Rouché's Theorem for a polynomial 
Show that $P(z) = z^{47} − z^{23}+ 2z^{11} − z^5 + 4z^2 + 1$ has at least one zero in the unit disk $D(0,1)$.

Here's my attempt:
Let $f(z)=z^{47} − z^{23}+ 2z^{11} − z^5 + 4z^2 + 1$ and $g(z)=− z^5 + 4z^2 + 1$
Using Rouché's Theorem, on $|z|=1, $ $|f(z)-g(z)|=|z^{47}-z^{23}+2z^{11}| \leq|z|^{47}+|-z|^{23}+2|z|^{11}=1+1+2=4.$ So $2<4.$ $(|f(z)-g(z)|\leq|f(z)|+|g(z)|)$ using the theorem.
So now how do I say how many zeros are there? I mean, by looking at what do we deduce the number of zeros in this function? And is this the correct way or must I change my assumptions to do it in a better way?
 A: This particular problem doesn't need complex analysis (beyond the fundamental theorem of algebra) as the product of the roots is $1$ in absolute value, so either at least one is in the open unit disc or all $47$ roots are on the unit circle and the latter is obviously not true since the non-real roots are conjugate so even in number and $\pm 1$ are manifestly not roots.
A: Using the notation from the Wikipedia article on Rouché's theorem.
Let $f(z)=4z^2+1$ and $g(z)=z^{47}−z^{23}+2z^{11}−z^5$, and consider them on the disk $D(0,\frac{2}{3})$. On its boundary,
$$\begin{multline*}
|g(z)|=|z^{47}−z^{23}+2z^{11}−z^5|\leq|z^{47}|+|z^{23}|+|2z^{11}|+|z^5|=\\
=\left(\frac{2}{3}\right)^{47}+\left(\frac{2}{3}\right)^{23}+2\cdot\left(\frac{2}{3}\right)^{11}+\left(\frac{2}{3}\right)^5\approx0.155\end{multline*}$$
and
$$|f(z)|=|4z^2+1|\geq|4z^2|-|1|=4\cdot\left(\frac{2}{3}\right)^2-1=\frac{7}{9}.$$
Since $|g(z)|<|f(z)|$ on the boundary, $f$ and $f+g$ have the same number of zeroes in this disk. Now, how many zeroes in this disk does $f(z)$ have?
