Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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)

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com