InterviewSolution
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InterviewSolution
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Find all possible values of `sqrt(i)+sqrt(-i)dot` |
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Answer» `sqrt(i) + sqrt(-i) = sqrt(0+1.i)+sqrt(0-1.i)` Now`sqrt(a+i.b) = {sqrt((1)/(2){sqrt(a^(2)+b^(2))+a})+ isqrt((1)/(2){sqrt(a^(2)+b^(2))-a})}` `and sqrt(a-i.b) =om{sqrt((1)/(2){sqrt(a^(2)+b^(a))+a}-i)sqrt((1)/(2){sqrt(a^(2)+b^(2))-a})}` `rArr sqrt(0+1.i) = pm{sqrt((1)/(2){sqrt(0 +1^(2)) +0})+isqrt((1)/(2){sqrt(0+1^(2)) -0})}` `= pm (1)/(sqrt(2)) (1-i)` ` and sqrt(0-1.i)= pm {sqrt((1)/(2) {sqrt(0+1^(2))+0})-isqrt((1)/(2){sqrt(0+1^(2))-0})}` ` = pm (1)/(sqrt(2)) (1-i)` Now ` sqrt(i) + sqrt(-i) = pm (1)/(sqrt(2)) (1+i) pm (1)/(sqrt(2))(1-i)` `or sqrt(i) + sqrt(-i) =pmsqrt(2) + 0.ior 0pm sqrt(2i)` |
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