Author Topic: FE Sample--Problem 5A  (Read 1292 times)

Victor Ivrii

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FE Sample--Problem 5A
« 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.


Meng Wu

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Re: FE Sample--Problem 5A
« Reply #1 on: November 27, 2018, 10:36:20 AM »
$(a)$ $\\$
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 $(5-1=4)$  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 $\{z\colon |z|>2\}$.
$\\$
$\\$
$\\$



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.

Victor Ivrii

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Re: FE Sample--Problem 5A
« Reply #2 on: November 27, 2018, 10:49:14 AM »
And how do you prove that two last equations are incompatible?

Meng Wu

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Re: FE Sample--Problem 5A
« Reply #3 on: November 27, 2018, 11:19:47 AM »
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 $$f'(z_0)=10z^4+4=0 \\ \Rightarrow z_0^4=-\frac{2}{5} \Rightarrow z_0=\frac{2}{5}e^{i\frac{\pi + 2k\pi}{4}}$$
$$\begin{align}f(z_0)&=2z_0\cdot z_0^4+4z_0+1=0\\&=2z_0\cdot(-\frac{2}{5})+4z_0+1=0\\&=\frac{16}{5}z_0+1=0\\ \Rightarrow z_0=-\frac{16}{5}\neq \frac{2}{5}e^{i\frac{\pi + 2k\pi}{4}}\end{align}$$

Victor Ivrii

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Re: FE Sample--Problem 5A
« Reply #4 on: November 30, 2018, 03:59:03 AM »
Indeed, this is the proof.