Question Number 48272 by gunawan last updated on 21/Nov/18
$$\mathrm{z}^{\mathrm{5}} =\mathrm{32} \\ $$$$\mathrm{find}\:\mathrm{all}\:{root}\:{z} \\ $$
Commented by maxmathsup by imad last updated on 21/Nov/18
![let z=r e^(iθ) so z^5 =32 ⇔r^5 e^(5iθ) =2^5 e^(i2kπ) ⇔r=2 and 5θ =2kπ ⇒ θ_k =((2kπ)/5) so the roots are z_k =2 e^(i((2kπ)/5)) and k∈[[0,4]] z_0 =2 ,z_1 =2 e^(i((2π)/5)) ,z_2 =2 e^(i((4π)/5)) , z_3 =2 e^(i((6π)/5)) , z_(4 ) =2 e^(i((8π)/5)) . z_1 =2(cos(((2π)/5))+isin(((2π)/5))) but cos(((2π)/5))=2cos^2 ((π/5))−1 =2(((1+(√5))/4))^2 −1 =2((6+2(√5))/(16)) −1 =((12+4(√5)−16)/(16)) =((4(√5)−4)/(16)) =(((√5)−1)/4) sin(((2π)/5))=(√(1−((((√5)−1)/4))^2 ))=(√(1−((6−2(√5))/(16))))=(√((10+2(√5))/(16)))=((√(10+2(√5)))/4) ⇒ z_1 =(((√5)−1)/2) +i((√(10+2(√5)))/2) z_2 =2 e^(i(π−(π/5))) =−2 e^(−i(π/5)) =−2(cos((π/5))−isin((π/5)))but cos((π/5))=((1+(√5))/4) sin((π/5))=(√(1−(((1+(√5))/4))^2 ))=(√(1−((6+2(√5))/(16))))=(√((10−2(√5))/(16)))=((√(10−2(√5)))/4) ⇒ z_2 =−((1+(√5))/2) +i((√(10−2(√5)))/2) z_3 =2 e^(iπ) e^(i(π/5)) =−2(cos((π/5))+isin((π/5))) =−((1+(√5))/2) −i((√(10−2(√5)))/2) z_4 =2 e^(i(2π−((2π)/5))) =2 e^(−i((2π)/5)) =2(cos(((2π)/5))−isin(((2π)/5))) =(((√5)−1)/2) −i((√(10+(√5)))/2)](Q48274.png)
$${let}\:{z}={r}\:{e}^{{i}\theta} \:\:\:{so}\:{z}^{\mathrm{5}} =\mathrm{32}\:\Leftrightarrow{r}^{\mathrm{5}} \:{e}^{\mathrm{5}{i}\theta} =\mathrm{2}^{\mathrm{5}} \:{e}^{{i}\mathrm{2}{k}\pi} \:\Leftrightarrow{r}=\mathrm{2}\:{and}\:\mathrm{5}\theta\:=\mathrm{2}{k}\pi\:\Rightarrow \\ $$$$\theta_{{k}} =\frac{\mathrm{2}{k}\pi}{\mathrm{5}}\:\:{so}\:{the}\:{roots}\:{are}\:{z}_{{k}} =\mathrm{2}\:{e}^{{i}\frac{\mathrm{2}{k}\pi}{\mathrm{5}}} \:\:{and}\:{k}\in\left[\left[\mathrm{0},\mathrm{4}\right]\right] \\ $$$${z}_{\mathrm{0}} =\mathrm{2}\:,{z}_{\mathrm{1}} =\mathrm{2}\:{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{5}}} \:,{z}_{\mathrm{2}} =\mathrm{2}\:{e}^{{i}\frac{\mathrm{4}\pi}{\mathrm{5}}} \:,\:{z}_{\mathrm{3}} =\mathrm{2}\:{e}^{{i}\frac{\mathrm{6}\pi}{\mathrm{5}}} \:,\:{z}_{\mathrm{4}\:} =\mathrm{2}\:{e}^{{i}\frac{\mathrm{8}\pi}{\mathrm{5}}} \:. \\ $$$${z}_{\mathrm{1}} =\mathrm{2}\left({cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)+{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\right)\:{but}\:{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)=\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{5}}\right)−\mathrm{1} \\ $$$$=\mathrm{2}\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{\mathrm{2}} −\mathrm{1}\:=\mathrm{2}\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}}\:−\mathrm{1}\:=\frac{\mathrm{12}+\mathrm{4}\sqrt{\mathrm{5}}−\mathrm{16}}{\mathrm{16}}\:=\frac{\mathrm{4}\sqrt{\mathrm{5}}−\mathrm{4}}{\mathrm{16}}\:=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}} \\ $$$${sin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)=\sqrt{\mathrm{1}−\left(\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{1}−\frac{\mathrm{6}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}}}=\sqrt{\frac{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}}}=\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\:\Rightarrow \\ $$$${z}_{\mathrm{1}} =\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{10}+\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}} \\ $$$${z}_{\mathrm{2}} =\mathrm{2}\:{e}^{{i}\left(\pi−\frac{\pi}{\mathrm{5}}\right)} =−\mathrm{2}\:{e}^{−{i}\frac{\pi}{\mathrm{5}}} =−\mathrm{2}\left({cos}\left(\frac{\pi}{\mathrm{5}}\right)−{isin}\left(\frac{\pi}{\mathrm{5}}\right)\right){but}\:{cos}\left(\frac{\pi}{\mathrm{5}}\right)=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}} \\ $$$${sin}\left(\frac{\pi}{\mathrm{5}}\right)=\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{4}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{1}−\frac{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}}}=\sqrt{\frac{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{16}}}=\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{4}}\:\Rightarrow \\ $$$${z}_{\mathrm{2}} =−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}} \\ $$$${z}_{\mathrm{3}} =\mathrm{2}\:{e}^{{i}\pi} \:{e}^{{i}\frac{\pi}{\mathrm{5}}} =−\mathrm{2}\left({cos}\left(\frac{\pi}{\mathrm{5}}\right)+{isin}\left(\frac{\pi}{\mathrm{5}}\right)\right) \\ $$$$=−\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\:−{i}\frac{\sqrt{\mathrm{10}−\mathrm{2}\sqrt{\mathrm{5}}}}{\mathrm{2}} \\ $$$${z}_{\mathrm{4}} =\mathrm{2}\:{e}^{{i}\left(\mathrm{2}\pi−\frac{\mathrm{2}\pi}{\mathrm{5}}\right)} =\mathrm{2}\:{e}^{−{i}\frac{\mathrm{2}\pi}{\mathrm{5}}} =\mathrm{2}\left({cos}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)−{isin}\left(\frac{\mathrm{2}\pi}{\mathrm{5}}\right)\right) \\ $$$$=\frac{\sqrt{\mathrm{5}}−\mathrm{1}}{\mathrm{2}}\:−{i}\frac{\sqrt{\mathrm{10}+\sqrt{\mathrm{5}}}}{\mathrm{2}} \\ $$