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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?


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@Crystal: I typeset the equations. – user17762 Nov 26 '10 at 20:34

Hint: This kind of questions are usually handled using Rouche's Theorem. I suggest you look it up in the wikipedia article, where you can see an example of its usage. Also here's an example.

The key is choosing wisely another function $f(z)$ with which to compare in the inequality in Rouche's theorem and such that you can easily decide how many zeroes does $f(z)$ have inside the region you are considering, which in your case is the circle $|z| < 2$.

About your other question, you don't need to factor the polynomial in order to answer this.

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Chhose $f(z)=z^4$ Then on $|z|=2$, $|f(z)-p(z)|<|p(z)|$ so by Rouche's Theorem $f(z)$ and $p(z)$ has same number of zeroes inside $|z|<2$ and we are done

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You surely meant $|f(z)-p(z)|<|p(z)|$. – Philippe Malot Apr 27 '13 at 12:29
@girianshiido Thank you for notifying – La Belle Noiseuse Apr 27 '13 at 12:38

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