Toronto Math Forum
MAT3342018F => MAT334Tests => End of Semester Bonussample problem for FE => Topic started by: Victor Ivrii on November 27, 2018, 07:20:10 AM

Determine the number of zeros of
$$
2z^5 + 4z + 1.
$$
(a) in the disk $\{z\colon z<1\}$;
(b) in the annulus $\{z\colon 1 <z < 2\}$.
(c) in the domain $\{z\colon z>2\}$.
Show that they are all distinct.

$(a)$ $\\$
At $z=1$, $$\begin{align}2z^5+4z+1+(4z)&=2z^5+1\\&\leq 2z^5+1\\&=3\\&<4=4z\end{align}$$
By Rouche's Theorem,
$2z^5+4z+1$ and $4z$ has the same number of zeros.$\\$ Since $4z$ has $1$ zero, therefore $2z^5+4z+1$ has $1$ zero in the disk $\{z\colon z<1\}$.
$\\$
$\\$
$(b)$ $\\$
At $z=2$, $$\begin{align}2z^5+4z+1+(2z^5)&=4z+1\\&\leq 4z+1\\&=5\\&<64=2z^5\end{align}$$
By Rouche's Theorem, $2z^5+4z+1$ and $2z^5$ has the same number of zeros.$\\$
Since $2z^5$ has $5$ zeros inside $z=2$, therefore $2z^5+4z+1$ has $4$ zeros $(51=4)$ in the annulus $\{z\colon 1 <z < 2\}$.
$\\$
$\\$
$(c)$ $\\$
Notice that the degree of $f(z)=2z^5+4z+1$ is $5$. Which means it has at most $5$ roots. Now from part(b), all the roots are inside $z=2$, therefore there are no roots/zeros in the domain $\{z\colon z>2\}$.
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Show distinct:
$$f(z_0)=2z_{0}^5+4z_0+1=0 \\ f'(z_0)=10z_0^4+4\neq0$$
Thus the multiplicity is $1$, therefore they are all distinct.

And how do you prove that two last equations are incompatible?

And how do you prove that two last equations are incompatible?
I don't know if I fully understood what you mean but here's my attempt:
Suppose $$f'(z_0)=10z^4+4=0 \\ \Rightarrow z_0^4=\frac{2}{5} \Rightarrow z_0=\frac{2}{5}e^{i\frac{\pi + 2k\pi}{4}}$$
$$\begin{align}f(z_0)&=2z_0\cdot z_0^4+4z_0+1=0\\&=2z_0\cdot(\frac{2}{5})+4z_0+1=0\\&=\frac{16}{5}z_0+1=0\\ \Rightarrow z_0=\frac{16}{5}\neq \frac{2}{5}e^{i\frac{\pi + 2k\pi}{4}}\end{align}$$

Indeed, this is the proof.