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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?

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    $\begingroup$ Sounds OK to me. Please use Mathjax to format your questions. Mostly just use $ $ around the equations and $\pi$ is $\pi$ $\endgroup$ – Andrei Mar 26 at 4:00
  • $\begingroup$ 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. $\endgroup$ – Tyler Rhodes Mar 26 at 4:19
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    $\begingroup$ \operatorname works better than \mathrm, e.g. \operatorname{cis}\theta $\operatorname{cis}\theta$ $\endgroup$ – gen-z ready to perish Mar 26 at 5:38
  • $\begingroup$ All good, thank you $\endgroup$ – Tyler Rhodes Mar 26 at 5:48

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