Show that an image of a Schlicht function contains ##\Delta(0,1/2)##

In summary, the conversation discusses a problem from a past exam where the last point involves using the Cauchy mean value theorem to construct a function that satisfies a certain condition. The solution involves dividing the integral into three parts and using the Cauchy mean value theorem to find a point that satisfies the condition. The hint provided can also be used to construct the desired function.
  • #1
Westlife5
1
0
Homework Statement
Let ##f:\Delta\to \mathbb{C}## be an injective holomorphic function such that ##f(0)=0## and ##f'(0)=1##

(i.e ##f\in\, S## is schlicht).

Assume that ##A=f(\Delta)## is convex and let ##r\in(0,1)## and ##e^{i\theta}\in\partial\Delta##


##1## Prove that

$$\frac{1}{2}re^{i\theta}=\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z}(1+\frac{z}{2re^{i\theta}}+\frac{re^{i\theta}}{2z})dz $$


##2## using part #1#, show that

$$\frac{1}{2}re^{i\theta}=\frac{1}{\pi}\int_{-\pi}^{\pi}f(re^{ix})\cos^2(\frac{\theta-x}{2})dx$$


##3##Show that ##\frac{1}{2}re^{i\theta}\in A##

##hint##: use that ##\int_{-\pi}^{\pi}\cos^2(\frac{\theta}{2})=\pi## and that ##A## is convex. In particular use the fact that given ##z_0\not\in A##, there exsits ##z\in\mathbb{C}## such that ##\langle z,z_0\rangle >\langle z,w \rangle## for every ##w\in A##, where ##\langle z,w\rangle ## denotes the standard scalar product between ##w,z\in\mathbb{C}## seen as elements of ##\mathbb{R}^2##
Relevant Equations
Not sure what to write here
Hello everyone this was a problem on one of the exams from last year and I'm having trouble with the last point ##3##

my solution for ##1##
$$\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z}(1+\frac{z}{2re^{i\theta}}+\frac{re^{i\theta}}{2z})dz =$$
I divided this integral into 3 different ones and got.
$$\int_{|z|=r}\frac{f(z)}{z}dz=f(0)=0$$
$$\int_{|z|=r}\frac{f(z)}{2re^{i\theta}}=\frac{1}{2re^{i\theta}}\int_{|z|=r}f(z)dz=0$$ since f is a holomorphic function
$$\frac{re^{i\theta}}{2}\int_{|z|=r}\frac{f(z)}{z^2}dz=\frac{re^{i\theta}}{2}f'(0)2\pi i=\frac{re^{i\theta}}{2}2\pi i$$so if we put everything together we get

$$\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z}(1+\frac{z}{2re^{i\theta}}+\frac{re^{i\theta}}{2z})dz =\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)}{z^2}=\frac{1}{2\pi i}re^{i\theta}\pi i=\frac{1}{2}re^{i\theta}$$

my solution for ##2##
First I parametrized the curve ##|z|=r## by setting ##z=re^{i\psi}## therefore ##dz=re^{i\psi}id\psi## substituting this into the equation I get
$$
\frac{1}{2\pi }\int_{-\pi}^{\pi}f(re^{i\psi})(1+\frac{1}{2}e^{i(\psi-\theta)}+\frac{1}{2}e^{i(\theta-\psi}))d\psi=
$$
$$
\frac{1}{2\pi }\int_{-\pi}^{\pi}f(re^{i\psi})(1+\frac{\cos(\psi-\theta)+\cos(\theta-\psi)}{2}+i\frac{\sin(\theta-\psi)+\sin(\psi-\theta)}{2}d\psi=
$$
sines cancel, cosines add up+ half angle formula
$$
\frac{1}{2\pi }\int_{-\pi}^{\pi}f(re^{i\psi})(1+cos(\psi-\theta))d\psi=\frac{1}{2\pi }\int_{-\pi}^{\pi}f(re^{i\psi})2\cos^2(\frac{\theta-\psi}{2})d\psi
$$
which is what we wanted.

##3##

This is the one I'm having trouble with. I know that I have to write ##\frac{1}{2}re^{i\theta}## as an element of the image ##f(z)=\frac{1}{2}re^{i\theta}## for some z. I've though about using something like the Cauchy mean value theorem, however as far as I'm aware that theorem can only be used if we are mapping from ##\mathbb{R}\to\mathbb{R}## (unless it can be used in this case too?). But I'm unsure of how to use the hint.
Could somebody give me a hint of how to use the hint?
 
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  • #2
The hint can be used to construct a function ##g(z)## such that ##f(z)=\frac{1}{2}re^{i\theta}## if and only if ##g(z)=0##. You can use the fact that ##f(z)-\frac{1}{2}re^{i\theta}## is holomorphic, so by the Cauchy mean value theorem, there exists a point ##z_0## such that $$f(z_0)-\frac{1}{2}re^{i\theta}=\frac{1}{2\pi i}\int_{|z|=r}\frac{f(z)-\frac{1}{2}re^{i\theta}}{z-z_0}dz=0$$Therefore, let ##g(z)=f(z)-\frac{1}{2}re^{i\theta}##, then you can solve for ##z_0## to get your desired result.
 

Related to Show that an image of a Schlicht function contains ##\Delta(0,1/2)##

1. What is a Schlicht function?

A Schlicht function is a complex-valued function that is analytic and injective in a given region of the complex plane.

2. What is the significance of an image of a Schlicht function containing ##\Delta(0,1/2)##?

The image of a Schlicht function containing ##\Delta(0,1/2)## is a visual representation of the function's behavior in the complex plane, specifically at the point (0,1/2). It can provide insights into the function's properties and relationships with other functions.

3. How is the image of a Schlicht function related to the concept of conformal mapping?

A Schlicht function is a type of conformal mapping, meaning it preserves angles and local shapes. Therefore, the image of a Schlicht function can be used to map a given region in the complex plane to another region with different properties while maintaining the same angles and shapes.

4. What is the role of the delta function in the image of a Schlicht function?

The delta function, denoted by ##\Delta(0,1/2)##, represents a point in the complex plane. In the context of a Schlicht function, this point is often significant as it can provide insights into the function's behavior and properties at that specific point.

5. How can one analyze the image of a Schlicht function containing ##\Delta(0,1/2)##?

There are various techniques for analyzing the image of a Schlicht function, such as using geometric methods, computing the function's derivatives, or using complex analysis tools. The specific approach will depend on the properties and behavior of the function and the intended analysis.

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