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{\bf Question}

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

$$z^6+(1+i)z+1=0$$

lie in the annulus $\frac{1}{2}\leq|z|<\frac{5}{4}.$

Use the argument principle to determine how many of these roots
lie in the quadrant $0<\arg z<\frac{1}{2}\pi$.


\vspace{0.25in}

{\bf Answer}

Rouche's Theorem states that if $f(z)$ and $g(z)$ are both
analytic inside and on the closed contour $C$, and if
$|g(z)|<|f(z)|$ on $C$ then $f(z)$ and $F(z)+g(z)$ have the same
number of zeros inside $C$.

\begin{itemize}
\item[i)]
Let $f(z)=1$, $g(z)=z^6+(1+i)z$.

Then for $|z|=\frac{1}{2}$,
$|g(z)|\leq(\frac{1}{2})^6+\frac{1}{2}\sqrt2<1=|f(z)|$

$f(z)$ has no zeros inside $|z|=\frac{1}{2}$, and so

$f(z)+g(z)$ has none inside $|z|=\frac{1}{2}$.


\item[ii)]
Let $f(z)=z^6$, $g(z)=(1+i)z+1$

Then for $|z|=\frac{5}{4}$,
$|g(z)|\leq\sqrt2\frac{5}{4}+1\approx2.77$

$|f(z)|=(\frac{5}{4})^6\approx3.81$

$f(z)$ has six zeros inside $|z|=\frac{5}{4}$, and so

$f(z)+g(z)$ has all six inside $|z|=\frac{5}{4}$.

\end{itemize}


Now consider the contour $C$ in the first quadrant.


DIAGRAM


\begin{itemize}
\item[I.]
On $OA$ $f=x^6+x+1+ix$ and $\ds\tan\arg z=\frac{x}{x^6+x+1}$.
This is continuous for $x>0$, it is zero at 0 and tends to zero as
$R\to\infty$.  So $[\arg f(z)]_{OA}=\epsilon$ (something small)


\item[II.]
On $BO$ $z=iy$ so $f=-y^6-y+1+iy$ and $\ds\tan\arg
z=\frac{y}{1-y-y^6}$.  Now the derivative of $1-y-y^6$ is
$-1-6y^5$ which is negative for all $y>0$.  So $1-y-y^6$ has just
one positive root.  Thus the graph of $\tan\arg z$ is


DIAGRAM


Hence $[\arg z]_{BO}=-\pi+\delta$ ($\delta$ is small)


\item[III.]
On $AB$ $z=Re^{i\theta}$ and $f(z)=R^6e^{6i\theta}(1+w)$, $|w|$ is
small.

So as $\theta$ goes from 0 to ${\pi}{2}$, $[\arg
f(z)]_{AB}=3\pi+\eta$, $\eta$ is small.

\end{itemize}

Thus $\ds\frac{1}{2\pi}[\arg f(z)]_C=1$ since it must be an
integer.

Thus the equation has 1 root in the first quadrant.

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