160/MR00. problem / Math - Solved Math Problems

MR-160. problem

Simplify the following expression:

${(\frac{1 + i \sqrt{3}}{2})}^{3000} + {(\frac{1 - i \sqrt{3}}{2})}^{3000}$

Source: MathReference: Generic Math Textbook - 160.)

Algebra → Complex numbers → Exponentiation of complex numbers

{(\frac{1 + i \sqrt{3}}{2})}^{3000} + {(\frac{1 - i \sqrt{3}}{2})}^{3000}

{(\frac{1 + i \sqrt{3}}{2})}^{3000} + {(\frac{1 - i \sqrt{3}}{2})}^{3000}

= {[{(\frac{1 + i \sqrt{3}}{2})}^{3}]}^{1000} + {[{(\frac{1 - i \sqrt{3}}{2})}^{3}]}^{1000}

$= {[\frac{1^{3} + 3 \cdot 1^{2} \cdot (i \sqrt{3}) + 3 \cdot 1 \cdot {(i \sqrt{3})}^{2} + {(i \sqrt{3})}^{3}}{2^{3}}]}^{1000} + {[\frac{1^{3} - 3 \cdot 1^{2} \cdot (i \sqrt{3}) + 3 \cdot 1 \cdot {(i \sqrt{3})}^{2} - {(i \sqrt{3})}^{3}}{2^{3}}]}^{1000}$

= {[\frac{1 + 3 \sqrt{3} i - 9 - 3 \sqrt{3} i}{8}]}^{1000} + {[\frac{1 - 3 \sqrt{3} i - 9 + 3 \sqrt{3} i}{8}]}^{1000}

= {[\frac{- 8}{8}]}^{1000} + {[\frac{- 8}{8}]}^{1000}

= {[- 1]}^{1000} + {[- 1]}^{1000}

$= 2$

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