We need to find the $z$ from the given 2 equations; $z^3 + \bar{w}^{7} = 0$ and $z^5 w^{11} = 1$ where $w$ is the cube root of unity.
My Try -:
From the first equation, $z^3 + w^{14} = 0 \rightarrow z^3 + w^{2} = 0 \ as \ \bar{w} = w^2 \ and \ w^3 = 1$
From the second equation, $z^5 = w$
Using above equations, we can write $z^2(-w^2) = w$ from using value of $z$ from equation $1$ and putting in equation $2$, we get $ z^2 + w^2 = 0 \rightarrow z = \pm iw$
But $z = \pm iw$ does not seem to be the correct answer as the textbook shows the correct answer as $\pm i$.
Attached is the solution from where this question is taken:
If complex numbers satisfy the system of equations $z^3+\overline{\omega^7}=0$ and $z^5 \omega^{11}=1$, then find $z$. $$\text{Sol. } z^3+\bar{\omega}^7=0$$ $$ \begin{array}{ll} \Rightarrow & z^3=-\bar{\omega}^7 \\ \Rightarrow & |z|^3=|-\bar{\omega}|^7=|\omega|^7 \\ \Rightarrow & |z|^{15}=|\omega|^{35} \qquad \qquad \qquad \,\,\,(1)\\ \text { Alśo, } & z^5 \omega^{11}=1 \\ \Rightarrow & |z|^5|\omega|^{11}=1 \\ \Rightarrow & |z|^{15}|\omega|^{33}=1 \qquad \qquad \qquad (2)\\ \end{array} $$ From $(1)$ and $(2)$, we have $$ |z|=|\omega|=1 $$ Again, $\bar{\omega}^7=-z^3 \quad$ and $\quad \omega^{11}=z^{-5}$ $$ \begin{array}{ll} \Rightarrow & \bar{\omega}^{77} \cdot \omega^{77}=-z^{33} \cdot z^{-35} \\ \Rightarrow & z^2=-1=i^2 \Rightarrow z= \pm i \end{array} $$
Please help me find the solutions to this problem. Thanks in advance!