Find The Zeros Of $z^3-2z^2+\frac{1}{4}=0$

Question

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 $|z|<1$?

Answer 1

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 $|z|>1$.

Answer 2

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 $$ -0.32772....\\ 0.39462....\\ 1.9331... $$

Answer 3

Writing $z = (4 \sin(s) +2)/3$, the equation becomes $$ 0 = 4 \sin(s)^3 - 3 \sin(s) - \frac{37}{64} = - \sin(3s) - \frac{37}{64}$$ This is $0$ for $s = -\arcsin(37/64)/3+2 \pi k/3$.