Canonical URI: https://w3id.org/kmath/textbook_solution/chunjae_hong_solution_p151_p092_performance_03
| rdf:type | math:TextbookSolution |
|---|---|
| rdfs:label | 천재홍 p092 수행 과제 3 풀이 |
| rdfs:comment | 천재홍 공통수학1 정답 및 풀이 p151의 p092 수행 과제 3 풀이. |
| math:answerText | (1) \(-1\) (2) \(-1\) (3) \(1\) |
| math:explanationText | (1) \(\omega^3=1,\ \omega^2+\omega+1=0\)임을 이용하면 \(\omega^{10}+\omega^5=(\omega^3)^3\omega+\omega^3\omega^2 =\omega+\omega^2=-1\). (2) \(\omega^3=\overline{\omega}^{\,3}=1,\ \omega^2+\omega+1= \overline{\omega}^{\,2}+\overline{\omega}+1=0\), \(\omega+\overline{\omega}=-1,\ \omega\overline{\omega}=1\)임을 이용하면 \(\omega^{20}+\frac1{\omega^{20}}=\omega^{20}+\overline{\omega}^{\,20} =(\omega^3)^6\omega^2+(\overline{\omega}^{\,3})^6\overline{\omega}^{\,2} =\omega^2+\overline{\omega}^{\,2}=\omega^2+\omega=-1\). (3) \(\omega^3=1,\ \omega^2+\omega+1=0\)임을 이용하면 \(\omega^{30}+\omega^{29}+\omega^{28}+\cdots+\omega+1 =(\omega^3)^{10}+(\omega^3)^9\omega^2+(\omega^3)^9\omega+\cdots+\omega+1 =1+(\omega^2+\omega+1)+(\omega^2+\omega+1)+\cdots+(\omega^2+\omega+1) =1+0+0+\cdots+0=1\). |
| math:mappingConfidence | 1.0 |
| math:pageStart | 151 |
| math:problem | textbook_problem:chunjae_hong_vision_p092_performance_03 |
| math:reviewStatus | reviewed |
| math:solutionKind | worked_solution |
| math:usesSolutionPattern | solution_pattern:complex_number_algebra |