Find the number of zeros of $z^{3}+2z^{2}-z-2+e^{z}$ which lies in the left half plane

Question

I am working on an exercise that

How many solutions of equation $$z^{3}+2z^{2}-z-2+e^{z}=0,$$ lie in the left half plane?

The first idea was definitely the Rouche's Theorem, but then we need to use a big circle (for $R$ large enough) centered at the origin, and then we use Rouche's theorem.

But the problem is it is really hard to argue that this function has the same number of zeros between left semi-circle and right semi-circle, so that we can just use half of the total number of zeros.

The difficulty of applying Argument Principle is also self-evident.

Without these two tools, what should I start with to compute the number of zeros of this function?

Thank you!

Answer 1

Note that $p(z)=z^3+2z^2-z-2=(z+2)(z-1)(z+1)$. For purely imaginary $z$ it follows that $\lvert p(z) \rvert \geq 2$. For some large enough fixed radius $R$ and $\lvert z \rvert = R$ also $\lvert p(z) \rvert \geq 2$. Since $\lvert \exp(z) \rvert \leq 1$ on the left half plane you can use Rouché to conclude that the number of zeroes on the left half plane equals the number of roots of $p(z)$ there.

Answer 2

Hint.

Making $z= x + i y$ and taking real and imaginary parts we have

$$ \cases{ x^3+2 x^2-3 x y^2+e^x \cos (y)-x-2 y^2-2=0\\ 3 x^2 y+4 x y+e^x \sin (y)-y^3-y=0 } $$

also

$$ \cos(y)^2+\sin(y)^2=1\Rightarrow e^{-2 x} \left((x-1)^2+y^2\right) \left((x+1)^2+y^2\right) \left((x+2)^2+y^2\right)-1=0 $$

contains all needed solutions.

Follows a plot showing the zero locations at the intersection of red and blue curves, as well in black, the last relationship.

and also a detailed plot.