Question #117380
given that w denotes either one of the non real roots of the equation z³=1. show that a. 1+w+w²=0 b. the other non real rootvis w². show that the non real root of the equation ((1-u)÷u)² can be expeessed in the form Aw and Bw², where A and B are real numbers and A and B
1
Expert's answer
2020-05-21T17:34:28-0400
z3=1z^3=1z31=0z^3-1=0(z1)(z2+z+1)=0(z-1)(z^2+z+1)=0

z2+z+1=0z^2+ z+1=0


z=1±124(1)22=1±i32z={-1\pm\sqrt{1^2-4(1)^2}\over2}={-1\pm i\sqrt{3}\over2}

Cube Root of Unity Value


w1=1, realw_1=1,\ real

w2=1i32,complexw_2={-1- i\sqrt{3}\over2}, complex

w3=1+i32, complexw_3={-1+ i\sqrt{3}\over2}, \ complex

b)


z3u3=0z^3-u^3=0

z3u3=(zu)(z2+uz+u2)z^3-u^3=(z-u)(z^2+u z+u^2)

Then

(zu)(z2+uz+u2)=0(z-u)(z^2+u z+u^2)=0

z2+uz+u2=0z^2+u z+u^2=0


z=u±u24u22=u(1±i32)z={-u\pm\sqrt{u^2-4u^2}\over2}=u({-1\pm i\sqrt{3}\over2})

z1=uw1=u1=1u+i0, realz_1=u w_1=u\cdot1=1\cdot u+i\cdot0, \ real

z2=uw2=u1i32=12uiu32, complexz_2=u w_2=u\cdot{-1- i\sqrt{3}\over2}=-{1\over 2}u-i\cdot{u\sqrt{3}\over 2},\ complex

z3=uw3=u1+i32=12u+iu32, complexz_3=u w_3=u\cdot{-1+ i\sqrt{3}\over2}=-{1\over 2}u+i\cdot{u\sqrt{3}\over 2},\ complex


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