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 $|z|<1$?

It is not too difficult to see that all three roots are real. 
The roots are the same as roots of $$ x^3 -2x^2 +1/4 =0$$ 
Zeros could be approximated by Newton's method as $$ -0.32772....\\ 0.39462....\\ 1.9331... 
 $$
