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Question Number 165620 by mathlove last updated on 05/Feb/22

solve z^7 =−4

$${solve} \\ $$$${z}^{\mathrm{7}} =−\mathrm{4} \\ $$

Answered by aleks041103 last updated on 05/Feb/22

z^7 =−4=4e^((2k+1)iπ) ⇒z=(4)^(1/7) e^(((2k+1)/7)iπ) , k∈Z z=(4)^(1/7) e^((iπ)/7) ;(4)^(1/7) e^((3iπ)/7) ;(4)^(1/7) e^((5iπ)/7) ;(4)^(1/7) e^(iπ) ;(4)^(1/7) e^((9iπ)/7) ;(4)^(1/7) e^((11iπ)/7) ;(4)^(1/7) e^(((13iπ)/7) )

$${z}^{\mathrm{7}} =−\mathrm{4}=\mathrm{4}{e}^{\left(\mathrm{2}{k}+\mathrm{1}\right){i}\pi} \\ $$$$\Rightarrow{z}=\sqrt[{\mathrm{7}}]{\mathrm{4}}\:{e}^{\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{7}}{i}\pi} ,\:{k}\in\mathbb{Z} \\ $$$${z}=\sqrt[{\mathrm{7}}]{\mathrm{4}}{e}^{\frac{{i}\pi}{\mathrm{7}}} ;\sqrt[{\mathrm{7}}]{\mathrm{4}}{e}^{\frac{\mathrm{3}{i}\pi}{\mathrm{7}}} ;\sqrt[{\mathrm{7}}]{\mathrm{4}}{e}^{\frac{\mathrm{5}{i}\pi}{\mathrm{7}}} ;\sqrt[{\mathrm{7}}]{\mathrm{4}}{e}^{{i}\pi} ;\sqrt[{\mathrm{7}}]{\mathrm{4}}{e}^{\frac{\mathrm{9}{i}\pi}{\mathrm{7}}} ;\sqrt[{\mathrm{7}}]{\mathrm{4}}{e}^{\frac{\mathrm{11}{i}\pi}{\mathrm{7}}} ;\sqrt[{\mathrm{7}}]{\mathrm{4}}{e}^{\frac{\mathrm{13}{i}\pi}{\mathrm{7}}\:} \\ $$

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