I want to show that the function $f(z) = 1 + 2z + 7z^2 + 3z^5$ has exactly two zeroes in the unit disk counted with multiplicty.
My approach: Consider $g(z) = -7z^2-3z^5$. Then, $h(z) = f(z) + g(z) = 1 + 2z$ has the same number of zeroes as $f(z)$ in the unit disk by Rouché since $|f(z)| > |g(z)|$ on the boundary of the unit disk. Then, I can compute the number of zeroes of $h$ using the Argument principle since $h$ is holomorphic, doesn't have a pole on the boundary and is non-vanishing on the boundary of the unit disk.
Let $n$ denote the number of zeroes. Then, by the argument principle (also noticing that the number of poles of $h$ is zero), we have that \begin{align*} n = \frac{1}{2\pi i}\int_{0}^{2\pi}\frac{h'(\phi(t)}{h(\phi(t))} \phi'(t) dt \end{align*} where $\phi(t) = e^{it}, t\in [0,2\pi]$ is a parametrization of the unit circle. When I evaluate this expression, however, it does not equal to 2. What am I missing? Is my approach wrong? I would greatly appreciate any help with this!