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This is probably a lame question, but what is the general approach to solving $\log z +z \sigma +1=0$ for $z$? Wolfram Alpha obtains a Lambert W-function, but I don't quite see how.

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Assuming $\sigma\neq 0$, you need something like Lambert's W function.

Taking exponentials of both sides, we have $\begin{align*} e^{\log z + z\sigma + 1} &= 1\\ e^{\log z}e^{\sigma z} e&= 1\\ ze^{\sigma z}&= \frac{1}{e}\\ \sigma ze^{\sigma z} &= \frac{\sigma}{e} \end{align*}$ Now let $x=\sigma z$. Then the equation is equivalent to $xe^x = \frac{\sigma}{e},$ which means that $x=W(\frac{\sigma}{e})$, hence $z = \frac{x}{\sigma} = \frac{1}{\sigma} W\left(\frac{\sigma}{e}\right).$

(If $\sigma=0$, then you just have $\log z + 1 = 0$, or $z=e^{-1}$.)

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    yes you are right. I mean $e^x=1+x+O(x^2)$2011-06-12