**Problem**: (1999-2000 Berkeley Math Circle Contest, Gabriel Carroll Original) Let $ z = \cos{\frac{2\pi}{n}}+i \sin{\frac{2\pi}{n}} $ where $ n $ is a positive integer. Prove that

$ \frac{1}{1+z}+\frac{1}{1+z^2}+\cdots+\frac{1}{1+z^n} = \frac{n}{2} $.

**Solution**: It should be clear that $ z $ is one of the $ n $th roots of unity. Then by DeMoivre's, we find that $ z, z^2, \ldots, z^n $ are exactly the $ n $th roots of unity. So we may write $ z^k $ as $ \cos{\frac{2\pi k}{n}}+i \sin{\frac{2\pi k}{n}} $.

Consider the sum $ \frac{1}{1+z^k}+\frac{1}{1+z^{n-k}} $. We have

$ \frac{1}{\left(1+\cos{\frac{2\pi k}{n}}\right)+i \sin{\frac{2\pi k}{n}}} + \frac{1}{\left(1+\cos{\frac{2\pi (n-k)}{n}}\right)+i \sin{\frac{2\pi (n-k)}{n}}} $.

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

$ \frac{\left(1+\cos{\frac{2\pi k}{n}}\right)-i \sin{\frac{2\pi k}{n}}}{2\left(1+\cos{\frac{2\pi k}{n}}\right)} + \frac{\left(1+\cos{\frac{2\pi (n-k)}{n}}\right)-i \sin{\frac{2\pi (n-k)}{n}}}{2\left(1+\cos{\frac{2\pi (n-k)}{n}}\right)} $.

Note that $ \cos{\frac{2\pi k}{n}} = \cos{\frac{2\pi (n-k)}{n}} $ and $ \sin{\frac{2\pi k}{n}} = -\sin{\frac{2\pi (n-k)}{n}} $, and it simplifies nicely to (after applying the Pythagorean identity)

$ \frac{2\left(1+\cos{\frac{2\pi k}{n}}\right)}{2\left(1+\cos{\frac{2\pi k}{n}}\right)} = 1 $.

Since $ n $ is odd, there are $ \frac{n-1}{2} $ pairs that sum to $ 1 $ and the only term left is $ \frac{1}{1+z^n} = \frac{1}{1+1} = \frac{1}{2} $, so the sum is $ \frac{n-1}{2}+\frac{1}{2} = \frac{n}{2} $ 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 $ z^k $ and $ z^{n-k} $, giving the approach above.

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Practice Problem: (2004 AIME1 - #13) The polynomial $P(x)=(1+x+x^2+\cdots+x^{17})^2-x^{17}$ has $34$ complex roots of the form $z_k=r_k[\cos(2\pi a_k)+i\sin(2\pi a_k)]$, $k=1, 2, 3,\ldots, 34$, with $0

1/(1 + z^k) + 1/(1 + z^{n - k})

ReplyDelete= (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.

Nice. I was thinking there might be a better solution... I need practice with complex numbers!

ReplyDelete