<s>[INST] Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$ a. at $\frac{1}{4}<|z|<1$ b. $|z|>1$
I know that $\mathbb{C}$ is a closed algebraic field so we can write the polynomial has a product of first degree polynomials, so we will have to guess one root and divide and find the others. but it is hard to guess here.
So I set $z=x+iy$ and got
$$(x^3-y^2x+2xy^2-2x^2+2y^2+\frac{1}{4})+i(-2x^2y+yx^2-y^3+4xy)=0$$
Looking at the imaginary part we get
$y(-x^2+4x-y^2)=0$ so or $y=0$ or $-x^2+4x-y^2=0\iff (x-2)^2+y^2=4$
But how I continue from here? and what does 
I read and now I am given it a try:
for a. we look at $|z|<1$ and $\frac{1}{4}<|z|$ we look at the boundary 
so fo $|z|=1$ $|z^3+\frac{1}{4}|<|-2z^2|$ so we choose $g(z)=z^3+\frac{1}{4}$ and $f(z)=-2z^2$
So $|z^3+\frac{1}{4}|\leq 1+\frac{1}{4}\leq 2=|-2z^2|$
So we can conclude that there are $2$ zeros in $|z|<1$?
 [/INST] Using Rouche's theorem, in $|z|\leq\dfrac14$ we consider $f(z)=z^3-2z^2$ and $g(z)=\dfrac14$ then 
$$|f(z)|=|z^3-2z^2|\leq|z|^3+2|z|^2=\dfrac{9}{64}<|g(z)|$$
then $z^3-2z^2+\frac14=0$ hasn't zero in $|z|\leq\dfrac14$.
In $|z|\leq1$ with $f(z)=z^3+\dfrac14$ and $g(z)=-2z^2$ then 
$$|f(z)|=|z^3+\dfrac14|\leq\dfrac54<2=|g(z)|$$
and $g(z)$ has two zeros in $\dfrac14<|z|<1$, then $z^3-2z^2+\frac14=0$ has two zeros there.
The third root will be in $|z|>1$.
</s>