Find the number of zeros for $f(z)=z^{10} + 10ze^{z+1} - 9$ in the unit disc.
It seems Rouche's theorem must be used. Trying out the variety of possibilities of $g$ doesn't seem to work. For example, let $g(z)=10ze^{z+1}$, then $ \lvert f(z) - g(z) \rvert = \lvert z^{10} - 9 \rvert \le 10 $ and $\lvert g(z) \rvert = \lvert 10ze^{z+1} \rvert = 10e \lvert e^z \rvert \ge 10$, so I only have $\lvert f(z) - g(z) \rvert \le \lvert g(z) \rvert$ and I find that $\lvert f(z) \rvert \ge 0$. So I cannot make the inequality strict, other combinations also seem to present this issue. Any suggestions would be great, thanks.
As is already proved, $|f(z) - g(z)| \leqslant 10 \leqslant |g(z)|$ for $|z| = 1$. Suppose that $|z_0| = 1$ and $|f(z_0) - g(z_0)| = |g(z_0)|$, then $|f(z_0) - g(z_0)| = 10 = |g(z_0)|$, which implies$$ 10 = |10z \mathrm{e}^{z_0 + 1}| = 10 |\mathrm{e}^{z_0 + 1}| = 10 \mathrm{e}^{\mathrm{Re}(z_0 + 1)}. $$ Thus$$ \mathrm{Re} z_0 + 1 = \mathrm{Re}(z_0 + 1) = 0. $$ Note that $|z_0| = 1$, thus $\mathrm{Re} z_0 = -1 \Rightarrow z_0 = -1$. But$$ |f(-1) - g(-1)| = 8 < 10, $$ a contradiction. Therefore,$$ |f(z) - g(z)| < |g(z)|. \quad \forall |z| = 1 $$ By Rouché's theorem, $f(z)$ and $g(z)$ have the same number of zeros inside the unit disk. Since the number of zeros of $g(z) = 10z \mathrm{e}^{z + 1}$ inside the unit disk is $1$, then the number of zeros of $f(z)$ inside the unit disk is also $1$.