Toronto Math Forum
MAT3342020S => MAT334Tests and Quizzes => Quiz 7 => Topic started 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: $z^33z+1$ in ${1<z<2}$
Answer: $f(z)=z^33z+1$
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 $z=1$.
$$31=2$$
Therefore, $f(z)=z^33z+1$ has two zeroes in ${1<z<2}$.