<s>[INST] For $z \in \mathbb{C}$, if $(z^a,z^c,z^b \bar{z}^d,z^{b+d})=(1,1,1,1)$, then $z=1$ under certain gcd conditions for $a,b,c,d$. 
Let $z \in \mathbb{C}$. I would like to show that if $(z^a,z^c,z^b \bar{z}^d,z^{b+d})=(1,1,1,1)$ and gcd$(a,b,c,d)=1$ and gcd$(a^2-c^2,b^2-d^2)=4$, then $z=1$. 

Note that I am not sure whether this is true or not, but I believe it is. My argument starts as follows:
Assume that $z$ is a primitive $n$th root of unity, that is, $z^n=1$ and $n$ is the smallest such positive integer. Then 
$z^a=1 \implies n |a$
$z^c=1 \implies n |c$
$z^b \bar{z}^d=1 \implies z^b=z^d \implies n |(b-d)$
$z^{b+d}=1 \implies n|(b+d)$.
Using the division conditions that n divides $b-d$ and $b+d$ it follows that $n|2b$ and $n|2d$. Moreover, since $n$ divides $a$ and $n$ divides $c$ it follows that $n$ divides $2a, 2b,2c$ and $2d$. Hence $n=1$ or $n=2$ because gcd$(2a,2b,2c,2d)=2$. If $n=2$ then since $2|a$ and $2|c$, then $4$ divides $a^2-c^2$. Also, since $2$ divides $(b-d)$ and $(b+d)$, we have that $4$ divides $b^2-d^2$. 

Now, this does not seem to give me a contradiction, so I am stuck.

 [/INST] The given condition is false. Take $z = -1$, and $a=4,c=2$, with $b=3,d=1$. Then, of course $\mbox{gcd}(a,b,c,d) = 1$, and $a^2-c^2 = 12$ with $b^2-d^2 = 8$ whose $\gcd$ is $4$. Furthermore, $\bar z = z$, so $z^a = z^c= 1$ and $z^b\bar z^d = z^bz^d=z^{b+d} = z^4=1$.
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