MHB How Does Complex Function Behavior Constrain Within and Outside the Unit Disk?

  • Thread starter Thread starter Markov2
  • Start date Start date
  • Tags Tags
    Complex
Markov2
Messages
149
Reaction score
0
Denote $D=\{z\in\mathbb C:|z|<1\}.$

1) Let $a\in\mathbb C$ with $|a|<1$ and $p(z)=\dfrac a2+(1-|a|^2)z-\dfrac{\overline a}2z^2.$ Show that for all $z\in D$ is $|p(z)|\le1.$

2) Characterize all the $f$ entire functions so that for each $z\in\overline D^c$ satisfy $\left| {f(z)} \right| \le {\left| z \right|^5} + \dfrac{1}{{{{\left| z \right|}^5}}} + \dfrac{1}{{{{\left| {z - 1} \right|}^3}}}.$

3) Let $w_1,w_2\in\mathbb C$ two $\mathbb R-$linearly independent numbers. Show that if $f\in\mathcal H(\mathbb C)$ is so that for each $z\in\mathbb C$ and $f(z+w_1)=f(z)=f(z+w_2),$ then $f$ is constant.

4) Let $\mathcal U\subset\mathbb C$ open and $z_0\in\mathcal U.$ Suppose that $f$ is continuous on $\mathcal U$ and analytic on $\mathcal U-\{z_0\}.$ Show that $f$ is analytic on $\mathcal U.$

Attempts:

1) I think I need to use the Maximum Modulus Principle, but I don't see how.

2) If I let $|z|=R$ then $\left| {f(z)} \right| \le {\left| R \right|^5} + \dfrac{1}{{{{\left| R \right|}^5}}} + \dfrac{1}{{{{\left| {R - 1} \right|}^3}}},$ but $f$ was given entire so it has convergent Taylor series and by using Cauchy's integral formula I can conclude that $f^{(k)}(0)=0$ for some $k\ge n,$ and then functions $f$ are polynomials of degree $n-1,$ does this make sense?

3) I think I could use Liouville here, but I don't have that $f$ is entire, but $f$ is periodic, right?, and a periodic entire function is bounded so I could conclude by using Liouville, but I don't have that $f$ is entire. Perhaps there's another way on doing this.

4) I think I should use a remarkable theorem here but I don't remember, it looks hard.
 
Last edited:
Physics news on Phys.org
3) Indeed, Liouville is the key. First we solve the case $\omega_1\in\mathbb R$. Show that the range of $f$ is the same as $f(Q)$, where $Q$ is diamond-shaped.
 
girdav said:
Show that the range of $f$ is the same as $f(Q)$, where $Q$ is diamond-shaped.
How would you do it? I don't see how, and what is "diamond-shaped?" I was looking at it but I didn't find the definition.

Can you help me with other problems please?
 
I need help for 1), am I on the right track? But I can't continue. Can anybody check my work for 2). Is 4) bad written? Because I see it contradicts itself.
 
I can't edit now but on problem 4) it's actually $\mathcal U\backslash\{z_0\}.$

I think we can apply Morera's Theorem here, but I don't know how.
 
For problem 1) mostly you're given with an inequality so that could apply the maximum modulus principle, but in this case I have $p(z)$ equal to something, so I don't see how to apply the MMP here, any help?

girdav could you please help me more on problem 3), and can anybody help for problem 4)?
 
If $\omega_1\in\mathbb R$ and $\omega_2=a+bi\in \mathbb C$ with $b\neq 0$ hen for $z=x+iy$, write $z=x+iy$, then choose an integer $n$ such that $y=nb+\xi$, where $\xi<|b|$, so $z=x+i(nb+\xi)=x+inb+i\xi+na-na$ and $f(z)=f(x+\xi i-na)$. Now choose an integer $m$ such that $x-na=m\omega_1+\xi'$ with $\xi'<|b|$.
 
Okay but what's the direction you're pointing at? Are you trying to prove that $f$ is bounded? But I don't get the procedure, or trying to prove that $f(\mathbb C)$ equals to $f(A\times A)$ where $A$ is a compact set?
 
Yes that's it. Putting $M:=\max(|b|,|\omega_1|)$, we can show that for each $z\in\mathbb C$, we can find two integers $m$ and $n$ such that $z=m\omega_1+n\omega_2+\xi_1+i\xi_2$ where $\xi_1,\xi_2\in [0,M]$.
 
  • #10
Okay so since $[0,M]$ is compact and $f$ is entire, we have that $f$ is constant by Liouville's Theorem. Is it okay or do we have to work with the other case? I mean the $w_2$ ?
 
  • #11
What do you mean by the other case? By commodity, I supposed that $\omega_1$ is a real number. So we just have to show that it's without lose of generality.
 
  • #12
Oh yes, yes, but is it okay by saying that since $[0,M]$ is compact and $f$ is entire, we have that $f$ is constant by Liouville's Theorem?

girdav, I need help with problem 1, I don't see how to use the maximum modulus principle, can you give me a hand?
 
  • #13
For the first problem, write $P(z)=\frac a2(1-z^2)+(1-|a|^2)z+\frac{a-\bar a}2z^2$.

In order to clarify the thread, maybe you can edit the first message and write which problems have already been solved.
 
  • #14
So I have $\displaystyle\left| {p(z)} \right| \le \frac{1}{2}\left| {1 - {z^2}} \right| + \left| {1 - {{\left| a \right|}^2}} \right|z + \frac{1}{2}\left| {a - \overline a } \right|{z^2} \le \frac{1}{2}(1 + 1) + (1 + 1) \cdot 1 + \operatorname{Im} (a) \cdot 1,$ but I don't get yet that $|p(z)|\le1,$ how to finish it?
 
  • #15
Your bound is too large, you can write $|p(z)|\leq |a|+|1-|a|^2|+|a|=-|a|^2+2|a|+1=-(|a|-1)^2+1\leq 1$.
 
Back
Top