My homework question: Show that all zeros of $p(z)=z^4 + 6z + 3$ lie in the circle of radius $2$ centered at the origin.
I know $p(z)$ has a zero-count of $4$ by using the Fundamental Theorem of Algebra. Then using the Local Representation Theorem the $\int \frac{n}{z+a} = 4(2 \pi i).$ I am assuming $a=0$ since we are centered at the origin. I apologize for my lack of math-type. What does $= 8 \pi i$ mean? Am I going around the unit circle $4$ times? Or is it even relevant to my final answer. Which I am assuming is finding the coordinates to the $4$ singularities. I have always looked for my singularities in the values that make the denominator zero, but in this question my denominator is $z$. $z=0$ doesn't seem right. So the question is, am I suppose to factor the polynomial $z^4 + 6z + 3$ to find the zeros?
Thanks