Application of Rouche's Theorem to find the no. of roots Find the no. of zeros of the function $\phi(z)=\alpha+z-e^z,$where $1<\alpha<\infty$ in the region where $Re(z)\le0.$
I have considered a semicircle in left half plane and considered the two functions as $f(z)=\alpha+z,g(z)=-e^z$ Now the difficulty is to show that $|f(z)|>|g(z)|$ on the boundary of the semicircle, I am not sure about my approach. Please help me .

Consider a semicircle contour in the left half-plane, with $R > 2\alpha$ (the distance from $0$ to the endpoints of the line segment of our contour along the imaginary axis).
If you take $f(z) = z +a$ as you did, and $g(z) = z + a - e^{z}$, then we can show that $|f(z) - g(z)| < |f(z)|$ (equivalent formulation of Rouche's Theorem) along the contour.
You'll have to do this for both the line segment $[-iR, iR]$ and the half-circle. 
I'll show you how do it for the line segment. Parametrize $z = it$ on $[-iR, iR]$, then:
$$ |f(z) - g(z)| = |e^{it}| = 1 < a \leq \sqrt{a^{2} + t^{2}} = |a + it| = |f(z)|$$
Using the hints given in the comments, you should be able to show $|f(z) - g(z)| < |f(z)|$ on $C_{R}$.
After that, you'll be able to conclude that $f$ and $g$ have the same number of zeros in the interior of our contour (in particular, just one since $f$ only has one).Then if we take $R \rightarrow \infty$ then it $g$ can be seen to have only one zero in the left half-plane.
