# Solution of transcendental complex function

I'm trying to solve this function

$$z^2 + 2z + 2\exp(-z) = 0$$

In one solution I found out the exponential function has been approximated to $\exp(-z)=1-z$, which makes it and easy polynomial to solve.

But is it the complete solution of this equation? Or are there better methods to solve this equation?

• Of course replacing $\exp(-z)$ with $1-z$ is only an approximation. Valid for $z$ near $0$. So you only expect an approximate solution to the original equation. And for solutions not close to $0$, this will not work at all. Jul 19, 2017 at 17:59
• google.com/… Jul 19, 2017 at 18:00
• set $$z=x+iy$$ to solve your equation Jul 19, 2017 at 18:04
• @Dr.SonnhardGraubner Huh? How in the world is this useful? It hardly does anything to simplify the problem. Jul 19, 2017 at 18:47
• hm ok i will post it Jul 19, 2017 at 18:50

HINT: you will have $$(x+iy)^2+2(x+iy)+2e^{-x}(\cos(y)+i\sin(y))=0$$ and you must solve the System $$x^2-y^2+2x+2e^{-x}\cos(y)=0$$ $$2xy+2y+2e^{-x}\sin(y)=0$$ by a numerical method we obtain $$x \approx -3.293017589, y \approx 6.912667999$$
HINT: Multiply both side of your equation $$z^2+2z+2e^{-z}=0$$ by $e^z$, from which it turns into $$(e^{z/2}z)^2+4ze^{z/2}+4=-2(ze^z-2ze^{z/2}-1)$$ i.e. $$\left(ze^{z/2}+2\right)^2=-2(ze^z-2ze^{z/2}-1)$$ and starting from this, you can try to manipulate RHS in order to get something useful.