Welcome to our community

Be a part of something great, join today!

Complex stuff questions (3)

Markov

Member
Feb 1, 2012
149
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:

girdav

Member
Feb 1, 2012
96
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.
 

Markov

Member
Feb 1, 2012
149
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?
 

Markov

Member
Feb 1, 2012
149
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.
 

Markov

Member
Feb 1, 2012
149
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.
 

Markov

Member
Feb 1, 2012
149
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)?
 

girdav

Member
Feb 1, 2012
96
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 $\xi'<|b|$.
 

Markov

Member
Feb 1, 2012
149
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?
 

girdav

Member
Feb 1, 2012
96
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 $\xi_1,\xi_2\in [0,M]$.
 

Markov

Member
Feb 1, 2012
149
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 $w_2$ ?
 

girdav

Member
Feb 1, 2012
96
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.
 

Markov

Member
Feb 1, 2012
149
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?
 

girdav

Member
Feb 1, 2012
96
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.
 

Markov

Member
Feb 1, 2012
149
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?
 

girdav

Member
Feb 1, 2012
96
Your bound is too large, you can write $|p(z)|\leq |a|+|1-|a|^2|+|a|=-|a|^2+2|a|+1=-(|a|-1)^2+1\leq 1$.