Problem 7

Exercise 10.7 (Rouché's theorem for a half-plane)   Let $a>1$. Show that the equation

$$a-z-e^{-z}=0$$ (10.35)

has exactly one solution in the right half-plane.

Proof. This can be shown with Rouché's theorem, whose setup we now perform. Let $f=a-z$ and $g=-e^{-z}$ and consider the semicircular contour with side $[Ri,-Ri]$ and arc $Re^{i\theta}$ for $\theta\in[-\pi/2,\pi/2]$ where $R>0$ is arbitrary.

For the side, de Moivre's theorem implies $\vert g\vert=1$. A quick computation can exactly determine the modulus of $f=a-z$, where $z\in[Ri,-Ri]$

$$\vert f\vert = \vert a-z\vert = \sqrt{a^2+\vert z\vert^2} \geq a > 1 = \vert g\vert$$ (10.36)

The inequalities follow by the monotonicity of the square root and the given information about $a$. Now for the arc, we shall have $z=Re^{i\theta}$. This means

$$\vert g\vert = \vert e^{-Re^{i\theta}}\vert = \vert e^{-R\cos(\theta) - iR\sin(\theta)}\vert$$ (10.37)

$$= \vert e^{-R\cos(\theta)} e^{-iR\sin(\theta)}\vert$$ (10.38)

$$= e^{-R\cos(\theta)}$$ (10.39)

The domain $\theta\in[-\pi/2,\pi/2]$ indicates $-R\cos(\theta)>0$, so that as $g$ vanishes as $R\to\infty$, showing trivially that $\vert f\vert>\vert g\vert$.

Therefore all along the semicircular contour, we have that $\vert f\vert>\vert g\vert$, which indicates $a-z-e^{-z}$ has the same number of zeros as $f$, a linear function with only zero. Sending $R\to\infty$, we can decide that the equation has only one root in the right half-plane. $\qedsymbol$