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The problem requires us to solve $w^{10} = 1$ beforehand ($w=cis(\frac{2k\pi}{5})$ for $k = 0,1,2,3,\ldots, 9)$ to solve $(\frac{z-1}{z})^{10}$ = 1 and show it is z = $\frac{1}{2}$ $(1+i\cot(\frac{r\pi}{10}))$ for r = 1,2...9$.

Using the prior steps I have reached z = $\frac{1}{1-cis(\frac{r\pi}{5})}$ but am unsure how to apply double angle to reach the required answer.

edit 1: sorry, just realized I typed the wrong numbers in some of my terms and have corrected. Additionally there was a subpart requiring us to explain why this new equation had only $9$ sols which I haven't listed but solved

I reached the answer I listed there through these steps: (lazy formatting sorry)

$\frac{z-1}{z} = cis(\frac{rpi}{5})$ (from the previous solution)

$$z-1 = z cis(\frac{r\pi}{5})$$

$$z(1-cis(\frac{r\pi}{5}) = 1$$

$$z = \frac{1}{(1-cis(\frac{r\pi}{5})}$$

I have gone further and gotten to

$$z = \frac{1}{(2\sin^2(\frac{r\pi}{10}) - 2i\sin(\frac{r\pi}{10})\cos(\frac{r\pi}{10})}$$

$$= \frac{1}{(2\sin\frac{r\pi}{10})(2\sin\frac{r\pi}{10}+i\cos\frac{r\pi}{10})}$$

should I try convert the 2nd term into cis form using angle identities?

Edit2: Thank you for the answers!!!! I have tried to edit through this post for clarity for future readers

Ophe
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    hi, can you use mathjax? – TShiong Apr 21 '23 at 15:20
  • The solutions of $w^{10}=1$ are still not correct. Either replace "over 5" with "over 10", or remove the 2 from the numerator. Btw, here is a link about formatting mathematics: https://math.stackexchange.com/help/notation – Anne Bauval Apr 21 '23 at 16:57

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In view of the answer you were given, and of one of your steps you listed, there is a misprint in your question. The equation to solve was not "(z-1/z)^10 = 1" (i.e. $\left(z-\frac1z\right)^{10}=1$), but $$\left(\frac{z-1}z\right)^{10}=1$$ and you were almost done! Let us set $t=\pi/10$ and start from your penultimate equation: $$z =\frac1{2\sin(rt)\left(\sin(rt)-i\cos(rt)\right)}$$ (for $r=1,2,\dots,9$ since $r=0$ is excluded because it would corresponds to $\frac{z-1}z=1$, which is impossible).

It is equivalent to $$z =\frac{\sin(rt)+i\cos(rt)}{2\sin(rt)},$$ i.e. $$z =\frac{1+i\cot(rt)}2.$$

Anne Bauval
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