Conformal mapping and its application in finding roots of polynomial So for a polynomial, if we want to find the roots in a complex plane. Rouche's theorem is the first tool in my head. However, I saw several problems of finding the roots in the first quadrant or upper plane. Sometimes we need to observe using the relation of root and its conjugate. Is there a common method of solving this kind of problem? Can we apply conformal mapping in this? Thanks.
 A: Alot of the time Vieta's formulas are used.
For example, if we want to find where the roots of $f(z):=az^3-z+b$ ($a>0$ and $b>2$) are, we first note that $f(0)=b>2$, and $\lim_{x\to -\infty}f(z)=-\infty$ (for real $x$), so we must have at least one root of the negative real line. Now If $\alpha_1,\alpha_2,\alpha_3$ are the three roots (with $\alpha_1$ being the aforementioned one on the negative real line), Vieta's formulas show that
$$
\alpha_1\alpha_2\alpha_3=-\frac{b}{a}\hspace{.5 in}\text{and}\hspace{.5 in}\alpha_1+\alpha_2+\alpha_3=0.
$$
There are couple possibilities here:
If $\alpha_2,\alpha_3$ are real as well then they must have the same sign (the product of the roots is a negative number) and the must be positive (else $\alpha_1+\alpha_2+\alpha_3<0$), so the two roots would lie on the positive real line.
If $\alpha_2,\alpha_3$ are not purely real then it must be the case that $\Re\alpha_2=\Re\alpha_3$ (conjugate roots). Again this shows that
$$
\Re\alpha_2=\Re\alpha_3=-\frac{1}{2}\Re\alpha_1=-\frac{1}{2}\alpha_1>0
$$
so $\alpha_2,\alpha_3$ both lie the right half-plane.
I hope this helps, and is what your question was asking.
