InterviewSolution
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InterviewSolution
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Let `s , t , r`be non-zerocomplex numbers and `L`be the setof solutions `z=x+i y (x , y in RR, i=sqrt(-1))`of the equation`s z+t z +r=0`, where ` z =x-i y`. Then,which of the following statement(s) is (are) TRUE?If `L`has exactlyone element, then `|s|!=|t|`(b) If `|s|=|t|`, then `L`hasinfinitely many elements(c) Thenumber of elements in `Lnn{z :|z-1+i|=5}`is at most2(d) If `L`has morethan one element, then `L`hasinfinitely many elementsA. If L has exactly one element, then `|s| ne |t|`B. If `|s| = |t|` then L has infinitely many elementsC. The number of elements in `L nn {z :|z-1+i|=5}` is at most 2D. If L has most than one elements, then L has infinitely many elements. |
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Answer» Correct Answer - A::C::D Given `sz + bartz + z =0 , barz = x -iy" "(i)` Taking conjugate, we get `barz barz + bartz + barz = 0" "…(i)` Adding (i) and (ii), we get `(t + bars) barz + (bart + s)z+(r+barr) = 0` Clearly, this is the equations of a straight line . Now, eliminating `barz` form (i) and (ii) we,get `sbarsz + rbars + rbars -tbartz - barrt =0` `rArr (|s|^(2) - |t|^(2))z = barrt - rbars` If `|S| ne|t|` then L has unique solution. If `|s| = |t|` and `barrt = rbars`, then L has infinite solutions. If `|s| = |t|` and `barrt ne rbars`, then L has no solutions. Thus, L has either unique, infinite or no solution. Also, line can intersect circle in maximum two points. So, number of elements in L `nn{z:|z-1+i|=5}` is at most 2 |
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