ze^z =e Obviously z=1 Now find at least one solution for z∈C


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Question Number 203502 by Frix last updated on 20/Jan/24

$z e^{z} = e$ $Obviously z = 1$ $Now find at least one solution for z \in C$
Commented by Frix last updated on 22/Jan/24

$See question 203570$
Commented by a.lgnaoui last updated on 22/Jan/24
$I just have an idea for z=a+ib and to trasforme to( cosx+isinx) ze^z =e z=e^(1−z) a+ib=e^((1−a)−ib) e^(−ib) =cos b−isin b a+ib=e^(1−a) ×(cos b−isin b) a−(cos b)e^(1−a) +i[(b+e^(1−a) (sin b)] { ((z=a+ib)),((cos be^(1−a) =0 sin be^(1−1) =0)) :} { ((ze^z −e=[(a−c)),((b=)) :} z=e^(1−z) =e^((1−a)−ib) (e^(1−a) /(cos b−isin b))=(a−cos be^(1−a) )+ i(b+e^(1−a) sin b) { ((e^(1−a) cos b=a−cos be^(1−a) )),((e^(1−a) sin b=b+sin be^(1−a) )) :} { ((a=2cos be^(1−a) )),((b=2sin be^(1−a) )) :} (a/b)=(1/(tan b)) a=(b/(tan b)) z=((b+ib)/(tan b))=e^((1−a)−ib) ⇒ { ((((b(1+i))/(tan b×e^(1−a) ))=e^(−ib) )),((=cos b−isin b)) :} { (((b/(tan be^(1−a) ))=cos b=−sin b=sin (−b))),((b=(𝛑/4))) :} donc (((π/4)(1+i))/e^(1−a) )=((√2)/2)(1−i) e^(1−a) =(π/(2(√2))) ⇒a=1−ln((π/(2(√2)))) (√(a^2 +b^2 )) (acos b−isin b) { (((b/( (√(a^2 +b^2 ))))=−(√2))),() :} ((a/b))^2 +1=4 (a/b)=(√3) b=a((√3)/3) z=a+ib z=[1−ln((𝛑/(2(√2))))]+[((√3)/3)(1−ln(π/(2(√2))))]i S={1;(1−ln((π/(2(√2)))))+i((√3)/3)(1−ln((π/( 2(√2))))} It is clear that there is many roots if we transcorme the[ expressiin in Argθ= { ((cos θ=((1−ln(π/2(√2) ))/( (√((1−ln(π/2(√2))^2 +1/3(1−lnπ/2(√2))^2 )))))),((sin θ=(((√3) /3)/( (√((2−ln(π/2(√(2)]^2 )) +1/3(2−lnπ/2(√2) )^2 )))))) :}$
$I just have an idea for$ $z = a + ib and to trasforme to (cosx + isinx)$ ${ze}^{z} = e z = e^{1 - z}$ $a + ib = e^{(1 - a) - ib}$ $e^{- ib} = \cos b - i \sin b$ $a + ib = e^{1 - a} \times (\cos b - i \sin b)$ $a - (\cos b) e^{1 - a} + i [(b + e^{1 - a} (\sin b)]$ ${\begin{cases} z = a + ib \\ \cos {be}^{1 - a} = 0 \sin {be}^{1 - 1} = 0 \end{cases}$ ${\begin{cases} {ze}^{z} - e = [(a - c \\ b = \end{cases}$ $z = e^{1 - z} = e^{(1 - a) - ib}$ $\frac{e^{1 - a}}{\cos b - i \sin b} = (a - \cos {be}^{1 - a}) +$ $i (b + e^{1 - a} \sin b)$ ${\begin{cases} e^{1 - a} \cos b = a - \cos {be}^{1 - a} \\ e^{1 - a} \sin b = b + \sin {be}^{1 - a} \end{cases}$ ${\begin{cases} a = 2 \cos {be}^{1 - a} \\ b = 2 \sin {be}^{1 - a} \end{cases}$ $\frac{a}{b} = \frac{1}{\tan b} a = \frac{b}{\tan b}$ $z = \frac{b + ib}{\tan b} = e^{(1 - a) - ib} \Rightarrow$ ${\begin{cases} \frac{b (1 + i)}{\tan b \times e^{1 - a}} = e^{- ib} \\ = \cos b - i \sin b \end{cases}$ ${\begin{cases} \frac{b}{\tan {be}^{1 - a}} = \cos b = - \sin b = \sin (- b) \\ b = \frac{π}{4} \end{cases}$ $donc \frac{\frac{π}{4} (1 + i)}{e^{1 - a}} = \frac{\sqrt{2}}{2} (1 - i)$ $e^{1 - a} = \frac{π}{2 \sqrt{2}} \Rightarrow a = 1 - \ln (\frac{π}{2 \sqrt{2}})$ $\sqrt{a^{2} + b^{2}} (acos b - isin b)$ ${\begin{cases} \frac{b}{\sqrt{a^{2} + b^{2}}} = - \sqrt{2} \end{cases}$ ${(\frac{a}{b})}^{2} + 1 = 4 \frac{a}{b} = \sqrt{3} b = a \frac{\sqrt{3}}{3}$ $z = a + ib$ $z = [1 - \ln (\frac{π}{2 \sqrt{2}})] + [\frac{\sqrt{3}}{3} (1 - \ln \frac{π}{2 \sqrt{2}})] i$ $S = {1; (1 - \ln (\frac{π}{2 \sqrt{2}})) + i \frac{\sqrt{3}}{3} (1 - \ln (\frac{π}{2 \sqrt{2}})}$ $I t i s c l e a r t h a t t h e r e i s m a n y$ $r o o t s i f w e t r a n s c o r m e t h e [$ $e x p r e s s i i n i n A r g θ =$ ${\begin{cases} \cos θ = \frac{1 - \ln (π / 2 \sqrt{2})}{\sqrt{(1 - \ln {(π / 2 \sqrt{2})}^{2} + 1 / 3 {(1 - \ln π / 2 \sqrt{2})}^{2}}} \\ \sin θ = \frac{\sqrt{3} / 3}{\sqrt{(2 - \ln (π / 2 \sqrt{{2)]}^{2}} + 1 / 3 {(2 - \ln π / 2 \sqrt{2})}^{2}}} \end{cases}$
	Answered by a.lgnaoui last updated on 20/Jan/24

	${ze}^{z} - e = 0$ $\ln z + z = 1$ $Solution est donne par graphe$ ${ze}^{z} - e = (z - 1) Q (z)$ $Q (z) = \frac{{ze}^{z} - e}{z - 1}$ $solugiin pour \frac{{ze}^{z} - e}{z - 1} est [donne$ $par graphe$
	Commented by Frix last updated on 20/Jan/24

	$There is only one real solution but there$ $are \infty complex solutions .$
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