Analyzing the Complex Function $g(z)$

[Dustinsfl]

$$
g(z) = z^{87} + 36z^{57} + 71z^{4} + z^3 - z + 1
$$

For $|z|<1$.

Let $f(z) = 71z^4$.

Then $|f(z) - g(z)| = |-z^{87} - 36z^{57} - z^3 + z - 1| \leq |z|^{87} + 36|z|^{57} + |z|^3 + |z| + 1 < 71|z^4|$

So g has the same number of zeros as f which is 0 with multiplicity of 4.

Correct?

 

[Prove It]

$$
g(z) = z^{87} + 36z^{57} + 71z^{4} + z^3 - z + 1
$$

For $|z|<1$.

Let $f(z) = 71z^4$.

Then $|f(z) - g(z)| = |-z^{87} - 36z^{57} - z^3 + z - 1| \leq |z|^{87} + 36|z|^{57} + |z|^3 + |z| + 1 < 71|z^4|$

So g has the same number of zeros as f which is 0 with multiplicity of 4.

Correct?

Isn't the number of roots the same as the order of the polynomial? Or do you want non-repeated roots?

 

[CaptainBlack]

$$
g(z) = z^{87} + 36z^{57} + 71z^{4} + z^3 - z + 1
$$

For $|z|<1$.

Let $f(z) = 71z^4$.

Then $|f(z) - g(z)| = |-z^{87} - 36z^{57} - z^3 + z - 1| \leq |z|^{87} + 36|z|^{57} + |z|^3 + |z| + 1 < 71|z^4|$

So g has the same number of zeros as f which is 0 with multiplicity of 4.

Correct?

Your last inequality cannot be right, the right hand side goes to zero as z goes zero, while the left hand side goes to 1.

CB

 

[Opalg]

$$
g(z) = z^{87} + 36z^{57} + 71z^{4} + z^3 - z + 1
$$

For $|z|<1$.

Let $f(z) = 71z^4$.

Then $|f(z) - g(z)| = |-z^{87} - 36z^{57} - z^3 + z - 1| \leq |z|^{87} + 36|z|^{57} + |z|^3 + |z| + 1 < 71|z^4|$ You should make it clear that you are applying Rouché's theorem for the disc $\color{red}|z|<1$ and that you therefore want to show that $\color{red}|f(z)-g(z)|<|f(z)|$ on the boundary of the disc. So you should have said that $\color{red}|f(z) - g(z)|<71|z^4|$ when $\color{red}|z|=1.$

So g has the same number of zeros as f which is 0 with multiplicity of 4. The number of zeros is 4. The fact that the four zeros of $\color{red}f(z)$ all occur at $\color{red}z=0$ is irrelevant.

Correct? Since you haven't said what you are being asked to prove, it's hard to know whether the answer is correct. If the question was asking for the number of zeros of $\color{red}g(z)$ inside the unit disc, then yes, you have provided the ingredients for showing that the answer is 4. But the solution could do with a good deal more in the way of explanation.

...

 

[blue_raver22]

Yes, your analysis is correct. By comparing the functions $f(z)$ and $g(z)$, we can see that they have the same number of zeros within the given region $|z|<1$. Furthermore, since $f(z)$ has no zeros within that region, we can conclude that $g(z)$ also has no zeros within that region. This is a useful way to analyze complex functions and determine their behavior within a given region.