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{\bf Question}

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

$$z^5+\alpha z^2+1=0,$$

where the constant $\alpha$ satisfies $|\alpha|\leq7$, lie inside
the circle $|z|=2$.

Assume that $\alpha=1+ib$ where $b>1$.  Use the argument principle
to show that just two of these roots lie in the first quadrant.


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{\bf Answer}

Rouche's Theorem

Let $f(z)=z^5$ and $g(z)=\alpha z^2+1$

For $|z|=2 \,\,\, |f(z)|=32, \,\,\,
|g(z)|\leq|\alpha||z|^2+1\leq7.4+1=29$  By Rouche's theorem $f$
and $f+g$ have the same number of roots inside $|z|=2$  i.e. all
the 5 roots of $f+g$ are in $C$.

${}$

Suppose $\alpha=1+ib \,\,\, b>1$


DIAGRAM


The number of zeros =$\ds\frac{1}{2\pi}[\arg g(z)]_C$

On the real axis $f(x)=x^5+x^2+1+ibx^2$

$\ds\tan\arg g(z)=\frac{bx^2}{x^5+x^2+1}$  this is continuous for
$x\geq0$, zero when $x=0$ and $\to0$ as $x\to\infty$.  So the
change in $\arg g(z)$ along $OA$ is $<\epsilon_1$.

On the imaginary axis

$g(z)=(iy)^5+(iy)^2+1+ib(iy)^2=iy^5-iby^2+y^2+1$

So $\ds\tan\arg f(z)=\frac{y^5-by^2}{y^2+1}$  this is continuous,
zero at $y=0$ and $\to\infty$ as $y\to\infty$.  So the change in
$\arg g(z)$ along $BO$ is $-\frac{\pi}{2}+\epsilon_2$.

On the circle $C$, $z=Re^{i\theta}$

$\ds g(z)=R^5e^{5i\theta}\left(1+\frac{e^{-3i\theta}}{R^2}+
\frac{e^{-5i\theta}}{R^5}+\frac{ibe^{-2i\theta}}{R^2}\right)$

$\arg g(z)=5\theta+arg(1+w)$

$w$ is small if $R$ is large, so $\arg(1+w)$ varies little,

so $\ds[\arg g(z)]_{{\rm arc\ }AB}=\frac{5\pi}{2}+\epsilon_3$

So $\ds\frac{1}{2\pi}[\arg g(z)]=
\frac{1}{2\pi}\left[\frac{5\pi}{2}-\frac{\pi}{2}+\epsilon\right]=2$
as it is an integer

So $g(z)$ has 2 roots in the first quadrant.




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