Prove that $p(z) = 2z^5 + 6z - 1 $ have roots (in two sets)

Question

Prove that $p(z) = 2z^5 + 6z - 1 $ have one real root in $(0,1)$ and four root in $\left\{ z \in \mathbb{C} : 1<|z|<2 \right\}$.

I suppose that we should use Rouché's theorem but I have no idea how. Thanks in advance for help.

Answer 1

First, on the circle $z=e^{it}$, you have

$$|6z-1|\geq |6z|-1 = 5 > |2z^5| = 2$$

So, Rouché's theorem tells us $p(z)$ and $6z-1$ have the same number of roots in the unit disc, that is $1$ root. Hence it must be real (complex roots come in conjugate pairs).

Now, on the circle $z=2e^{it}$,

$$|6z-1|\leq |6z|+1=13$$ $$|2z^5|=64>13$$

So, the same theorem tells us $p(z)$ and $2z^5$ have the same number of roots in the disc $|z|<2$.

Hence there are $4$ roots in the annulus $1<|z|<2$. (notice the case $|z|=1$ was eliminated by the first application of Rouché's theorem)

Finally, is there more than one real root? We have $p'(x)=10x^4+6>0$ so $p$ is increasing, and there is exactly one real root. This real root is also positive, since it's obvious from coefficients that $p(x)<0$ when $x\leq0$, so the real root is actually in $]0,1[$.

Answer 2

Hint: The derivative of $p(z)$ is always positive on the real line, so can there be more than or less than one real root? Based on the value of the polynomial at $0$ and $1$, you can confirm the real zero is in the interval $(0,1)$. As for the other part, Rouche's theorem is indeed the way to go. See http://en.wikipedia.org/wiki/Rouch%C3%A9%27s_theorem#Usage for an example of how to use it.

Answer 3

Prove that equation $2z^5 + 8z – 1 = 0$ has one real root in $| z | < 1$ and four roots lie between the circles |$ z | = 1$, and $| z | = 2$