Rouche’s theorem Application right half plane

Question

Show that there are two roots of $(z^4+2)e^z+z+1=0$ which lies in the right half plane (that is, Rez > 0).

I have been stuck in this problem for a week, and I only know that $|e^z|$>1

Can somebody give me some hints? Thanks a lot!

Answer 1

Along the imaginary axis, $|(z^4 + 2)e^z| > |z+1|$ (note that $z^4$ is real and positive in this case), while on a large enough semicircle centered at the origin going into the right half-plane, we clearly have $|(z^4 + 2)e^z| > |z+1|$. Using that semicircle together with its diameter, Rouche's theorem implies that $(z^4 + 2)e^z + z + 1$ has as many zeroes within this semicircle as does $(z^4 + 2)e^z$, which clearly has two. Since we can choose the semicircle to be as large as we want, there can be only two zeroes in the right half-plane.