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Question Number 61652 by maxmathsup by imad last updated on 05/Jun/19

$$solveinsideC{z}^4=\frac{1-i}{1+i\sqrt{3}}$$

Commented by maxmathsup by imad last updated on 06/Jun/19

$$wehave\mid 1-i\mid =\sqrt{2}\Rightarrow 1-i=\sqrt{2}{e}^{-\frac{i\pi }{4}}$$$$\mid 1+i\sqrt{3}\mid =2\Rightarrow 1+i\sqrt{3}=2\{\frac{1}{2}+\frac{i\sqrt{3}}{2}\}=2e^{\frac{i\pi }{3}}\Rightarrow \frac{1-i}{1+i\sqrt{3}}=\frac{\sqrt{2}}{2}{e}^{-\frac{i\pi }{4}}{e}^{-\frac{i\pi }{3}}$$$$=\frac{\sqrt{2}}{2}{e}^{-i(\frac{\pi }{4}+\frac{\pi }{3})}=\frac{\sqrt{2}}{2}{e}^{-\frac{i7\pi }{12}}letz=r{e}^{i\theta }$$$$\left(e\right)\Rightarrow r^4{e}^{i4\theta }=2^{-\frac{1}{2}}{e}^{-i\left(\frac{7\pi }{12}\right)}\Rightarrow r=2^{-\frac{1}{8}}and4\theta =-\frac{7\pi }{12}+2k\pi \Rightarrow$$$${\theta }_k=-\frac{7\pi }{48}+\frac{k\pi }{2}and0⩽k⩽3sothesolutionsare$$$$Z_k=2^{-\frac{1}{8}}{e}^{i(\frac{k\pi }{2}-\frac{7\pi }{48})}with0⩽k⩽3.$$

Answered by MJS last updated on 06/Jun/19

$$\frac{1-\mathrm{i}}{1+\mathrm{i}\sqrt{3}}=\frac{1-\sqrt{3}}{4}-\mathrm{i}\frac{1+\sqrt{3}}{4}=\frac{\sqrt{2}}{2}{\mathrm{e}}^{-\mathrm{i}\frac{7\pi }{12}}$$$$z^4=\frac{\sqrt{2}}{2}{\mathrm{e}}^{-\mathrm{i}\frac{7\pi }{12}}$$$$z=\frac{1}{\sqrt[8]{2}}{\mathrm{e}}^{\mathrm{i}\pi \times \{-\frac{31}{48},-\frac{7}{48},\frac{17}{48},\frac{41}{48}\}}$$

Commented by maxmathsup by imad last updated on 06/Jun/19

$$thankssirmjs$$

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