MHB Understanding Complex Functions and Polynomials in the Complex Plane

[Markov2]

1) Let $f(z)=e^{-z}+z^2-4z+4.$ Show that $f$ has exactly two zeroes on $\{z\in\mathbb C:|z-2|<1\}.$ Show that these zeroes are distinct, that is, it's not a zero of order two.

2) Let $f,g\in\mathcal H(\mathbb C)$ be no constant. Prove that $(f\circ g)(z)$ is a polynomial iff $f$ and $g$ are polynomials.

Attempts:

1) I know I have to use Rouché's Theorem, but I don't know how for this case, it annoys me the $|z-2|<1.$

2) The $\Longleftarrow$ implication is trivial, but not the another one. I don't know how to proceed though.

 

[Fernando Revilla]

1) Let $f(z)=e^{-z}+z^2-4z+4.$ Show that $f$ has exactly two zeroes on $\{z\in\mathbb C:|z-2|<1\}.$ Show that these zeroes are distinct, that is, it's not a zero of order two.

Hint Using the substitution $u=z-2$ we obtain $f(z)=e^{-u-2}+u^2$ and $|\;e^{-u-2}\;|<|u^2|$ on $|u|=1$ .

 

[Markov2]

Very nice Fernando! Now let $|f(u)-u^2|=|e^{-u-2}|\le|u^2|=1\le|f(u)|+|u|^2,$ since $f(u)$ and $u^2$ have no zeroes for $|u|=1,$ then by Rouché's Theorem $f$ has two zeroes for $|u|<1,$ but how to justify that those zeroes are distincts? (Second part of the problem.)

Could you help me with second problem please?

 

[Fernando Revilla]

but how to justify that those zeroes are distincts? (Second part of the problem.)

Suppose $a$ is a double root of $g(u)=e^{-u-2}+u^2$ in $|u|<1$ then, $a$ is also a root of $g'(u)=-e^{-u-2}+2u$, but $g(a)=0$ and $g'(a)=0$ implies $a^2+2a=0$ that is, $a=0$ or $a=-2$ (contradiction) .

 

[Markov2]

Fernando last thing: is my solution correct by applying Rouché's Theorem?

 

[Markov2]

Can anydoby check if my solution by using Rouché is correct?

 

[Fernando Revilla]

Can anydoby check if my solution by using Rouché is correct?

I don't see any solution on your part. :)