Question

If w is a complex cube root of unity then show that (1-w) (1-w2 ) (1-w4) ( 1-w5) =9​

Answer 1

Step-by-step explanation:

$(1 - w)(1 - w ^{2} )(1 - w ^{4} )(1 - w ^{5} ) \\ \\ = (1 - {w}^{2} - w + {w}^{3} )(1 - {w}^{5} - {w}^{4} + {w}^{9} ) \\ \\ = (1 - {w}^{2} - w + 1)(1 - {w}^{2} - w + 1) \: \: \: \: \: ( {w}^{3} = 1. \: \: \: {w}^{9} = ({w}^{3}) ^{3} = (1) ^{3} = 1 )\\ \\= (2 - {w}^{2} - w)(2 - w - {w}^{2} ) \\ \\ = (2 -( {w}^{2} + w))(2 - (w + {w}^{2} )) \\ \\ =( 2 - ( - 1))(2 - ( - 1)) \\ \\ = (2 + 1)(2 + 1) \\ \\ = 3 \times 3 \\ \\ = 9 \: \: \: (proved) \\ \\ \\ formula \: \: \: - \: \: {w}^{3} = 1 \\ \\ 1 + w + {w}^{2} = 0 \\ = > w + {w}^{2} = - 1$