### Author Topic: Q7: Wednesday Sitting  (Read 1185 times)

#### Ziyi Wang

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##### Q7: Wednesday Sitting
« 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 ${1<|z|<2}$.
« Last Edit: May 03, 2020, 09:01:11 PM by Ziyi Wang »