1984 AIME Problems/Problem 8
The equation has complex roots with argument between and in the complex plane. Determine the degree measure of .
We shall introduce another factor to make the equation easier to solve. If is a root of , then . The polynomial has all of its roots with absolute value and argument of the form for integer (the ninth degree roots of unity). Now we simply need to find the root within the desired range that satisfies our original equation .
This reduces to either or . But can't be because if , then . (When we multiplied by at the beginning, we introduced some extraneous solutions, and the solution with was one of them.) This leaves .
The substitution simplifies the equation to . Applying the quadratic formula gives roots , which have arguments of and respectively. We can write them as and . So we can use De Moivre's theorem (which I would suggest looking at if you never heard of it before) to find the fractional roots of the expressions above! For we have and Similarly for , we have and The only argument out of all these roots that fits the description is
Note: We can add to the angles of the previous solutions to get new solutions because De Moivre's formula says that and . ~programmeruser
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