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{\bf Question}

State Rouche's theorem and use it to show that all the roots of
the equation

$$z^6+\alpha z+1=0,$$

where the constant $\alpha$ satisfies $|\alpha|=1$, lie in the
annulus $\frac{1}{2}<|z|<2$.

Use the argument principle to show that, if $Re(\alpha)>0$, then
just one of these roots lies in the first quadrant.



\vspace{0.25in}

{\bf Answer}

If $|z|=\frac{1}{2}, \hspace{0.2in} |z^6+\alpha z|\leq
|z|^6+|\alpha||z|<(\frac{1}{1})^6+\frac{1}{2}<1$

So $z^6+\alpha z+1$ has no roots in $|z|\leq\frac{1}{2}$

If $|z|=2, \hspace{0.2in} \alpha
z+1|\leq|\alpha||z|+1\leq3<2^6=|z|^6$

so all the roots are inside $|z|=2$


DIAGRAM


Let $\alpha=a+ib, \hspace{0.2in} a>0$

$f(z)=z^6+\alpha z+1$

On $OA, \hspace{0.2in} f=x^6+ax+1+ibx$

$\ds\tan\arg f(z)=\frac{bz}{x^6+ax+1}$ continuous on $OA$ as $a>0$

This is zero when $x=0$ and $\to0$ as $x\to\infty$, so the total
change of $\arg f(z)$ on $OA$ is $\epsilon_1$ - small.

${}$

On $BO \hspace{0.2in} z=iy \hspace{0.2in} f=-y^6+by+1+iay$

Consider the real parts $-y^6-by+1$, the derivative is $-6y^5-b$
which is always negative if $b>0$, and which has just one positive
root for $b>0$.  So $-y^6-by+1$ has one positive root.

So the graph of $\tan\arg f(z)$ is of the form


DIAGRAM


So as $y$ goes from $\infty$ to 0, the change of argument of
$f(z)$ is $-\pi$.

On the semicircle $Re^{i\theta} \hspace{0.2in}
f(z)=R^6e^{i6\theta}(1+w)$ - $w$ small.

So as $\theta:0\to\frac{\pi}{2} \hspace{0.2in} \arg f(z)$
increases by approximately $3\pi$.

Thus $[\arg f(z)]_C=2\pi$ and thus there is just one root in the
first quadrant.


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