The equation...
$\displaystyle e^{z}=z$ (1)
... in the complex domain is equilvalent to a couple of equations in real domain. Setting $\displaystyle z= \sigma + i\cdot \omega$ the (1) became...
$\displaystyle e^{\sigma} \cdot \cos \omega = \sigma$
$\displaystyle e^{\sigma}\cdot \sin \omega = \omega$ (2)
If $\displaystyle \sigma^{*} + i\cdot \omega^{*}$ is a solution of (2) then the proposed D.E. becomes...
$\displaystyle y^{'}= \sigma^{*} + i\cdot \omega^{*} $ (3)
... and its solution is...
$\displaystyle y^{'}= \{\sigma^{*} + i\cdot \omega^{*}\} \cdot x + c$ (4)
The question is now: does (2) have solutions?...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
The answers is done by the follwing lemma of complex variable functions theory...
Given the complex variable function...
$\displaystyle w = f(z) = c_{1} z + c_{2} z^{2} + c_{3} z^{3} + ... $ , $\displaystyle c_{1} \ne 0$ (1)
... holomorphic in a circle $\displaystyle |z|< \epsilon$ , its inverse function is done by the Taylor expansion...
$\displaystyle z= f^{-1} (w) = d_{1} w + d_{2} w^{2} + d_{3} w^{3} + ...$ , $\displaystyle d_{n} = \frac{1}{n!} \lim _{z \rightarrow 0} \frac {d^{n-1}}{dz^{n-1}} \{\frac{z}{f(z)}\}^{n}$ (2)
Now we use this lemma to find the solution $\displaystyle z=z(w)$ of the trascendental equation...
$\displaystyle w= z\cdot e^{-z}$ (3)
With the formulas (2) we obtain the Taylor expansion...
$\displaystyle z= \sum_{n=1}^{\infty} \frac {n^{n-1}}{n!} w^{n} $ (4)
... which converges only for $\displaystyle w=0$. But our equation was...
$\displaystyle z\cdot e^{-z}=1$ (5)
... and then it has been demonstrated it has no solutions...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$