2017 AMC 12B Problems/Problem 12

Revision as of 21:18, 16 February 2017 by Vedadehhc (talk | contribs) (Solution)

Problem 12

What is the sum of the roots of $z^{12}=64$ that have a positive real part?

$\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 4 \qquad \textbf{(C)}\ \sqrt{2}+2\sqrt{3} \qquad \textbf{(D)}\ 2\sqrt{2}+\sqrt{6} \qquad \textbf{(E)}\ (1+\sqrt{3}) + (1+\sqrt{3})i$


The root of any polynomial of the form $z^n = a$ will have all $n$ of it roots will have magnitude $\sqrt[n]{a}$ and be the vertices of a regular $n$-gon in the complex plane (This concept is known as the Roots of Unity). For the equation $z^{12} = 64$, it is easy to see $\pm\sqrt{2}$ and $\pm {i} \sqrt{2}$ as roots. Graphing these in the complex plane, we have four vertices of a regular dodecagon. Since the roots must be equally spaced, besides $\sqrt{2}$, there are four more roots with positive real parts lying in the first and fourth quadrants. We also know that the angle between these roots is $30^{\circ}$. We only have to find the real parts of the roots lying in the first quadrant, because the imaginary parts would cancel out with those from the fourth quadrant. We have two $30-60-90$ triangles (the triangles formed by connecting the origin to the roots, and dropping a perpendicular line from each root to the real-axis), both with hypotenuse $\sqrt{2}$. This means that one has base $\frac{\sqrt{2}}{2}$ and the other has base $\frac{\sqrt{6}}{2}$. Adding these and multiplying by two, we get the sum of the four roots as $\sqrt{2} + \sqrt{6}$. However, we have to add in the original solution of $\sqrt{2}$, so the answer is $\boxed{\textbf{(D)}2\sqrt{2} + \sqrt{6}}$.

Solution by: vedadehhc

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