ze-z-e-Obviously-z-1-Now-find-at-least-one-solution-for-z-C-

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 - 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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