Is cube root of unity is a complex number I know the sum is 0 and product is -1 but I am somewhat confused please give me some idea. Thanks in advance

  • $\begingroup$ Thanks for your question. Can you expand your question with some details? You said "I know the sum is $0$ and the product is $-1$." The sum of what is $0$? The product of what is $-1$? $\endgroup$
    – layman
    Jun 19 '16 at 17:26
  • $\begingroup$ What does this have to do with linear algebra or mathematicians? This is a very bad stated question. Please read the following article and restate your question: math.stackexchange.com/help/how-to-ask $\endgroup$
    – Noam
    Jun 19 '16 at 18:02
  • $\begingroup$ Basically I mean to say cube root of unity 1,w,w^2 $\endgroup$ Jun 19 '16 at 18:13
  • $\begingroup$ @noam I m really sorry but I couldn't understand in which category to post this type of question so I think the mathematician are great who always know every solution that's why I posted.. $\endgroup$ Jun 19 '16 at 18:17

There are 3 distinct complex numbers $z_1,z_2,z_3$ that satisfy $z^3=1$. In other words, $z_1,z_2$, and $z_3$ are zeros of the polynomial $z^3-1$. So we infer that $$ z^3-1=(z-z_1)(z-z_2)(z-z_3)=z^3-(z_1+z_2+z_3)z^2+z(z_1z_2+z_1z_3+z_2z_3)-z_1z_2z_3. $$ From this, you can see that $\sum_jz_j=0$ and $\prod_jz_j=1$ (and, as a bonus, $\sum_{i<j}z_iz_j=0$). Of course, you can also get these results directly if you know the exact expressions for $z_1,z_2,z_3$: $$ \{z_1,z_2,z_3\}=\Big\{1,-\frac{1}{2}+\frac{\sqrt{3}}{2}i,-\frac{1}{2}-\frac{\sqrt{3}}{2}i\Big\}. $$


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