# Show that if $z^7 = 3$, then $w = \mathrm{cis}(2\pi/7) \cdot z$ also satisfies $w^7 = 3$

I need to show if $$z^7 = 3$$, then $$w = \mathrm{cis}(2\pi/7) \cdot z$$ also satisfies $$w^7 = 3$$.

This is what I have done:

Calculate $$w^7$$ and simplify the expression to show that it is equal to $$3$$

\begin{align}w^7 &= (\mathrm{cis}(2\pi/7))^7 \cdot z^7\\ &= \mathrm{cis}(2\pi) \cdot z^7\\ &= 1 \cdot z^7\\ &= z^7\end{align} Since $$w^7 = z^7$$ and we know $$z^7 = 3$$, this means $$w^7 = 3$$ and shows that if $$z^7 = 3$$, then $$w = \mathrm{cis}(2\pi/7) \cdot z$$ also satisfies $$w^7 = 3$$.

Would this be the correct way to solve this question?

• Sounds OK to me. Please use Mathjax to format your questions. Mostly just use  around the equations and $\pi$ is $\pi$ – Andrei Mar 26 at 4:00
• Thank you for formatting my question better and your help. I was unaware of that program but will make use of it in the future. – Tyler Rhodes Mar 26 at 4:19
• \operatorname works better than \mathrm, e.g. \operatorname{cis}\theta $\operatorname{cis}\theta$ – gen-z ready to perish Mar 26 at 5:38
• All good, thank you – Tyler Rhodes Mar 26 at 5:48