Complex Plane Delivery
During the weekends, Eli delivers milk in the complex plane. On Saturday, he begins at $ z $ and delivers milk to houses located at $ z^3,z^5,z^7,\ldots,z^{2013} $ in that order; on Sunday, he begins at $ 1 $ and delivers milk to houses located at $ z^2,z^4,z^6,\ldots,z^{2012} $ in that order. Eli always walks directly (in a straight line) between two houses. If the distance he must travel from his starting point to the last house is $ \sqrt{2012} $ on both days, find the real part of $ z^2 $.
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- $\frac{a}{b}$
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- $\infty$
Solution
The sum of the Saturday distances is
\[|z^3 - z| + |z^5 - z^3| + \dots + |z^{2013} - z^{2011}| = \sqrt{2012}.\]The sum of the Sunday distances is
\[|z^2 - 1| + |z^4 - z^2| + \dots + |z^{2012} - z^{2010}| = \sqrt{2012}.\]Note that
\[|z^3 - z| + |z^5 - z^3| + \dots + |z^{2013} - z^{2011}| = |z| (|z^2 - 1| + |z^4 - z^2| + \dots + |z^{2012} - z^{2010}|),\]so $ |z| = 1 $.
Then
\begin{align*}
|z^2 - 1| + |z^4 - z^2| + \dots + |z^{2012} - z^{2010}| &= |z^2 - 1| + |z^2| |z^2 - 1| + \dots + |z^{2010}| |z^2 - 1| \\
&= |z^2 - 1| + |z|^2 |z^2 - 1| + \dots + |z|^{2010} |z^2 - 1| \\
&= 1006 |z^2 - 1|,
\end{align*}so
\[|z^2 - 1| = \frac{\sqrt{2012}}{1006}.\]We have that $ |z^2| = |z|^2 = 1 $. Let $ z^2 = a + bi, $ where $ a $ and $ b $ are real numbers, so $ a^2 + b^2 = 1 $. From the equation $ |z^2 - 1| = \frac{\sqrt{2012}}{1006}, $
\[(a - 1)^2 + b^2 = \frac{2012}{1006^2} = \frac{1}{503}.\]Subtracting these equations, we get
\[2a - 1 = 1 - \frac{1}{503} = \frac{502}{503},\]so $ a = \boxed{\frac{1005}{1006}} $.
Freeflows