Solving a transcendental equation with complex exponential 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.  
 A: 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 
A: 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 $z$...
