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True. For three terms $1+i-i=1$, all of which are $4^{th}$-root of $1$. For two terms you can also write $-\frac{1+\sqrt{3}i}{2}-\frac{1-\sqrt{3}i}{2}=-1$, all of which are $6^{th}$-root of $1$. And so on and so on…

Edit: Now that you edited the question and $p$ becomes prime, there is a more general answer. As Douglas pointed out the key word here is cyclotomic polynomials $P_p(z)=\sum_{n=0}^{p-1}z^n$, whose roots are precisely the $p^{th}$-root of unity except $1$. In that case

$1+P(z_j)=1$

gives you such a relation as long as $z_j\neq 1$.

Ah, but now you'd perhaps like that the indices on the RHS and LHS do not overlap (although it is not stated)! Well, you can try to work out other relations by yourself, but the answer to your question is definitly «true».

True. For three terms $1+i-i=1$, all of which are $4^{th}$-root of $1$. For two terms you can also write $-\frac{1+\sqrt{3}i}{2}-\frac{1-\sqrt{3}i}{2}=-1$, all of which are $6^{th}$-root of $1$. And so on and so on…

Edit: Now that you edited the question and $p$ becomes prime, there is a more general answer. As Douglas pointed out the key word here is cyclotomic polynomials $P_p(z)=\sum_{n=0}^{p-1}z^n$, whose roots are precisely the $p^{th}$-root of unity except $1$. In that case

$1+P(z_j)=1$

gives you such a relation as long as $z_j\neq 1$.

True. For three terms $1+i-i=1$, all of which are $4^{th}$-root of $1$. For two terms you can also write $-\frac{1+\sqrt{3}i}{2}-\frac{1-\sqrt{3}i}{2}=-1$, all of which are $6^{th}$-root of $1$. And so on and so on…

True. For three terms $1+i-i=1$, all of which are $4^{th}$-root of $1$. For two terms you can also write $-\frac{1+\sqrt{3}i}{2}-\frac{1-\sqrt{3}i}{2}=-1$, all of which are $6^{th}$-root of $1$. And so on and so on…

Edit: Now that you edited the question and $p$ becomes prime, there is a more general answer. As Douglas pointed out the key word here is cyclotomic polynomials $P_p(z)=\sum_{n=0}^{p-1}z^n$, whose roots are precisely the $p^{th}$-root of unity except $1$. In that case

$1+P(z_j)=1$

gives you such a relation as long as $z_j\neq 1$.

Ah, but now you'd perhaps like that the indices on the RHS and LHS do not overlap (although it is not stated)! Well, you can try to work out other relations by yourself, but the answer to your question is definitly «true».

True. For three terms $1+i-i=1$, all of which are $4^{th}$-root of $1$. For two terms you can also write $-\frac{1+\sqrt{3}}{2}-\frac{1-\sqrt{3}}{2}=-1$, all of which are $6^{th}$-root of $1$. And so on and so on…

True. For three terms $1+i-i=1$, all of which are $4^{th}$-root of $1$. For two terms you can also write $-\frac{1+\sqrt{3}i}{2}-\frac{1-\sqrt{3}i}{2}=-1$, all of which are $6^{th}$-root of $1$. And so on and so on…