Problem:
Let be real numbers. Suppose that all the roots of are complex numbers lying on a circle in the complex plane centered at and having radius . The sum of the reciprocals of the roots is necessarily
Answer Choices:
A.
B.
C.
D.
E.
Solution:
If is a complex root of a polynomial equation with real coefficients, then so is . (In fact, if is a multiple root, so is , with the same multiplicity. This is important below because in any sum involving roots it is meant that each root should be counted once for each multiplicity.) Now if lies on the unit circle centered at , then the reciprocal of is . Hence the reciprocal of each root is again a root. Thus the sum of the reciprocals of the roots is the same as the sum of the roots. In general, the sum of the roots of is , so in our case the sum is simply .
Note. is also a correct expression for the sum, but it wasn't listed. To see that it is correct, first let
Next observe that for
Thus is a nonzero root of iff is a root of . Furthermore, all the roots of are nonzero, and hence their reciprocals exist, iff . (Why?) Thus, whenever all the roots of are nonzero (not just when they are on the unit circle), the sum of their reciprocals equals the sum of the roots of . By the general fact stated in the last sentence of the solution, this latter sum is .
When the roots are on the unit circle, we can say more: . (Why?) Thus one of and is also correct, but neither alone is always correct. [Exercises: with all roots on the unit circle, find a specific with and another with . For any such show that the coefficients are either symmetric ( ) or antisymmetric ( ). Finally, extend everything in this Note to general degree polynomials.]