Adv Alegbra HW Solutions 20

Adv Alegbra HW Solutions 20 - w = n r [cos (/ n ) + i sin...

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20 If n = 1, then the equation is precisely the product rule of Calculus: ( fg ) 0 = f 0 g + fg 0 . For the inductive step, we have ( fg ) n + 1 =[ ( fg ) n ] 0 = h n X k = 0 ³ n k ´ f k g n k i 0 = n X k = 0 ³ n k ´h f k g n k i 0 = n X k = 0 ³ n k ´h f k + 1 g n k + f k g n k + 1 i = n X k = 0 ³ n k ´ f k + 1 g n k + n X k = 0 ³ n k ´ f k g n k + 1 . Rewrite this last expression without the sigma notation: ³ n 0 ´ f 0 g n + 1 + ³ n 1 ´ f 1 g n +···+ ³ n k ´ f k g n k + 1 +··· + ³ n 0 ´ f 1 g n +···+ ³ n k 1 ´ f k g n k + 1 ··· . The coef f cient of f k g n k + 1 is thus ( n k 1 ) + ( n k ) = ( n + 1 k ) , by Lemma 1.17, as desired. 1.43 Find i . Solution. By De Moivre s theorem, since i = e i π/ 2 ,wehave i = e i π/ 4 = cos π/ 4 + i sin π/ 4 = 2 2 + i 2 2 . 1.44 (i) If z = r [ cos θ + i sin θ ] , show that
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Unformatted text preview: w = n r [cos (/ n ) + i sin (/ n ) ] is an n th root of z , where r 0. Solution. By De Moivre s theorem, w n = ( n r ) n ( [cos (/ n ) + i sin (/ n ) ] ) n = r [ cos () + i sin () ] . (ii) Show that every n th root of z has the form k w , where is a primitive n th root of unity and k = , 1 , 2 , . . . , n 1....
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