InterviewSolution
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InterviewSolution
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If `n in N >1`, then the sum of real part of roots of`z^n=(z+1)^n`is equal to`n/2`b. `((n-1))/2`c. ` n/2`d. `((1-n))/2`A. `(n)/(2)`B. `((n-1))/(2)`C. `-(n)/(2)`D. `((1-n))/(2)` |
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Answer» Correct Answer - D The equations `z^(n) = (z+ 1)^(n)` will have excactly n - 1 roots. We have `((z+1)/(z^(n)))= 1 or |(z+1)/(z)|= 1` or `|z+1|= |z|` Therefore,z lies on the the right bisector of the segment connecting the points (0,0) and (-1,0). Thus `Re(z) = - 1//2`. Hence roots are colliner and will have their real parts of roots is `(-1//2)(n-1)` |
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