InterviewSolution
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InterviewSolution
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If `omega` is a complex cube root of unity, then the equation `|z- omega|^2+|z-omega^2|^2=lambda` will represent a circle, if |
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Answer» `|z-w|^2+|z-w^2|^2=lambda`......(1) `barz xx z=|z|^2` so (1) turns into =>`bar|z-w|xx |z-w| +bar|z-w^2| xx |z-w^2|=lamda` `|barz-barw|xx |z-w| +|barz-bar(w^2)| xx |z-w^2|=lamda` opening and simplifying we get=>`2zbarz-(w+w^2)barz-(barw+bar(w^2))z+w^2+(w^2)^2=lamda` `2|z|^2-(w+w^2)barz-(barw+bar(w^2))z+|w^2|+(|w^2|)^2=lamda` `|w|=1 and 1+w+w^2=0` substituting these we get =>`2|z|^2+barz+z+2=lamda` if `z=x+iotay+barz=x-iotay ` then z+barz=2x substituting values=>`x^2+y^2+x+((2-lamda)/2)=0` centre of circle=`(-1/2,0)` radius=`sqrt(1/4-(2-lamda)/2)=sqrt((lamda)/2-3/4` for circle to exist `radius>=0` `(lamda>=3/2)` so `lamda` belongs to`(3/2,oo)` |
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