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2000 SMT/Advanced Topics Problems/Problem 2

Revision as of 08:23, 24 July 2020 by Orca25 (talk | contribs) (Created page with "==Problem== Simplify <math>\left(\frac{-1+i\sqrt{3}}{2}\right)^6+\left(\frac{-1-i\sqrt{3}}{2}\right)^6</math> to the form <math>a+bi</math>. ==SMT Solution== Since <math>...")
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Problem

Simplify $\left(\frac{-1+i\sqrt{3}}{2}\right)^6+\left(\frac{-1-i\sqrt{3}}{2}\right)^6$ to the form $a+bi$.


SMT Solution

Since $\cos\frac{2\pi}{3}=-\frac{1}{2}$ and $\sin\frac{2\pi}{3}=\frac{\sqrt{3}}{2},$ we can write the first term as $\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)^6$. Since $\cos\frac{4\pi}{3}=-\frac{1}{2}$ and $\sin\frac{4\pi}{3}=-\frac{\sqrt{3}}{2},$ we can write the second term as $\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)^6.$ Now, we apply DeMoivre's Theorem to simplify the first expression to $\left(\cos6\cdot\frac{2\pi}{3}+\sin6\cdot\frac{2\pi}{3}\right)=\left(\cos4\pi+\sin4\pi\right)=1+0=1.$ Similarly, we simplify the second expression to $\left(\cos6\cdot\frac{4\pi}{3}+\sin6\cdot\frac{4\pi}{3}\right)=\left(\cos8\pi+\sin8\pi\right)=1+0=1.$ Thus, the total sum is $1+1=\mathbf{2}.$



Credit

Problem and solution were taken from https://sumo.stanford.edu/old/smt/2000/

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