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Question Number 95259 by mhmd last updated on 24/May/20
find all roots ((√6) −(√2)i)^(1/3) by using demover theorem ?
$${find}\:{all}\:{roots}\:\left(\sqrt{\mathrm{6}}\:−\sqrt{\mathrm{2}}{i}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} {by}\:{using}\:{demover}\:{theorem}\:? \\ $$
Commented by mhmd last updated on 24/May/20
help me ?
$${help}\:{me}\:? \\ $$
Answered by mathmax by abdo last updated on 24/May/20
let find z / z^2 =((√6)−i(√2))^(1/3) we have (√6)−i(√2)=2(√2){((√6)/(2(√2)))−((i(√2))/(2(√2)))} =2(√2)e^(iarctan(−((√2)/( (√6))))) ⇒ ((√6)−i(√2))^(1/3) =(2(√2))^(1/3) e^(−(i/3)arctan(((√2)/( (√6))))) let z =re^(iθ) (e)⇒r^2 e^(2iθ ) =(2(√2))^(1/3) e^(−(i/3)arctan(((√2)/( (√6))))) ⇒r =(2(√2))^(1/6) and 2θ =−(1/3)arctan(((√2)/( (√6))))+2kπ ⇒θ_k =−(1/6) arctan(((√2)/( (√6))))+kπ so the roots are z_z =^(12) (√8) e^(i(−(1/6)arctan(((√2)/( (√6))))+kπ)) k ∈{0,1}
$$\mathrm{let}\:\mathrm{find}\:\mathrm{z}\:\:/\:\mathrm{z}^{\mathrm{2}} \:=\left(\sqrt{\mathrm{6}}−\mathrm{i}\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\:\mathrm{we}\:\mathrm{have}\: \\ $$$$\sqrt{\mathrm{6}}−\mathrm{i}\sqrt{\mathrm{2}}=\mathrm{2}\sqrt{\mathrm{2}}\left\{\frac{\sqrt{\mathrm{6}}}{\mathrm{2}\sqrt{\mathrm{2}}}−\frac{\mathrm{i}\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{\mathrm{2}}}\right\}\:=\mathrm{2}\sqrt{\mathrm{2}}\mathrm{e}^{\mathrm{iarctan}\left(−\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{6}}}\right)} \:\Rightarrow \\ $$$$\left(\sqrt{\mathrm{6}}−\mathrm{i}\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:=\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{e}^{−\frac{\mathrm{i}}{\mathrm{3}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{6}}}\right)} \:\mathrm{let}\:\mathrm{z}\:=\mathrm{re}^{\mathrm{i}\theta} \:\: \\ $$$$\left(\mathrm{e}\right)\Rightarrow\mathrm{r}^{\mathrm{2}} \:\mathrm{e}^{\mathrm{2i}\theta\:} \:=\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{3}}} \:\mathrm{e}^{−\frac{\mathrm{i}}{\mathrm{3}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{6}}}\right)} \:\Rightarrow\mathrm{r}\:=\left(\mathrm{2}\sqrt{\mathrm{2}}\right)^{\frac{\mathrm{1}}{\mathrm{6}}} \:\mathrm{and} \\ $$$$\mathrm{2}\theta\:=−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{6}}}\right)+\mathrm{2k}\pi\:\Rightarrow\theta_{\mathrm{k}} =−\frac{\mathrm{1}}{\mathrm{6}}\:\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{6}}}\right)+\mathrm{k}\pi \\ $$$$\mathrm{so}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{are}\:\mathrm{z}_{\mathrm{z}} =^{\mathrm{12}} \sqrt{\mathrm{8}}\:\mathrm{e}^{\mathrm{i}\left(−\frac{\mathrm{1}}{\mathrm{6}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{6}}}\right)+\mathrm{k}\pi\right)} \: \\ $$$$\mathrm{k}\:\in\left\{\mathrm{0},\mathrm{1}\right\} \\ $$
Commented by mhmd last updated on 24/May/20
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by turbo msup by abdo last updated on 24/May/20
you are welcome
$${you}\:{are}\:{welcome} \\ $$

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