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Question Number 143194 by ZiYangLee last updated on 11/Jun/21

Suppose z^(50) +z^(25) +m=0, where z=((1+i)/( (√2))) find the value of m.

$$\mathrm{Suppose}\:{z}^{\mathrm{50}} +{z}^{\mathrm{25}} +{m}=\mathrm{0},\:\mathrm{where}\:{z}=\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{m}. \\ $$

Answered by Olaf_Thorendsen last updated on 11/Jun/21

z = ((1+i)/( (√2))) = e^(i(π/4)) z^(50) = e^(i(π/4)×50) = e^(i((25π)/2)) = e^(i(π/2)) = i z^(25) = e^(i(π/4)×25) = e^(i((25π)/4)) = e^(i(π/4)) = ((1+i)/( (√2))) = z m = −z^(25) −z^(50) = −(z+i) = −((1+(1+(√2))i)/( (√2)))

$${z}\:=\:\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\:=\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \\ $$$${z}^{\mathrm{50}} \:=\:{e}^{{i}\frac{\pi}{\mathrm{4}}×\mathrm{50}} \:=\:{e}^{{i}\frac{\mathrm{25}\pi}{\mathrm{2}}} \:=\:{e}^{{i}\frac{\pi}{\mathrm{2}}} \:=\:{i} \\ $$$${z}^{\mathrm{25}} \:=\:{e}^{{i}\frac{\pi}{\mathrm{4}}×\mathrm{25}} \:=\:{e}^{{i}\frac{\mathrm{25}\pi}{\mathrm{4}}} \:=\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:=\:\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}}}\:=\:{z} \\ $$$${m}\:=\:−{z}^{\mathrm{25}} −{z}^{\mathrm{50}} \:=\:−\left({z}+{i}\right)\:=\:−\frac{\mathrm{1}+\left(\mathrm{1}+\sqrt{\mathrm{2}}\right){i}}{\:\sqrt{\mathrm{2}}} \\ $$

Commented by ZiYangLee last updated on 11/Jun/21

thanks sir

$$\mathrm{thanks}\:\mathrm{sir} \\ $$

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