# 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

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.

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