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Good day. I want to solve the following equation for $z\in\mathbb{C}$, $$e^{\alpha zi}+\beta z+\gamma=0$$ for some constants $\alpha,\beta,\gamma\in\mathbb{C}$. My question is: is there a way to solve it getting the explicit values of $z$? Is it even possible? I tried to expand the exponential in its Taylor series but it led me nowhere. Thanks for your help.

Tom Builder
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2 Answers2

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Your equality is of the form $(a,b,c\ne0)$: $$ae^x+bx+c=0$$

Let us find the general solution .

First it's equivalent to : $$\frac{a}{b}e^x+x+\frac{c}{b}=0$$ Or $$\frac{a}{b}e^x=-\left(x+\frac{c}{b}\right)$$ Or: $$\frac{b}{a}e^{-x}\left(-\left(x+\frac{c}{b}\right)\right)=1$$ Or: $$e^{-x-\frac{c}{b}}\left(-\left(x+\frac{c}{b}\right)\right)=\frac{a}{b}e^{-\frac{c}{b}}$$

Put $u=-x-\frac{c}{b}$ and $v=\frac{a}{b}e^{-\frac{c}{b}}$ wich gets :

$$ue^u=v$$

Now you can solve it using the definition of the Lambert's function

Miss and Mister cassoulet char
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  • Thanks! It seems correct to me! I'm accepting this answer. One quick question. I have to admit I didn't know the Lambert's Function. I saw it on Wikipedia, but do you know a book where I can find it in an easy way to understand it? – Tom Builder May 04 '20 at 15:24
  • See here https://math.stackexchange.com/questions/2969478/book-on-the-lambert-w-function – Miss and Mister cassoulet char May 04 '20 at 15:28
  • When everything is complex numbers. Is there a way to obtain a series representation of the value of W at that point?, is it possible to obtain where is the range of W at that point? I am particularly interested in knowing if the solutions of the initial equation could be near the real axis – Tom Builder May 04 '20 at 23:37
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The general solution can be written in terms of the so-called $W$ function (see https://en.wikipedia.org/wiki/Lambert_W_function), $$ z = \frac{-\beta \, W\left(\frac{i \alpha }{\beta} e^{-\frac{i \alpha \gamma }{\beta}} \right)-i \alpha \gamma }{i \alpha \beta } $$ but you cannot express $W$ in terms of "elementary functions".

Expanding in Taylor series does not help because it is a very high degree (infinite) polynomial in $z$, which you cannot solve. Hence, the expansion is quite useless, unless you want to expand for $|\alpha|\approx 0 $ and find an approximate solution. For example, the solution up to the second order in $\alpha$ is $$ z = -\frac{1+\gamma}{\beta} + i \alpha \frac{1+\gamma}{\beta^2} +O(\alpha^2) $$ You can easily find the solution up to the third order, but you have to expand the exponential by keeping the term in $\alpha^2$. If you are patient enough you can expand the exponential to the third order as well and solve a cubic equation in $z$...

Quillo
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