Using Rouche's Theorem, find the radius of the roots of $z^{3}+z^{2}-2z+3$
To answer this question, I have set functions $f\left(z\right)=z^{3}$ and $g\left(z\right)=z^{2}-2z+3$, to try and show that $\left|f\left(z\right)\right|\geq \left|g\left(z\right)\right|$. Then, by Rouche's Theorem $f+g$ and $f$ have the same number of roots.
I have so far that $\left|f\left(z\right)\right|=|z|^{3}$ and $\left|g\left(z\right)\right|=|z|^{2}+2|z|+3$ but I'm unsure where to go from here to show that $\left|f\left(z\right)\right|\geq \left|g\left(z\right)\right|$
To apply Rouche theorem, you have to consider a closed contour on which $|g|<|f|$, on and inside the contour $f$ and $g$ are holomorphic, then $f$ and $f+g$ have the same number of zeros counting multiplicity inside the contour.
If you consider the circle of center $0$ and of radius $3$, and chose $f$ and $g$ the same way you did, you have $|g|<|f|$ on this circle, so $f$ and $f+g$ have the same number of roots counting multilicity inside the considered circle, and since $z^{3}+z^{2}-2z+3$ has the same numbers of zeros than $z^3$, all the roots of $z^{3}+z^{2}-2z+3$ are inside the considered circle.
So the radius of the roots of $z^{3}+z^{2}-2z+3$ is $\le 3$.