Rouché's Theorem counterexample. In Rouché's Theorem. 
Question . If $|f(z)+g(z)|=|f(z)|+|g(z)|$ instead of hypothesis $|f(z)+g(z)|<|f(z)|+|g(z)|$
. Exists a counterexample?
 A: $|a+b|=|a|+|b|$ implies that $\frac  a b$ is a non-negatiave real number. Using this we see that if $|f(z)+g(z)| =|f(z)|+|g(z)|$ then $|f(z)+ig(z)| \neq |f(z)|+|ig(z)|$ which implies $|f(z)+ig(z)| <|f(z)|+|ig(z)|$. Apply the theorem to $f$ and $ig$.
A: One can also use the following version of Rouche which I usually find the most helpful:
assume $f,g$ with no zeroes and poles on a circle, meromorphic on a neighborhood of the respective closed disc; then if there is a direction $e^{it}$ s.t. $f+ce^{it}g$ has no zeroes on the given circle for any $c \ge 0$, then Rouche applies
(proof is just noticing that $\lambda f +(1-\lambda)e^{it}g, 0 \le \lambda \le 1$ is a non-zero homotopy on the circle between $f$ and $e^{it}g$ from which the argument principle gives the result)
In particular if $|f+g|=|f|+|g|$ on the whole circle and $f,g$ no zeros/poles, one gets $f+cg$ has no zeroes on the circle for $c \ge 0$ since otherwise $c>0$ and $|f+g|(w)=(1-c)|g(w)| < (1+c)|g(w)|=|f(w)|+|g(w)|$ at such a zero, so we are done!
Note that the result is non-trivial since any positive trigonometric polynomial on the circle gives rise to such a pair where $f,g$ are usual polynomials
(eg $3+2\cos\theta=3+e^{i\theta}+e^{-i\theta}=3+z+1/z=\frac{z^2+3z+1}{z}$ so $z^2+3z+1, z$ are a pair of $f,g$ as in the OP)
