# Use Rouche’s Theorem to show that all roots of $z ^6 + (1 + i)z + 1 = 0$ lines inside the annulus  $\frac{1}{2} \leq |z| \leq \frac{5}{4}$

Use Rouche’s Theorem to show that all roots of $z ^6 + (1 + i)z + 1 = 0$ lines inside the annulus $\frac{1}{2} \leq |z| \leq \frac{5}{4}$.

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$. The answer is accepted. Join Matchmaticians Affiliate Marketing Program to earn up to 50% commission on every question your affiliated users ask or answer. • answered • 443 views •$10.00