Solutions to $e^z+z-a=0$ with $Re(z)>0$ and $a$ is a real number larger than 1 In an exercise I am required to show that the only solution to $e^z+z-a=0$ in the right half plane is a real one, where $a$ is fixed a real number larger than 1. I am not able to work it out. I tried to used the Argument Principle to compute the number of zeros of this function, but I cannot make my integral converge. Could anybody give me a hand?
Thanks very much!

Let $a=\log(3\pi/2)$, so $a$ is real and $a\gt1$. Let $z=\log(3\pi/2)+(3\pi/2)i$, so the real part of $z$ is positive, and $z$ is not real. Then $$\eqalign{e^z+z&=e^{\log(3\pi/2)}e^{(3\pi/2)i}+z=(3\pi/2)(\cos(3\pi/2)+i\sin(3\pi/2))+z\cr&=-(3\pi/2)i+\log(3\pi/2)+(3\pi/2)i=a\cr}$$
So, for $a=\log(3\pi/2)$, there is a nonreal solution in the right half-plane. 
