Find how many roots of $z^5+2z^2+7z+1=0$ lie inside $|z|=2$? How many are real? Find how many roots of $z^5+2z^2+7z+1=0$ lie inside $|z|=2$? How many are real?
Let $f(z)=z^5+2z^2+7z+1$, and $g(z)=z^5$. For $|z|=2,$ we note that
$$
|f(z)-g(z)|\leq2|z|^2+7|z|+1=23<32=2^5=|g(z)|.
$$
Since $g$ has $5$ zeros in $|z|\leq 2$, Rouche's theorem implies that $f$ also has $5$ zeros lie inside $|z|=2$. 
However, I have no idea to find how many of them are real. Can anyone tell me how to do it? Thanks.  

I'd never seen Sturm's theorem before, but it was interesting enough that I'll add the computation:
\begin{align}
p_0(x) &= x^5+2x^2+7x+1\\
p_0'(x) = p_1(x) &= 5x^4+4x+7\\
\text{remainder of $-p_0/p_1$:}\quad p_2(x) &= -\tfrac{6}{5}x^2 - \tfrac{28}{5}x - 1\\
\text{remainder of $-p_1/p_2$:}\quad p_3(x) &= \tfrac{12562}{27}x + \tfrac{8669}{108}\\
\text{remainder of $-p_2/p_3$:}\quad p_4(x) &= -\tfrac{87841935}{1262430752}\\
\end{align}
Computing at the end points, 
$$p_0(-2)=-37,\;p_1(-2)=79,\;p_2(-2)=\tfrac{27}{5},\;p_3(-2)=-\tfrac{3401}{4},\;p_4(-2)=-\tfrac{87841935}{1262430752}$$
$$p_0(2)=55,\;p_1(2)=95,\;p_2(2)=-17,\;p_3(2)=\tfrac{109165}{108},\;p_4(2)=-\tfrac{87841935}{1262430752}$$
Hence at $-2$ we have the sign sequence $-++--$, which has $2$ sign changes and at $2$ we have $++-+-$, which has $3$ sign changes and so there are $3-2=1$ real roots in $[-2,2]$.
Alternatively, we may note that the second derivative $4(5x^3+1)$ is zero at $x = -\tfrac{1}{\sqrt[3]{5}}$, which is a global minimum of the derivative, which is positive there, hence $x^5+2x^2+7x+1$ is monotone increasing everywhere. As $p_0(-2) < 0 < p_0(2)$, by continuity of polynomials and intermediate value theorem there is precisely one root in $[-2,2]$.
