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

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

Answered by Olaf_Thorendsen last updated on 11/Jun/21

$$z=\frac{1+i}{\sqrt{2}}=e^{i\frac{\pi }{4}}$$$$z^{50}=e^{i\frac{\pi }{4}\times 50}=e^{i\frac{25\pi }{2}}=e^{i\frac{\pi }{2}}=i$$$$z^{25}=e^{i\frac{\pi }{4}\times 25}=e^{i\frac{25\pi }{4}}=e^{i\frac{\pi }{4}}=\frac{1+i}{\sqrt{2}}=z$$$$m=-z^{25}-z^{50}=-(z+i)=-\frac{1+(1+\sqrt{2})i}{\sqrt{2}}$$

Commented by ZiYangLee last updated on 11/Jun/21

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

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