Question-178810

Question Number 178810 by Ib last updated on 21/Oct/22

Commented by ARUNG_Brandon_MBU last updated on 21/Oct/22

Tu peux essayer de rogner l'image la prochaine fois pour améliorer la qualité car là c'est pas si agréable pour la vue.

Answered by ARUNG_Brandon_MBU last updated on 21/Oct/22

z=((1−e^(i(π/3)) )/(1+e^(i(π/3)) ))=((e^(i(π/6)) (e^(−i(π/6)) −e^(i(π/6)) ))/(e^(i(π/6)) (e^(−i(π/6)) +e^(i(π/6)) )))=−((e^(i(π/6)) −e^(i(π/6)) )/(e^(i(π/6)) +e^(−i(π/6)) )) =−((2isin((π/6)))/(2cos((π/6))))=−((√3)/3)i=((√3)/3)e^(((3π)/2)i) ⇒∣z∣=((√3)/3) , arg(z)=((3π)/2)

$${z}=\frac{\mathrm{1}−{e}^{{i}\frac{\pi}{\mathrm{3}}} }{\mathrm{1}+{e}^{{i}\frac{\pi}{\mathrm{3}}} }=\frac{{e}^{{i}\frac{\pi}{\mathrm{6}}} \left({e}^{−{i}\frac{\pi}{\mathrm{6}}} −{e}^{{i}\frac{\pi}{\mathrm{6}}} \right)}{{e}^{{i}\frac{\pi}{\mathrm{6}}} \left({e}^{−{i}\frac{\pi}{\mathrm{6}}} +{e}^{{i}\frac{\pi}{\mathrm{6}}} \right)}=−\frac{{e}^{{i}\frac{\pi}{\mathrm{6}}} −{e}^{{i}\frac{\pi}{\mathrm{6}}} }{{e}^{{i}\frac{\pi}{\mathrm{6}}} +{e}^{−{i}\frac{\pi}{\mathrm{6}}} } \\ $$$$\:\:\:=−\frac{\mathrm{2}{i}\mathrm{sin}\left(\frac{\pi}{\mathrm{6}}\right)}{\mathrm{2cos}\left(\frac{\pi}{\mathrm{6}}\right)}=−\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{i}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{e}^{\frac{\mathrm{3}\pi}{\mathrm{2}}{i}} \:\Rightarrow\mid{z}\mid=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\:,\:\mathrm{arg}\left({z}\right)=\frac{\mathrm{3}\pi}{\mathrm{2}} \\ $$

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Question-178810

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