Question Number 143194 by ZiYangLee last updated on 11/Jun/21
$$\mathrm{Suppose}\:{z}^{\mathrm{50}} +{z}^{\mathrm{25}} +{m}=\mathrm{0},\:\mathrm{where}\:{z}=\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{m}. \\ $$
Answered by Olaf_Thorendsen last updated on 11/Jun/21
$${z}\:=\:\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\:=\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$${z}^{\mathrm{50}} \:=\:{e}^{{i}\frac{\pi}{\mathrm{4}}×\mathrm{50}} \:=\:{e}^{{i}\frac{\mathrm{25}\pi}{\mathrm{2}}} \:=\:{e}^{{i}\frac{\pi}{\mathrm{2}}} \:=\:{i} \\ $$$${z}^{\mathrm{25}} \:=\:{e}^{{i}\frac{\pi}{\mathrm{4}}×\mathrm{25}} \:=\:{e}^{{i}\frac{\mathrm{25}\pi}{\mathrm{4}}} \:=\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:=\:\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\:=\:{z} \\ $$$${m}\:=\:−{z}^{\mathrm{25}} −{z}^{\mathrm{50}} \:=\:−\left({z}+{i}\right)\:=\:−\frac{\mathrm{1}+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){i}}{\:\sqrt{\mathrm{2}}} \\ $$
Commented by ZiYangLee last updated on 11/Jun/21
$$\mathrm{thanks}\:\mathrm{sir} \\ $$