Consider a quadratic equation with complex coefficients and roots , :
By comparing coefficients, we can see that
Generalizing this idea for a polynomial of degree , we have
Vieta's Formula. Let be a polynomial with complex coefficients and degree , having complex roots . Then for any integer :
This follows immediately by comparing coefficients on the polynomial . We leave the proof to the reader.
1. Find all triples of complex numbers which satisfy the system of equations:
Solution: Any 3 numbers can be considered as the roots of the monic cubic polynomial . Applying Vieta's Formula to the polynomial, we get . Hence, the only root of is (with multiplicity 3), which implies that is the only triple of complex numbers that satisfies the system of equations.
2. Let and be the roots of the polynomial . Evaluate
Solution: The expression is equal to . By Vieta's Formula, we know that
Note: A common approach would be to try and find each root of the polynomial, especially since we know that one of the roots must be real (why?). However, this is not necessarily a viable option, since it is hard for us to determine what the roots actually are.
3. Let and be the roots of the polynomial . Evaluate
Solution: In this problem, the application of Vieta's Formulas is not immediately obvious, and the expression has to be transformed. From Factorization 4, we have that
Now we only need to know how to calculate . Again, from factorization, we have that
so this allows us to conclude that
4. [IMO shortlist] Determine all real values of the parameter for which the equation
has exactly four distinct real roots that form a geometric progression.
Solution: Suppose that satisfies the requirements of the problem and that , , , are the roots of the given equation. Then and we may assume that , so that . Notice that the coefficients are symmetric,
namely the first coefficient is the same as the fifth one, the second is the same as the fourth and the third is the same as the third. It guarantees us that if, is a root, then its reciprocal (which is ) will also be a root. Hence, so and the roots are .
Now, by Vieta's formula we have and . On setting these equations become
Substituting in the second equation leads to . We observe that this polynomial factors as . Since , the only viable value is . Consequently
Rearranging the equation, we get . To simplify it, we can call and therefore , thus getting a new form , whose roots are and . We have to plug the two back into , leading us to two more quadratics, getting finally .
If are the roots to the equation , what is the quadratic equation whose roots are and ?
ARML 2012, Team Problems, #6 The zeros of are distinct complex numbers. Compute
the average value of over all possible permutations of these six numbers.
(Created by Mark Hennings. This is HARD) Show that for any integer . What is