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Question Number 131309 by shaker last updated on 03/Feb/21

Answered by Ar Brandon last updated on 03/Feb/21

$${\mathrm{z}}^5+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\mathrm{i}=0\Rightarrow {\mathrm{z}}^5+{\mathrm{e}}^{\frac{\pi }{4}\mathrm{i}}=0$$$${\mathrm{z}}^5=-{\mathrm{e}}^{\frac{\pi }{4}\mathrm{i}}={\mathrm{e}}^{\frac{\pi }{4}\mathrm{i}+(2\mathrm{k}+1)\pi \mathrm{i}}={\mathrm{e}}^{\frac{8\mathrm{k}+5}{4}\pi \mathrm{i}}$$$$\mathrm{z}={\mathrm{e}}^{\frac{8\mathrm{k}+5}{20}\pi \mathrm{i}},\mathrm{k}\in [0,4]$$

Answered by mathmax by abdo last updated on 03/Feb/21

$${\mathrm{z}}^5+\frac{1}{\sqrt{2}}+\frac{\mathrm{i}}{\sqrt{2}}=0\Rightarrow {\mathrm{z}}^5=-{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}={\mathrm{e}}^{\mathrm{i}\pi }.{\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}}={\mathrm{e}}^{\mathrm{i}(\pi +\frac{\pi }{4})}={\mathrm{e}}^{\mathrm{i}\frac{5\pi }{4}}={\mathrm{e}}^{\mathrm{i}(\frac{5\pi }{4}+2\mathrm{k}\pi )}\Rightarrow$$$$\mathrm{the}\mathrm{roots}\mathrm{are}{\mathrm{z}}_{\mathrm{k}}={\mathrm{e}}^{\frac{\mathrm{i}(\frac{5\pi }{4}+2\mathrm{k}\pi )}{5}}={\mathrm{e}}^{\frac{\mathrm{i}\pi }{4}+\mathrm{i}\frac{2\mathrm{k}\pi }{5}}\mathrm{with}\mathrm{k}\in \left[[0,4]\right]$$

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