Prove that the equation $z=2-e^{-z}$ has exactly one root in the right half-plane and why must this root be real?
Prove that the polynomial $P(z)=z^4+2z^3+3z^2+z+2$ has exactly two zeros in the right half-plane.
For $1$, I can rearrange it to $e^{-z}+z-2=0$ and can use Rouche's theorem, but I do not know which function $|f(z)|$ and $|g(z)|$, where $|g(z)|\le|f(z)|$ I should use to show that it has the specified roots and the same confusion holds for question number $2$.
Best Answer
We have: $$ \left|z - 2\right| = e^{-\Re z} $$
In the right half-plane, $\Re z > 0$ and hence $e^{-\Re z} < 1$. It follows that any roots in the right half-plane must satisfy $\left|z - 2\right| < 1$.
Apply Rouché's theorem to the circle centered at $2$ with radius $1$:
$$ \left|\left(e^{-z} + z - 2\right) - (z - 2)\right| = e^{-\Re z} < 1 = \left|z-2\right| $$
Thus, $e^{-z} + z - 2$ and $z - 2$ have the same number of roots inside the circle, which is one.
To show that the root is real, notice that the equation $e^{-x} + x - 2 = 0$ has a positive real solution. This can be shown by studying the derivative of $f(x) = e^{-x} + x - 2$.