MHB How Does Complex Analysis Explain Geometry and Entire Functions?

  • Thread starter Thread starter Markov2
  • Start date Start date
  • Tags Tags
    Complex
Click For Summary
The discussion focuses on complex analysis and its relation to geometry and entire functions. It begins with a proof that the angle between two complex numbers is π/2 if their quotient is pure imaginary or their products sum to zero. The second point proves that three points on the unit circle that sum to zero form an equilateral triangle. The third point characterizes entire functions bounded by a specific growth condition, concluding that such functions are polynomials of degree at most one. Lastly, participants seek assistance with the proofs and calculations related to these problems.
Markov2
Messages
149
Reaction score
0
Denote $D=\{z\in\mathbb C:|z|<1\}$

1) Given $z,w\in\mathbb C,$ prove that the angle between $z$ and $w$ equals $\dfrac\pi2$ iff $\dfrac zw$ is pure imaginary or $\overline zw+z\overline w=0$

2) Let $a,b,c\in\partial D$ so that $a+b+c=0.$ Prove that the triangle $\Delta(a,b,c)$ is equilateral.

3) Let $C>0$ be a fixed constant. Characterize all the $f$ entire functions so that for all $z\in\mathbb C$ with $|z|>1$ is $|f(z)|\le\dfrac{C|z|^3}{\log|z|}.$

4) Consider a function $f\in\mathcal H(D)\cap C(\overline D).$ If exists $a\in\mathbb C$ so that for all $t\in[0,\pi]$ is $f(e^{it})=a,$ prove that for all $z\in D$ is $f(z)=a.$

Attempts:

1) I don't know how to see the stuff of the angles, what's the way to prove it?

2) I don't see a way to work it analytically, how to start?

3) Since $f$ is entire, it has convergent Taylor series then $f(z)=\displaystyle\sum_{k=0}^\infty\frac{f^{(k)(0)}}{k!}z^k,$ now by using Cauchy's integral formula we have $\displaystyle\left| {{f}^{(k)}}(0) \right|\le \frac{k!}{2\pi }\int_{\left| z \right|=R}{\frac{\left| f(z) \right|}{{{R}^{k+1}}}\,dz},$ and $\dfrac{{\left| {{f^{(k)}}(0)} \right|}}{{k!}} \le \dfrac{{C{R^3}}}{{{R^{k + 1}}\log R}} = \dfrac{{C{R^{2 - k}}}}{{\log R}}$ this clearly goes to zero as $R\to\infty,$ but for $k\ge2,$ then $f^{(k)}(0)=0$ for $k\ge2$ so the functions are polynomials of degree 1.

4) I don't see how to do this one.
 
Physics news on Phys.org
Markov said:
Denote $D=\{z\in\mathbb C:|z|<1\}$

1) Given $z,w\in\mathbb C,$ prove that the angle between $z$ and $w$ equals $\dfrac\pi2$ iff $\dfrac zw$ is pure imaginary or $\overline zw+z\overline w=0$

Hint:Dot product of z with w must be 0.

If $z=x+iy$ and $w=u+iv$ then:

$z\cdot w=xu+yv=0$

or:

$xu=-yv$

Now, what is $\frac{z}{w}$?

(Multiply the numerator and denominator by $\bar{w}$ )
 
Also sprach Zarathustra said:
If $z=x+iy$ and $w=u+iv$ then:

$z\cdot w=xu+yv=0$
I don't get this, why is not $z\cdot w=xu-yv$ ? Now $\dfrac{z}{w} = \dfrac{{(x + yi)(u - vi)}}{{{u^2} + {v^2}}} = \dfrac{{xu + (uy - xv)i + vy}}{{{u^2} + {v^2}}} = \dfrac{{(uy - xv)i}}{{{u^2} + {v^2}}},$ so this shows that $\dfrac zw$ is pure imaginary. Now for the converse, how do I start? I assume that $\dfrac zw$ is pure imaginary or the other one?

Can you help me with the other problems please?
 
I need help with 2), and, can anybody check my work for 3) please?
 
Markov said:
I need help with 2), and, can anybody check my work for 3) please?
Very nice question (2) !
From the given $a,b,c\in\partial D$ we deduce: $|a|=|b|=|c|$.

Now, for any complex numbers $z$ and $w$ we wave:

$|z-w|^2+|z+x|^2=2(|z|^2+|w|^2)$ (Prove it)

Now,

$a+b=-c$

With the formula above we have:

$|a-b|^2=3|c|^2$

Similarly:

$a+c=-b$

and $|a-c|^2=3|b|^2$$b+c=-a$

and $|b-c|^2=3|a|^2$But, $|a|=|b|=|c|$, hence:$|a-b|=|c-a|=|b-c|$The end!
 
Last edited:
http://www.mathhelpboards.com/member.php?52-Also-sprach-Zarathustra, can you help me with problem 1), I posted some questions there.

Can anybody check my work on problem 3)?
 
Also sprach Zarathustra said:
(Prove it)

You rang? :P
 
Prove It said:
You rang? :P

To be honest, I thought about you when I writ it down... :)
 
Also sprach Zarathustra said:
To be honest, I thought about you when I writ it down... :)

You're only human, how could you NOT think of me? ;)
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 0 ·
Replies
0
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K