Question Number 203919 by Samuel12 last updated on 02/Feb/24
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Answered by Mathspace last updated on 02/Feb/24
![1=e^(2ikπ) let z=re^(iθ) z^5 =1 ⇔r^5 e^(i5θ) =e^(i2kπ) ⇔ r=1 and θ=((2kπ)/5) so the roots are z_k =e^(i((2kπ)/5)) and k∈[[0,4]] z_0 =1 , z_1 =e^((i2π)/5) z_2 =e^(i((4π)/5)) =−e^(−((iπ)/5)) z_3 =e^(i((6π)/5)) =−e^((iπ)/5) z_4 =e^(i((8π)/5)) =e^(i((10π−2π)/5)) =e^(−i((2π)/5)) e^((iπ)/5) =cos((π/5))+isin((π/5)) and cos((π/5))=((1+(√5))/4) .....](Q203932.png)
$$\mathrm{1}={e}^{\mathrm{2}{ik}\pi} \:\:{let}\:{z}={re}^{{i}\theta} \\ $$$${z}^{\mathrm{5}} =\mathrm{1}\:\Leftrightarrow{r}^{\mathrm{5}} {e}^{{i}\mathrm{5}\theta} ={e}^{{i}\mathrm{2}{k}\pi} \Leftrightarrow \\ $$$${r}=\mathrm{1}\:{and}\:\theta=\frac{\mathrm{2}{k}\pi}{\mathrm{5}} \\ $$$${so}\:{the}\:{roots}\:{are}\:{z}_{{k}} ={e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{5}}} \\ $$$${and}\:{k}\in\left[\left[\mathrm{0},\mathrm{4}\right]\right] \\ $$$${z}_{\mathrm{0}} =\mathrm{1}\:\:\:,\:{z}_{\mathrm{1}} ={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{5}}} \\ $$$${z}_{\mathrm{2}} ={e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{5}}} =−{e}^{−\frac{{i}\pi}{\mathrm{5}}} \:\:\:\:\:{z}_{\mathrm{3}} ={e}^{{i}\frac{\mathrm{6}\pi}{\mathrm{5}}} \:=−{e}^{\frac{{i}\pi}{\mathrm{5}}} \\ $$$${z}_{\mathrm{4}} ={e}^{{i}\frac{\mathrm{8}\pi}{\mathrm{5}}} ={e}^{{i}\frac{\mathrm{10}\pi−\mathrm{2}\pi}{\mathrm{5}}} ={e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{5}}} \\ $$$${e}^{\frac{{i}\pi}{\mathrm{5}}} ={cos}\left(\frac{\pi}{\mathrm{5}}\right)+{isin}\left(\frac{\pi}{\mathrm{5}}\right) \\ $$$${and}\:{cos}\left(\frac{\pi}{\mathrm{5}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\:..... \\ $$