<s>[INST] Find the number of roots of equation $z^5 -12z^2+14=0$ that lie in the region {$z \in \Bbb C : 2 \leq |z|< \frac {5}{2}$} 
Problem: Find the number of roots of equation $z^5 -12z^2+14=0$ that lie in the region
{$z \in \Bbb C  :  2 \leq |z|< \frac {5}{2}$}
Solution :
In  $f(z)=z^5 -12z^2+14=0$, there is $2$ change in sign
So it has two positive real roots
Now $f(2)f(2.5) <0$
So two positive real roots lie between $2$ and $2.5$
In  $-z^5 -12z^2+14=0$, there is $1$ change in sign
So it has one negative real roots (It will not lie  between $2$ and $2.5$)

I have no idea of other two conjugate complex roots
Please help me in finding the answer
 [/INST] The idea is to use Rouché's theorem.
On the outer circle $\lvert z\rvert = \frac{5}{2}$, we have
$$\lvert z\rvert ^5 = \frac{3125}{32} = 97 + \frac{21}{32} > 89 = 75 + 14 = 12\lvert z\rvert^2+14 \geqslant \lvert -12z^2 + 14\rvert,$$
so by Rouché's theorem, $f(z) = z^5 - 12z^2 + 14$ has as many zeros in the disk $\lvert z\rvert < \frac{5}{2}$ as $z^5$ has, counting multiplicity. So all five zeros of $f$ lie in the larger disk.
On the inner circle, we have
$$\lvert z^5+14\rvert \leqslant \lvert z\rvert^5+14 = 32+14 = 46 < 48 = 12\lvert z\rvert^2 = \lvert -12z^2\rvert,$$
so by Rouché's theorem, in the smaller disk $\lvert z\rvert < 2$, $f$ has as many zeros (counting multiplicity) as $-12z^2$ has, namely $2$.
Thus in the annulus $2 <\lvert z\rvert < \frac{5}{2}$, there are three zeros of $f$.

So two positive real roots lie between $2$ and $2.5$

That is wrong, on the interval $[2,2.5]$ the function is strictly increasing ($5x^4 - 24x > 0$ there), so $f$ has only one real root between $2$ and $2.5$.
</s>