z|z-1|+\bar{z}|z-1|+|z-1{{|}^{2}}\] \[|z+1{{|}^{2}}-|z-1{{|}^{2}}=(z+\bar{z})\,\left[ |z-1|+|z+1| \right]\] \[(z+1)(\bar{z}+1)-(z-1)(\bar{z}-1)=(z+\bar{z})\left[ |z-1|+|z+1| \right]\]\[\left( z\bar{z}+z+\bar{z}+1 \right)-\left( z\bar{z}-z-\bar{z}+1 \right)=\left( z+\bar{z} \right)\,\,\left[ |z-1|+|z+ \right.\]\[2\left( z+\bar{z} \right)\,=\,\left( z+\bar{z} \right)\,\,\left[ |z+1|+|z-1| \right]\] \[\left( z+\bar{z} \right)\,\,\left[ |z+1|+|z-1|-2 \right]=0\] Now, either \[z+\bar{z}=0\]which means that z is purely imaginary. So, z lies on y axis and thus,\[x=0\]. or \[|z+1|+|z-1|=2\] which means that z lies on the line segment joining \[(-1,0)\] and \[(1,0)\].
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