solve-sinz-2-zfromC-

Question Number 37296 by math khazana by abdo last updated on 11/Jun/18

$${solve}\:{sinz}\:=\mathrm{2}\:\:\:\:\:{zfromC} \\ $$$$ \\ $$

Commented by prof Abdo imad last updated on 17/Jun/18

sinz =2 ⇔((e^(iz) −e^(−iz) )/2) =2 ⇔e^(iz) −e^(−iz) =4⇒ let t =e^(iz) ⇒t−(1/t)=4 ⇒t^2 −1=4t ⇒ t^2 −4t−1=0 Δ^′ =4 +1=5 ⇒ t_1 =2+(√5) and t_2 =2−(√5) e^(iz) =2+(√5)⇒iz =ln(2+(√5)) ⇒z=−iln(2+(√5)) e^(iz) =2−(√5) ⇒iz =ln(2−(√5)) =ln(−1) +ln(−2 +(√5)) = iπ +ln(−2+(√5)) ⇒ z =π −iln(−2+(√5)) so the solutions for this equation are −iln(2+(√5)) and π −iln(−2+(√5)) .

$${sinz}\:=\mathrm{2}\:\Leftrightarrow\frac{{e}^{{iz}} \:−{e}^{−{iz}} }{\mathrm{2}}\:=\mathrm{2}\:\Leftrightarrow{e}^{{iz}} \:−{e}^{−{iz}} =\mathrm{4}\Rightarrow \\ $$$${let}\:{t}\:={e}^{{iz}} \:\Rightarrow{t}−\frac{\mathrm{1}}{{t}}=\mathrm{4}\:\Rightarrow{t}^{\mathrm{2}} −\mathrm{1}=\mathrm{4}{t}\:\Rightarrow \\ $$$${t}^{\mathrm{2}} −\mathrm{4}{t}−\mathrm{1}=\mathrm{0}\: \\ $$$$\Delta^{'} =\mathrm{4}\:+\mathrm{1}=\mathrm{5}\:\Rightarrow\:{t}_{\mathrm{1}} =\mathrm{2}+\sqrt{\mathrm{5}}\:\:\:{and}\:{t}_{\mathrm{2}} =\mathrm{2}−\sqrt{\mathrm{5}} \\ $$$${e}^{{iz}} =\mathrm{2}+\sqrt{\mathrm{5}}\Rightarrow{iz}\:={ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\:\Rightarrow{z}=−{iln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right) \\ $$$${e}^{{iz}} \:=\mathrm{2}−\sqrt{\mathrm{5}}\:\Rightarrow{iz}\:={ln}\left(\mathrm{2}−\sqrt{\mathrm{5}}\right) \\ $$$$={ln}\left(−\mathrm{1}\right)\:+{ln}\left(−\mathrm{2}\:+\sqrt{\mathrm{5}}\right)\:=\:{i}\pi\:+{ln}\left(−\mathrm{2}+\sqrt{\mathrm{5}}\right)\:\Rightarrow \\ $$$${z}\:=\pi\:−{iln}\left(−\mathrm{2}+\sqrt{\mathrm{5}}\right)\:{so}\:{the}\:{solutions}\:{for} \\ $$$${this}\:{equation}\:{are} \\ $$$$−{iln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\:{and}\:\pi\:−{iln}\left(−\mathrm{2}+\sqrt{\mathrm{5}}\right)\:. \\ $$$$ \\ $$

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