How many roots has $2z^4 -3z^3 + 3z^2 - z + 1 = 0$ in each quadrant?
What I've thought so far is that the polynomial has real coefficients.
So we can either have
There is one root in each quadrant.
There are two roots in the first and fourth quadrant.
There are two roots in the second and third quadrant.
However, I am stuck in finding which of these cases hold.
On
As one can "see", the polynomial is $z^4+1+z(z-1)^3$. On the real axis $z=x$, outside the interval $[0,1]$, all terms are positive. Inside that interval the last term is smaller than $1$ in absolute value, so that the argument of the function value stays at the angle $0$.
On the imaginary axis $z=iy$ the polynomial value can be decomposed into real and imaginary parts to get $$ 2y^4-3y^2+1+i\cdot (3y^3-y)=(2y^2-1)(y^2-1)+iy(3y^2-1) $$ The quadrants that are traversed are \begin{array}{r|l} y\in(-\infty,-1)&IV\\ (-1,-\frac1{\sqrt2})&III\\ (-\frac1{\sqrt2},-\frac1{\sqrt3}))&IV\\ (-\frac1{\sqrt3},0)&I\\ (0,\frac1{\sqrt3})&IV\\ (\frac1{\sqrt3},\frac1{\sqrt2})&I\\ \frac1{\sqrt2},1)&II\\ (1,\infty)&I \end{array} at no point is the real negative axis crossed, giving a winding number of $0$ along the half axes.
Thus the winding number for each quadrant is determined by the leading term $2z^4$ along large quarter circles which in the end means that there is exactly one root inside every quadrant.
$$2z^4-3z^3+3z^2-z+1=0$$ it's $$z^4-1.5z^3+1.5z^2-0.5z+0.5=0$$ or for all $k$ $$(z^2-0.75z+k)^2-((2k-0.6)z^2-(1.5k-0.5)z+k^2-0.5)=0.$$ Now, let $k$ be a root of the equation $$(1.5k-0.5)^2-4(2k-0.6)(k^2-0.5)=0$$ and $$2k-0.6>0.$$
Easy to see that $0.7<k<0.8$ is valid.
Now, $$-0.75+\sqrt{2k-0.6}>-0.75+\sqrt{2\cdot0.7-0.6}>0$$ and $$-0.75-\sqrt{2k-0.6}<0,$$ which says that $$z^4-1.5z^3+1.5z^2-0.5z+0.5=(z^2+az+b)(z^2+cz+d),$$ where $\{a,b,c,d\}\subset\mathbb R$ and $ac<0.$
Now, easy to see that $$2z^4-3z^3+3z^2-z+1=(2z^2-3z^3+2z^2)+(z^2-z+1)>0$$ for all $z\in\mathbb R$, which says that our equation has one root in each quadrant.
Done!