# Mathematical Food for Thought

Serves a Daily Special and an All-You-Can-Eat Course in Problem Solving. Courtesy of me, Jeffrey Wang.

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Zzz… Topic: Complex Numbers/Trigonometry. Level: AIME/Olympiad. February 21st, 2006

Problem: (1999-2000 Berkeley Math Circle Contest, Gabriel Carroll Original) Let where is a positive integer. Prove that

.

Solution: It should be clear that is one of the th roots of unity. Then by DeMoivre’s, we find that are exactly the th roots of unity. So we may write as .

Consider the sum . We have

.

This is ugly, but by multiplying the denominators by their complex conjugates, it comes out to

.

Note that and , and it simplifies nicely to (after applying the Pythagorean identity)

.

Since is odd, there are pairs that sum to and the only term left is , so the sum is as desired. QED.

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Comment: This is quite a bit of algebra to work through; the way I first saw it was geometrically. The reciprocal of a complex number is equivalent to the complex conjugate divided by the norm, so clearly adding the reciprocals of complex conjugates will clear the imaginary part. Since we have the roots of unity, the complex conjugates are and , giving the approach above.

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Practice Problem: (2004 AIME1 – #13) The polynomial has complex roots of the form , , with and . Given that , where and are relatively prime positive integers, find .

### 2 Responses to “Zzz… Topic: Complex Numbers/Trigonometry. Level: AIME/Olympiad.”

1. Anon Says:

1/(1 + z^k) + 1/(1 + z^{n – k})
= (1 + z^k + 1 + z^{n – k})/((1 + z^k)(1 + z^{n – k})
= (2 + z^k + z^{n – k})(2 + z^k + z^{n – k})
= 1.