香港介紹ppt

標題:

Complex Numbers

發問:

求救!!! 有無人學過Complex Numbers? 小女有幾條題目唔識做 1.) Use de Moivre's theorem to obtain solutions to the equation z^4 -1 = 0 and z^5 -1 =0 Comment on your results and try to formulate a conjecture. 2.) Factorize z^n -1 for n = 3, 4 and 5 謝謝謝謝謝謝謝謝!!!!!!

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1) z^4 - 1 = 0 ==> z^4 = 1 ==> z^4 = cos 2kπ + i sin 2kπ . . . . (where k = 0, 1, 2, 3) ==> z = cos (kπ/2) + i sin (kπ/2) z^5 - 1 = 0 ==> z^5 = 1 ==> z^5 = cos 2kπ + i sin 2kπ . . . . (where k = 0, 1, 2, 3, 4) ==> z = cos (2kπ/5) + i sin (2kπ/5) z^n - 1 = 0 ==> z = cos (2kπ/n) + i sin (2kπ/n) . . . . (where k = 0, 1, 2, ... (n-1)) 2) z^3 - 1 = (z - 1)[z - cos (2π/3) - i sin (2π/3)][z - cos (4π/3) - i sin (4π/3)] = (z - 1)[z + 1/2 - i (√3)/2][z + 1/2 + i (√3)/2] [or (z - 1)(z^2 + z + 1) with real factors] z^4 - 1 = (z - 1)[z - cos (2π/4) - i sin (2π/4)][z - cos (4π/4) - i sin (4π/4)][z - cos (6π/4) - i sin (6π/4)] = (z - 1)(z - i)(z + 1)(z + i) [or (z - 1)(z + 1)(z^2 + 1) with real factors] z^5 - 1 = (z - 1)[z - cos (2π/5) - i sin (2π/5)][z - cos (4π/5) - i sin (4π/5)][z - cos (6π/5) - i sin (6π/5)][z - cos (8π/5) - i sin (8π/5)] = (z - 1)[z - cos (2π/5) - i sin (2π/5)][z - cos (4π/5) - i sin (4π/5)][z - cos (4π/5) + i sin (4π/5)][z - cos (2π/5) + i sin (2π/5)] {or (z - 1)[z^2 - 2z cos (2π/5) + 1][z^2 - 2z cos (4π/5) + 1] with real factors}

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