Question Number 128080 by mohammad17 last updated on 04/Jan/21
Answered by mathmax by abdo last updated on 04/Jan/21
![z^4 −1=(√3)i ⇒z^4 =1+(√3)i =2((1/2)+i((√3)/2))=2e^((iπ)/3) let z =re^(iθ) z^4 =1+(√3)i ⇒r^4 e^(4iθ) =2e^(i(π/3)+2ikπ) ⇒r=^4 (√2) and 4θ =(π/3)+2kπ ⇒ θ=(π/(12))+((kπ)/2) so the roots are z_k =^4 (√2) e^(i((π/(12))+((kπ)/2))) with k∈[[0,3]] z_0 +z_2 =^4 (√2){ e^((iπ)/(12)) +e^((iπ)/(12)) .e^(iπ) } =^4 (√2){e^((iπ)/(12)) −e^((iπ)/(12)) }=0 z_1 +z_3 =^4 (√2){e^((iπ)/(12)) .e^((iπ)/2) +e^((iπ)/(12)) .e^(i((3π)/2)) } =^4 (√2){i e^((iπ)/(12)) −ie^((iπ)/(12)) } =0](https://www.tinkutara.com/question/Q128084.png)
$$\mathrm{z}^{\mathrm{4}} −\mathrm{1}=\sqrt{\mathrm{3}}\mathrm{i}\:\Rightarrow\mathrm{z}^{\mathrm{4}} \:=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\:=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)=\mathrm{2e}^{\frac{\mathrm{i}\pi}{\mathrm{3}}} \:\mathrm{let}\:\mathrm{z}\:=\mathrm{re}^{\mathrm{i}\theta} \: \\ $$$$\mathrm{z}^{\mathrm{4}} \:=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}\:\Rightarrow\mathrm{r}^{\mathrm{4}} \:\mathrm{e}^{\mathrm{4i}\theta} \:=\mathrm{2e}^{\mathrm{i}\frac{\pi}{\mathrm{3}}+\mathrm{2ik}\pi} \:\Rightarrow\mathrm{r}=^{\mathrm{4}} \sqrt{\mathrm{2}}\:\mathrm{and}\:\mathrm{4}\theta\:\:=\frac{\pi}{\mathrm{3}}+\mathrm{2k}\pi\:\Rightarrow \\ $$$$\theta=\frac{\pi}{\mathrm{12}}+\frac{\mathrm{k}\pi}{\mathrm{2}}\:\mathrm{so}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{z}_{\mathrm{k}} =^{\mathrm{4}} \sqrt{\mathrm{2}}\:\mathrm{e}^{\mathrm{i}\left(\frac{\pi}{\mathrm{12}}+\frac{\mathrm{k}\pi}{\mathrm{2}}\right)} \:\:\mathrm{with}\:\mathrm{k}\in\left[\left[\mathrm{0},\mathrm{3}\right]\right] \\ $$$$\mathrm{z}_{\mathrm{0}} \:+\mathrm{z}_{\mathrm{2}} =^{\mathrm{4}} \sqrt{\mathrm{2}}\left\{\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{12}}} \:\:+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{12}}} \:.\mathrm{e}^{\mathrm{i}\pi} \right\}\:=^{\mathrm{4}} \sqrt{\mathrm{2}}\left\{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{12}}} −\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{12}}} \right\}=\mathrm{0} \\ $$$$\mathrm{z}_{\mathrm{1}} \:+\mathrm{z}_{\mathrm{3}} =^{\mathrm{4}} \sqrt{\mathrm{2}}\left\{\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{12}}} .\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{2}}} \:+\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{12}}} \:.\mathrm{e}^{\mathrm{i}\frac{\mathrm{3}\pi}{\mathrm{2}}} \right\}\:\:=^{\mathrm{4}} \sqrt{\mathrm{2}}\left\{\mathrm{i}\:\mathrm{e}^{\frac{\mathrm{i}\pi}{\mathrm{12}}} \:−\mathrm{ie}^{\frac{\mathrm{i}\pi}{\mathrm{12}}} \right\}\:=\mathrm{0} \\ $$