Question Number 207408 by hardmath last updated on 13/May/24
$$\boldsymbol{\mathrm{z}}\:\:=\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:−\:\frac{\mathrm{1}}{\mathrm{2}}\:\boldsymbol{\mathrm{i}}\:\:\:\:\:\mathrm{find}:\:\:\boldsymbol{\mathrm{z}}^{\mathrm{11}} \:=\:? \\ $$
Answered by Frix last updated on 13/May/24
$${z}=\mathrm{e}^{−\frac{\pi}{\mathrm{6}}\mathrm{i}} \\ $$$${z}^{\mathrm{11}} =\mathrm{e}^{−\frac{\mathrm{11}\pi}{\mathrm{6}}\mathrm{i}} =\mathrm{e}^{\left(\mathrm{2}\pi−\frac{\mathrm{11}\pi}{\mathrm{6}}\right)\mathrm{i}} =\mathrm{e}^{\frac{\pi}{\mathrm{6}}\mathrm{i}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$
Answered by Rasheed.Sindhi last updated on 14/May/24
$$\mathrm{z}^{\mathrm{3}} =−\mathrm{i}\Rightarrow\mathrm{z}^{\mathrm{12}} =\left(\mathrm{z}^{\mathrm{3}} \right)^{\mathrm{4}} =\left(−\mathrm{i}\right)^{\mathrm{4}} =\mathrm{1} \\ $$$$\mathrm{z}^{\mathrm{11}} =\frac{\mathrm{z}^{\mathrm{12}} }{\mathrm{z}}=\frac{\mathrm{1}}{\mathrm{z}}=\frac{\overset{−} {\mathrm{z}}}{\mathrm{z}\:\overline {\mathrm{z}}} \\ $$$$\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}}{\left(\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\frac{\sqrt{\mathrm{3}}\:}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i}}{\mathrm{1}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{i} \\ $$