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##### Quiz 7 / Q7: Wednesday Sitting

« Last post by**Ziyi Wang**on

*March 22, 2020, 02:51:12 PM*»

**Question:**Using argument principle along line on the picture, calculate the number of zeroes of the following function in the given annulus: $z^3-3z+1$ in ${1<|z|<2}$

**Answer:**$f(z)=z^3-3z+1$

On the circle $|z|=1$

$$|z^3+1| \leq |z^3|+|1|=2$$

$$|-3z| = |3z| = 3$$

So $|z^3+1|<|-3z|$. By Roche's Theorem, $f(z)$ and $g(z)=-3z$ have the same number of zeroes with $|z|=1$.

Since $g(z)$ has one zero with $|z|=1$, we know that $f(z)$ has one zero with $|z|=1$.

On the circle $|z|=2$

$$|z^3|=8$$

$$|-3z+1| \leq |3z|+|1| = 7$$

So $|-3z+1|<|z^3|$. By Roche's Theorem, $f(z)$ and $g(z)=z^3$ have the same number of zeroes with $|z|=2$.

Since $g(z)$ has three zeroes with $|z|=1$, we know that $f(z)$ has three zeroes with $|z|=1$.

$$3-1=2$$

Therefore, $f(z)=z^3-3z+1$ has two zeroes in ${1<|z|<2}$.