Question Number 203502 by Frix last updated on 20/Jan/24
$${z}\mathrm{e}^{{z}} =\mathrm{e} \\ $$$$\mathrm{Obviously}\:{z}=\mathrm{1} \\ $$$$\mathrm{Now}\:\mathrm{find}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{solution}\:\mathrm{for}\:{z}\in\mathbb{C} \\ $$
Commented by Frix last updated on 22/Jan/24
$$\mathrm{See}\:\mathrm{question}\:\mathrm{203570} \\ $$
Commented by a.lgnaoui last updated on 22/Jan/24
![I just have an idea for z=a+ib and to trasforme to( cosx+isinx) ze^z =e z=e^(1−z) a+ib=e^((1−a)−ib) e^(−ib) =cos b−isin b a+ib=e^(1−a) ×(cos b−isin b) a−(cos b)e^(1−a) +i[(b+e^(1−a) (sin b)] { ((z=a+ib)),((cos be^(1−a) =0 sin be^(1−1) =0)) :} { ((ze^z −e=[(a−c)),((b=)) :} z=e^(1−z) =e^((1−a)−ib) (e^(1−a) /(cos b−isin b))=(a−cos be^(1−a) )+ i(b+e^(1−a) sin b) { ((e^(1−a) cos b=a−cos be^(1−a) )),((e^(1−a) sin b=b+sin be^(1−a) )) :} { ((a=2cos be^(1−a) )),((b=2sin be^(1−a) )) :} (a/b)=(1/(tan b)) a=(b/(tan b)) z=((b+ib)/(tan b))=e^((1−a)−ib) ⇒ { ((((b(1+i))/(tan b×e^(1−a) ))=e^(−ib) )),((=cos b−isin b)) :} { (((b/(tan be^(1−a) ))=cos b=−sin b=sin (−b))),((b=(𝛑/4))) :} donc (((π/4)(1+i))/e^(1−a) )=((√2)/2)(1−i) e^(1−a) =(π/(2(√2))) ⇒a=1−ln((π/(2(√2)))) (√(a^2 +b^2 )) (acos b−isin b) { (((b/( (√(a^2 +b^2 ))))=−(√2))),() :} ((a/b))^2 +1=4 (a/b)=(√3) b=a((√3)/3) z=a+ib z=[1−ln((𝛑/(2(√2))))]+[((√3)/3)(1−ln(π/(2(√2))))]i S={1;(1−ln((π/(2(√2)))))+i((√3)/3)(1−ln((π/( 2(√2))))} It is clear that there is many roots if we transcorme the[ expressiin in Argθ= { ((cos θ=((1−ln(π/2(√2) ))/( (√((1−ln(π/2(√2))^2 +1/3(1−lnπ/2(√2))^2 )))))),((sin θ=(((√3) /3)/( (√((2−ln(π/2(√(2)]^2 )) +1/3(2−lnπ/2(√2) )^2 )))))) :}](https://www.tinkutara.com/question/Q203580.png)
$$\mathrm{I}\:\mathrm{just}\:\mathrm{have}\:\mathrm{an}\:\mathrm{idea}\:\mathrm{for}\: \\ $$$$\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{ib}}\:\:\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{trasforme}}\:\boldsymbol{\mathrm{to}}\left(\:\boldsymbol{\mathrm{cosx}}+\boldsymbol{\mathrm{isinx}}\right) \\ $$$$\boldsymbol{\mathrm{ze}}^{\boldsymbol{\mathrm{z}}} =\boldsymbol{\mathrm{e}}\:\:\:\:\:\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{e}}^{\mathrm{1}−\boldsymbol{\mathrm{z}}} \\ $$$$\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{ib}}=\boldsymbol{\mathrm{e}}^{\left(\mathrm{1}−\boldsymbol{\mathrm{a}}\right)−\boldsymbol{\mathrm{ib}}} \\ $$$$\boldsymbol{\mathrm{e}}^{−\boldsymbol{\mathrm{ib}}} =\mathrm{cos}\:\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{i}}\mathrm{sin}\:\boldsymbol{\mathrm{b}} \\ $$$$\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{ib}}=\boldsymbol{\mathrm{e}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} ×\left(\mathrm{cos}\:\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{i}}\mathrm{sin}\:\boldsymbol{\mathrm{b}}\right) \\ $$$$\:\:\boldsymbol{\mathrm{a}}−\left(\mathrm{cos}\:\boldsymbol{\mathrm{b}}\right)\boldsymbol{\mathrm{e}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} +\boldsymbol{\mathrm{i}}\left[\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{e}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} \left(\mathrm{sin}\:\boldsymbol{\mathrm{b}}\right)\right]\right. \\ $$$$\begin{cases}{\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{ib}}}\\{\mathrm{cos}\:\boldsymbol{\mathrm{be}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} =\mathrm{0}\:\:\:\:\:\mathrm{sin}\:\boldsymbol{\mathrm{be}}^{\mathrm{1}−\mathrm{1}} =\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{ze}}^{\boldsymbol{\mathrm{z}}} −\boldsymbol{\mathrm{e}}=\left[\left(\boldsymbol{\mathrm{a}}−\mathrm{c}\right.