Question

8. Let w cos(2π/5) + isin(2π/5). Here we describe how to express w in terms of square roots. (a) Show that w is a root of the polynomial 24+2+22+21. Hint: 25-1-(-(24+23+22+2+1) (b) Show that w + is a root of the polynomial u2 + u-1 (c) Show that Ve, where V5 means the positive square root of Hint: Figure out the sign of w by adding the polar forms of w and 1/w. (d) Put β--12vS So in part (c), you showed w+ 1/w-3. Now show that 2 where v4 means the positive square root of 4-B2, ( e) Deduce from (d) that cos(2T/5) - and sin(2π/5) = 4-β2 2

Answer 1

We have (a) w = cos (2 pi/5) + i sin (2 pi/5) we know, [cos theta + i sin theta]^n = (cos n theta + i sin n theta) Therefore w^5 = (cos 2 pi/5 + i sin 2 pi/5)^5 = cos 5(2 pi/5) + i sin 5(2 pi/5) w^5 = cos 2 pi + i sin (2 pi) = 1 w^5 = 1 i.e. w^5 -1 = 0 i.e. w is the 5th root of unity. as w^5 -1 = (w -1) (w^4 + w^3 + w^2 + w + 1) = 0 as w doesn't satisfy the factor z -1 Hence w satisfies z^4 + z^3 + z^2 + z + 1 i.e. w is a root of polynomial z^4 + z^3 + z^2 + z + 1. B) T.S w + 1/w is a root of u^2 + u -1. Put w + 1/w in u^2 + u -1 (w + 1/w)^2 \r\n