This might be a silly question, but I can't find for any solution.
I want to solve $e^{2z}(1+z)=1-z$ over the complex numbers $\mathbb{C}$.
It looks like that the solutions are given by $z=0$ and $ir_{n}$, where $\{r_{n}\}$ are real roots of $\tan r = -r$.
Is there an easy way to prove that all solutions of this equation are purely imaginary so that the above solutions are all solutions for the given equation?
Thanks in advance.
Let $z=x+yi$ where $x,y\in\mathbb R$.
Then, the equation $e^{2z}(1+z)=1-z$ can be written as $$e^{2x+2yi}(1+x+yi)=1-x-yi\tag1$$ The magnitude squared of this expression is $$|e^{2x+2yi}|^2|1+x+yi|^2=|1-x-yi|^2$$ i.e. $$e^{4x}\bigg((1+x)^2+y^2\bigg)=(1-x)^2+y^2$$ i.e. $$e^{4x}\bigg((1+x)^2+y^2\bigg)=-4x+(1+x)^2+y^2\tag2$$
Since $(x,y)\not=(-1,0)$, dividing the both sides of $(2)$ by $(1+x)^2+y^2$ gives $$e^{4x}=\frac{-4x}{(1+x)^2+y^2}+1$$ i.e. $$\frac{4x}{(1+x)^2+y^2}=1-e^{4x}\tag3$$
If $x\gt 0$, then LHS of $(3)$ is positive while RHS is negative.
If $x\lt 0$, then LHS of $(3)$ is negative while RHS is positive.
Therefore, we have to have $x=0$.
From $(1)$, we get $$e^{2yi}(1+yi)=1-yi$$ i.e. $$\bigg(\cos(2y)+i\sin(2y)\bigg)(1+yi)=1-yi$$ from which we get $$\cos(2y)-y\sin(2y)=1\tag4$$ $$y\cos(2y)+\sin(2y)=-y\tag5$$
We have $$(4)\iff \cos y(\cos y-y\sin y)=1$$ $$(5)\iff \cos y(y\cos y+\sin y)=0$$
Since $\cos y\not=0$, we have $y\cos y+\sin y=0$, i.e. $y=-\tan y$.
On the other hand, if $y=-\tan y$, then $\cos y(\cos y-y\sin y)=\cos^2y+\tan y\cos y\sin y=\cos^2y+\sin^2y=1$.
So, we can say that $(4)(5)\iff y=-\tan y$.
Therefore, the solutions of the equation $e^{2z}(1+z)=1-z$ are $$\color{red}{z=iy\qquad\text{where $y$ is a real number such that $y=-\tan y$}}$$
