Question Number 68207 by mr W last updated on 07/Sep/19 | ||
$${solve}\:{for}\:{x}\in\mathbb{C} \\ $$$$\mathrm{sin}\:{x}={z}\:\:\:\left({z}={a}+{bi}={re}^{{i}\theta} \right) \\ $$ | ||
Commented by mathmax by abdo last updated on 07/Sep/19 | ||
$${sinx}\:={z}\:\Leftrightarrow\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}={z}\:\Leftrightarrow{e}^{{ix}} −{e}^{−{ix}} =\mathrm{2}{ir}\:{e}^{{i}\theta} \\ $$$${let}\:{t}\:={e}^{{ix}} \:\Rightarrow{ix}\:={ln}\left({t}\right)+{i}\mathrm{2}{k}\pi\:\:\:{with}\:{k}\in{Z} \\ $$$$\left({e}\right)\:\Rightarrow{t}−{t}^{−\mathrm{1}\:} =\mathrm{2}{ir}\:{e}^{{i}\theta} \:\Rightarrow{t}^{\mathrm{2}} −\mathrm{1}\:=\mathrm{2}{ir}\:{e}^{{i}\theta} \:{t}\:\Rightarrow \\ $$$${t}^{\mathrm{2}} −\mathrm{2}{ire}^{{i}\theta} {t}\:−\mathrm{1}\:=\mathrm{0}\:\rightarrow\Delta^{'} =\left({ire}^{{i}\theta} \right)^{\mathrm{2}} \:+\mathrm{1}\:=−{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} \:+\mathrm{1} \\ $$$${t}_{\mathrm{1}} ={ir}\:{e}^{{i}\theta} \:+\sqrt{\mathrm{1}−{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} \:}\:\:\:\:{and}\:{t}_{\mathrm{2}} ={ir}\:{e}^{{i}\theta} −\sqrt{\mathrm{1}−{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} } \\ $$$${ix}\:={ln}\left({t}_{\mathrm{1}} \right)+{i}\mathrm{2}{k}\pi\:\Rightarrow−{x}\:={ilnt}_{\mathrm{1}} −\mathrm{2}{k}\pi\:\Rightarrow{x}\:=−{ilnt}_{\mathrm{1}} +\mathrm{2}{k}\pi \\ $$$$=−{iln}\left({r}\:{e}^{{i}\theta} −{i}\sqrt{\mathrm{1}−{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} }\right)+\mathrm{2}{k}\pi \\ $$$${ix}\:={lnt}_{\mathrm{2}} +{i}\mathrm{2}{k}\pi\:\Rightarrow−{x}\:={ilnt}_{\mathrm{2}} −\mathrm{2}{k}\pi\:\Rightarrow{x}\:=−{ilnt}_{\mathrm{2}} +\mathrm{2}{k}\pi \\ $$$$=−{iln}\left({ir}\:{e}^{{i}\theta} −\sqrt{\mathrm{1}−{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\theta} }\right)+\mathrm{2}{k}\pi \\ $$ | ||
Answered by Smail last updated on 07/Sep/19 | ||
$$\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}}{e}^{−{i}\pi/\mathrm{2}} ={re}^{{i}\theta} \\ $$$${e}^{\mathrm{2}{ix}} −\mathrm{1}−\mathrm{2}{re}^{{ix}} ×{e}^{{i}\left(\theta+\pi/\mathrm{2}\right)} =\mathrm{0} \\ $$$$\left({e}^{{ix}} −{re}^{{i}\left(\theta+\pi/\mathrm{2}\right)} \right)^{\mathrm{2}} =\mathrm{1}+{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\left(\theta+\pi/\mathrm{2}\right)} \\ $$$${e}^{{ix}} =\underset{−} {+}\left(\mathrm{1}+{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\left(\theta+\pi/\mathrm{2}\right)} \right)^{\mathrm{1}/\mathrm{2}} +{re}^{{i}\left(\theta+\pi/\mathrm{2}\right)} \\ $$$${x}=−{iln}\left({re}^{{i}\left(\theta+\pi/\mathrm{2}\right)} \underset{−} {+}\sqrt{\mathrm{1}+{r}^{\mathrm{2}} {e}^{\mathrm{2}{i}\left(\theta+\pi/\mathrm{2}\right)} }\right) \\ $$$${x}=−{isinh}^{−\mathrm{1}} \left({re}^{{i}\left(\theta+\pi/\mathrm{2}\right)} \right) \\ $$ | ||