Question Number 131309 by shaker last updated on 03/Feb/21
Answered by Ar Brandon last updated on 03/Feb/21
![z^5 +(1/( (√2)))+(1/( (√2)))i=0 ⇒z^5 +e^((π/4)i) =0 z^5 =−e^((π/4)i) =e^((π/4)i+(2k+1)πi) =e^(((8k+5)/4)πi) z=e^(((8k+5)/(20))πi) , k∈[0, 4]](https://www.tinkutara.com/question/Q131314.png)
$$\mathrm{z}^{\mathrm{5}} +\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\mathrm{i}=\mathrm{0}\:\Rightarrow\mathrm{z}^{\mathrm{5}} +\mathrm{e}^{\frac{\pi}{\mathrm{4}}\mathrm{i}} =\mathrm{0} \\ $$$$\mathrm{z}^{\mathrm{5}} =−\mathrm{e}^{\frac{\pi}{\mathrm{4}}\mathrm{i}} =\mathrm{e}^{\frac{\pi}{\mathrm{4}}\mathrm{i}+\left(\mathrm{2k}+\mathrm{1}\right)\pi\mathrm{i}} =\mathrm{e}^{\frac{\mathrm{8k}+\mathrm{5}}{\mathrm{4}}\pi\mathrm{i}} \\ $$$$\mathrm{z}=\mathrm{e}^{\frac{\mathrm{8k}+\mathrm{5}}{\mathrm{20}}\pi\mathrm{i}} ,\:\mathrm{k}\in\left[\mathrm{0},\:\mathrm{4}\right] \\ $$
Answered by mathmax by abdo last updated on 03/Feb/21
![z^5 +(1/( (√2)))+(i/( (√2)))=0 ⇒z^5 =−e^((iπ)/4) =e^(iπ) .e^((iπ)/4) =e^(i(π+(π/4)) ) =e^(i((5π)/4)) =e^(i(((5π)/4)+2kπ)) ⇒ the roots are z_k =e^((i(((5π)/4)+2kπ))/(5 )) =e^(((iπ)/4) +i((2kπ)/5)) with k∈[[0,4]]](https://www.tinkutara.com/question/Q131352.png)
$$\mathrm{z}^{\mathrm{5}} \:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\frac{\mathrm{i}}{\:\sqrt{\mathrm{2}}}=\mathrm{0}\:\Rightarrow\mathrm{z}^{\mathrm{5}} \:=−\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:=\mathrm{e}^{\mathrm{i}\pi} .\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}} \:=\mathrm{e}^{\mathrm{i}\left(\pi+\frac{\pi}{\mathrm{4}}\right)\:} =\mathrm{e}^{\mathrm{i}\frac{\mathrm{5}\pi}{\mathrm{4}}} \:=\mathrm{e}^{\mathrm{i}\left(\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{2k}\pi\right)} \:\Rightarrow \\ $$$$\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{z}_{\mathrm{k}} =\mathrm{e}^{\frac{\mathrm{i}\left(\frac{\mathrm{5}\pi}{\mathrm{4}}+\mathrm{2k}\pi\right)}{\mathrm{5}\:}} =\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{4}}\:+\mathrm{i}\frac{\mathrm{2k}\pi}{\mathrm{5}}} \:\:\mathrm{with}\:\mathrm{k}\in\left[\left[\mathrm{0},\mathrm{4}\right]\right] \\ $$