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

solve inside C z^4 =((1−i)/(1+i(√3)))

$${solve}\:{inside}\:{C}\:\:{z}^{\mathrm{4}} \:=\frac{\mathrm{1}−{i}}{\mathrm{1}+{i}\sqrt{\mathrm{3}}} \\ $$

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

we have ∣1−i∣=(√2) ⇒1−i =(√2)e^(−((iπ)/4)) ∣1+i(√3)∣ =2 ⇒1+i(√3)=2{(1/2) +((i(√3))/2)} =2 e^((iπ)/3) ⇒((1−i)/(1+i(√3))) =((√2)/2) e^(−((iπ)/4)) e^(−((iπ)/3)) =((√2)/2) e^(−i((π/4)+(π/3))) =((√2)/2) e^(−((i7π)/(12))) let z =r e^(iθ) (e) ⇒ r^4 e^(i4θ) =2^(−(1/2)) e^(−i(((7π)/(12)))) ⇒r =2^(−(1/8)) and 4θ =−((7π)/(12)) +2kπ ⇒ θ_k =−((7π)/(48)) +((kπ)/2) and 0≤k≤3 so the solutions are Z_k = 2^(−(1/8)) e^(i(((kπ)/2)−((7π)/(48)))) with 0≤k≤3 .

$${we}\:{have}\:\mid\mathrm{1}−{i}\mid=\sqrt{\mathrm{2}}\:\Rightarrow\mathrm{1}−{i}\:=\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \\ $$$$\mid\mathrm{1}+{i}\sqrt{\mathrm{3}}\mid\:=\mathrm{2}\:\Rightarrow\mathrm{1}+{i}\sqrt{\mathrm{3}}=\mathrm{2}\left\{\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right\}\:=\mathrm{2}\:{e}^{\frac{{i}\pi}{\mathrm{3}}} \:\Rightarrow\frac{\mathrm{1}−{i}}{\mathrm{1}+{i}\sqrt{\mathrm{3}}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \:{e}^{−\frac{{i}\pi}{\mathrm{3}}} \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{e}^{−{i}\left(\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{3}}\right)} \:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{e}^{−\frac{{i}\mathrm{7}\pi}{\mathrm{12}}} \:\:\:{let}\:{z}\:={r}\:{e}^{{i}\theta} \\ $$$$\left({e}\right)\:\Rightarrow\:{r}^{\mathrm{4}} \:{e}^{{i}\mathrm{4}\theta} \:=\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:{e}^{−{i}\left(\frac{\mathrm{7}\pi}{\mathrm{12}}\right)} \:\Rightarrow{r}\:=\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{8}}} \:\:\:{and}\:\mathrm{4}\theta\:\:\:=−\frac{\mathrm{7}\pi}{\mathrm{12}}\:+\mathrm{2}{k}\pi\:\Rightarrow \\ $$$$\theta_{{k}} =−\frac{\mathrm{7}\pi}{\mathrm{48}}\:+\frac{{k}\pi}{\mathrm{2}}\:\:\:\:\:{and}\:\mathrm{0}\leqslant{k}\leqslant\mathrm{3}\:\:{so}\:{the}\:{solutions}\:{are}\: \\ $$$${Z}_{{k}} =\:\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{8}}} \:{e}^{{i}\left(\frac{{k}\pi}{\mathrm{2}}−\frac{\mathrm{7}\pi}{\mathrm{48}}\right)} \:\:\:\:\:{with}\:\mathrm{0}\leqslant{k}\leqslant\mathrm{3}\:. \\ $$

Answered by MJS last updated on 06/Jun/19

((1−i)/(1+i(√3)))=((1−(√3))/4)−i((1+(√3))/4)=((√2)/2)e^(−i((7π)/(12))) z^4 =((√2)/2)e^(−i((7π)/(12))) z=(1/(2)^(1/8) ) e^(iπ×{−((31)/(48)), −(7/(48)), ((17)/(48)), ((41)/(48))})

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

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

thanks sirmjs

$${thanks}\:{sirmjs} \\ $$

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