Use Rouche’s Theorem to show that all roots of $z ^6 + (1 + i)z + 1 = 0$ lines inside the annulus $ \frac{1}{2} \leq z \leq \frac{5}{4}$
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 $C$.
i) Let $f(z) = 1, g(z) = z^6 + (1 + i)z$. Then for $z = 1/2$
$g(z) ≤ ( 1/2 )^ 6 + 1/2 \sqrt{2} < 1 = f(z)$.
Since f(z) has no zeros inside $z = 1/2$, and so $z^6 + (1 + i)z+1=f(z) + g(z)$ has no zero inside $z = 1/2$ .
ii) Let $f(z) = z^6$ , $g(z) = (1 + i)z + 1$. Then for $z = 5/4$ ,
$ g(z) ≤ \sqrt{2}(5/4) + 1 ≈ 2.77$ < f(z) = ( 5/4 )^6 ≈ 3.81$.
Since $f(z)$ has six zeros inside $z = 5/4$, and so $z^6 + (1 + i)z+1=f(z) + g(z)$ has all its six zeros inside $z = 5/4$.
Erdos
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