\right.}\\{\mathrm{b}=}\end{cases} \\ $$$$\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{e}}^{\mathrm{1}−\boldsymbol{\mathrm{z}}} =\boldsymbol{\mathrm{e}}^{\left(\mathrm{1}−\boldsymbol{\mathrm{a}}\right)−\boldsymbol{\mathrm{ib}}} \\ $$$$\:\:\:\:\frac{\boldsymbol{\mathrm{e}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} }{\mathrm{cos}\:\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{i}}\mathrm{sin}\:\boldsymbol{\mathrm{b}}}=\left(\boldsymbol{\mathrm{a}}−\mathrm{cos}\:\boldsymbol{\mathrm{be}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} \right)+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{i}}\left(\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{e}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} \mathrm{sin}\:\boldsymbol{\mathrm{b}}\right) \\ $$$$\begin{cases}{\boldsymbol{\mathrm{e}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} \mathrm{cos}\:\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{a}}−\mathrm{cos}\:\boldsymbol{\mathrm{be}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} }\\{\boldsymbol{\mathrm{e}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} \mathrm{sin}\:\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{b}}+\mathrm{sin}\:\boldsymbol{\mathrm{be}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} }\end{cases} \\ $$$$\:\:\begin{cases}{\boldsymbol{\mathrm{a}}=\mathrm{2cos}\:\boldsymbol{\mathrm{be}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} }\\{\boldsymbol{\mathrm{b}}=\mathrm{2sin}\:\boldsymbol{\mathrm{be}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} }\end{cases} \\ $$$$\:\:\frac{\boldsymbol{\mathrm{a}}}{\boldsymbol{\mathrm{b}}}=\frac{\mathrm{1}}{\mathrm{tan}\:\boldsymbol{\mathrm{b}}}\:\:\:\:\boldsymbol{\mathrm{a}}=\frac{\boldsymbol{\mathrm{b}}}{\mathrm{tan}\:\boldsymbol{\mathrm{b}}} \\ $$$$\:\:\boldsymbol{\mathrm{z}}=\frac{\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{ib}}}{\mathrm{tan}\:\boldsymbol{\mathrm{b}}}=\boldsymbol{\mathrm{e}}^{\left(\mathrm{1}−\boldsymbol{\mathrm{a}}\right)−\boldsymbol{\mathrm{ib}}} \Rightarrow \\ $$$$\:\:\begin{cases}{\frac{\boldsymbol{\mathrm{b}}\left(\mathrm{1}+\boldsymbol{\mathrm{i}}\right)}{\mathrm{tan}\:\mathrm{b}×\mathrm{e}^{\mathrm{1}−\mathrm{a}} }=\mathrm{e}^{−\mathrm{ib}} }\\{=\mathrm{cos}\:\boldsymbol{\mathrm{b}}−\boldsymbol{\mathrm{i}}\mathrm{sin}\:\boldsymbol{\mathrm{b}}}\end{cases} \\ $$$$\:\:\begin{cases}{\frac{\boldsymbol{\mathrm{b}}}{\mathrm{tan}\:\boldsymbol{\mathrm{be}}^{\mathrm{1}−\boldsymbol{\mathrm{a}}} }=\mathrm{cos}\:\boldsymbol{\mathrm{b}}=−\mathrm{sin}\:\boldsymbol{\mathrm{b}}=\mathrm{sin}\:\left(−\boldsymbol{\mathrm{b}}\right)}\\{\boldsymbol{\mathrm{b}}=\frac{\boldsymbol{\pi}}{\mathrm{4}}}\end{cases} \\ $$$$\mathrm{donc}\:\:\:\frac{\frac{\pi}{\mathrm{4}}\left(\mathrm{1}+\mathrm{i}\right)}{\mathrm{e}^{\mathrm{1}−\mathrm{a}} }=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{1}−\mathrm{i}\right) \\ $$$$\:\:\:\: \\ $$$$\:\:\:\mathrm{e}^{\mathrm{1}−\mathrm{a}} =\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\:\:\Rightarrow\boldsymbol{\mathrm{a}}=\mathrm{1}−\mathrm{ln}\left(\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\right) \\ $$$$\:\:\:\:\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \:}\:\left(\mathrm{acos}\:\mathrm{b}−\mathrm{isin}\:\mathrm{b}\right) \\ $$$$\:\:\:\begin{cases}{\frac{\mathrm{b}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}=−\sqrt{\mathrm{2}}}\\{}\end{cases} \\ $$$$\:\:\:\left(\frac{\mathrm{a}}{\mathrm{b}}\right)^{\mathrm{2}} +\mathrm{1}=\mathrm{4}\:\:\:\:\:\frac{\mathrm{a}}{\mathrm{b}}=\sqrt{\mathrm{3}}\:\:\:\:\:\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{a}}\frac{\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$$\:\:\boldsymbol{\mathrm{z}}=\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{ib}} \\ $$$$ \\ $$$$\:\:\boldsymbol{\mathrm{z}}=\left[\mathrm{1}−\boldsymbol{\mathrm{ln}}\left(\frac{\boldsymbol{\pi}}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\right]+\left[\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{ln}\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\right]\boldsymbol{\mathrm{i}} \\ $$$$ \\ $$$$ \\ $$$$\:\:\: \\ $$$$\:\:\boldsymbol{\mathrm{S}}=\left\{\mathrm{1};\left(\mathrm{1}−\mathrm{ln}\left(\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\right)\right)+\mathrm{i}\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{ln}\left(\frac{\pi}{\:\mathrm{2}\sqrt{\mathrm{2}}}\right)\right\}\right. \\ $$$$\:\:\:{It}\:{is}\:{clear}\:{that}\:{there}\:{is}\:{many} \\ $$$$\:\:{roots}\:{if}\:{we}\:{transcorme}\:{the}\left[\right. \\ $$$${expressiin}\:{in}\:\:{Arg}\theta= \\ $$$$\:\:\:\begin{cases}{\mathrm{cos}\:\theta=\frac{\mathrm{1}−\mathrm{ln}\left(\pi/\mathrm{2}\sqrt{\mathrm{2}}\:\right)}{\:\sqrt{\left(\mathrm{1}−\mathrm{ln}\left(\pi/\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}/\mathrm{3}\left(\mathrm{1}−\mathrm{ln}\pi/\mathrm{2}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \right.}}}\\{\mathrm{sin}\:\theta=\frac{\sqrt{\mathrm{3}}\:/\mathrm{3}}{\:\sqrt{\left(\mathrm{2}−\mathrm{ln}\left(\pi/\mathrm{2}\sqrt{\left.\mathrm{2}\left.\right)\right]^{\mathrm{2}} }\:+\mathrm{1}/\mathrm{3}\left(\mathrm{2}−\mathrm{ln}\pi/\mathrm{2}\sqrt{\mathrm{2}}\:\right)^{\mathrm{2}} \right.\right.}}}\end{cases} \\ $$$$ \\ $$$$\: \\ $$
Answered by a.lgnaoui last updated on 20/Jan/24
$$\mathrm{ze}^{\mathrm{z}} −\mathrm{e}=\mathrm{0} \\ $$$$\mathrm{ln}\boldsymbol{\mathrm{z}}+\boldsymbol{\mathrm{z}}=\mathrm{1} \\ $$$$\boldsymbol{\mathrm{Solution}}\:\boldsymbol{\mathrm{est}}\:\boldsymbol{\mathrm{donne}}\:\boldsymbol{\mathrm{par}}\:\boldsymbol{\mathrm{graphe}} \\ $$$$ \\ $$$$\boldsymbol{\mathrm{ze}}^{\boldsymbol{\mathrm{z}}} −\boldsymbol{\mathrm{e}}=\left(\boldsymbol{\mathrm{z}}−\mathrm{1}\right)\boldsymbol{\mathrm{Q}}\left(\boldsymbol{\mathrm{z}}\right) \\ $$$$\boldsymbol{\mathrm{Q}}\left(\boldsymbol{\mathrm{z}}\right)=\frac{\boldsymbol{\mathrm{ze}}^{\boldsymbol{\mathrm{z}}} −\boldsymbol{\mathrm{e}}}{\boldsymbol{\mathrm{z}}−\mathrm{1}}\:\: \\ $$$$\mathrm{solugiin}\:\mathrm{pour}\:\:\frac{\boldsymbol{\mathrm{ze}}^{\boldsymbol{\mathrm{z}}} −\boldsymbol{\mathrm{e}}}{\boldsymbol{\mathrm{z}}−\mathrm{1}}\:\:\boldsymbol{\mathrm{est}}\left[\boldsymbol{\mathrm{donne}}\:\right. \\ $$$$\boldsymbol{\mathrm{par}}\:\boldsymbol{\mathrm{graphe}} \\ $$
Commented by Frix last updated on 20/Jan/24
$$\mathrm{There}\:\mathrm{is}\:\mathrm{only}\:\mathrm{one}\:\mathrm{real}\:\mathrm{solution}\:\mathrm{but}\:\mathrm{there} \\ $$$$\mathrm{are}\:\infty\:\mathrm{complex}\:\mathrm{solutions}. \\ $$