Prove that this function is holomorphic

In summary: If...If ##z = r e^{i \theta},## then ##|arg(z)| < \pi## means ## 0 \leq \theta < \pi##. Together with ##z \neq 0##, this ensures that you will never have ##y = 0## when ##x \leq 0##. Certainly we have ##y > 0## in the interior of the region ##\Omega##. Is that enough? How do your course notes and/or textbook deal with the concept of differentiability at a boundary point of a region? (Back in the Stone Age when I took complex variables we did not let "holomorhpic" look at boundary points.)
  • #1
GwtBc
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6

Homework Statement


Prove that the function ## f(z)= 1/\sqrt{2}(\sqrt{\sqrt{x^{2}+y^{2}}+x}+i*sgn(y)\sqrt{\sqrt{x^{2}+y^{2}}-x})## is holomorphic on the domain ## \Omega = \left \{ z: z \neq 0, \left | \arg{z} \right | <\pi\right \} ## and further that in this domain ##f(z)^{2} = z. ##

Homework Equations


if
## f(z) = u(x,y) + iv(x,y) ##
Then we have the Cauchy-Riemann relations for a holomorphic function
## \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\hspace{0.3cm} \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}##

The Attempt at a Solution



Essentially I've been trying to show that the function conforms with the Cauchy Riemann equations, but there are two issues. Firstly, the imaginary component seems not to be differentiable where y = 0. I tried invoking first principles but the limit doesn't exist. Second is that it's not immediately obvious that the Cauchy relations hold even where im{f} is differentiable. I could see this latter issue being resolved by some algebraic trick, but I'm kind of clueless about the first.

Thanks in advance everyone.
 
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  • #2
GwtBc said:

Homework Statement


Prove that the function ## f(z)= 1/\sqrt{2}(\sqrt{\sqrt{x^{2}+y^{2}}+x}+i*sgn(y)\sqrt{\sqrt{x^{2}+y^{2}}-x})## is holomorphic on the domain ## \Omega = \left \{ z: z \neq 0, \left | \arg{z} \right | <\pi\right \} ## and further that in this domain ##f(z)^{2} = z. ##

Homework Equations


if
## f(z) = u(x,y) + iv(x,y) ##
Then we have the Cauchy-Riemann relations for a holomorphic function
## \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y},\hspace{0.3cm} \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}##

The Attempt at a Solution



Essentially I've been trying to show that the function conforms with the Cauchy Riemann equations, but there are two issues. Firstly, the imaginary component seems not to be differentiable where y = 0. I tried invoking first principles but the limit doesn't exist. Second is that it's not immediately obvious that the Cauchy relations hold even where im{f} is differentiable. I could see this latter issue being resolved by some algebraic trick, but I'm kind of clueless about the first.

Thanks in advance everyone.

Have you forgotten the restrictions placed on ##z##?
 
  • #3
Ray Vickson said:
Have you forgotten the restrictions placed on ##z##?
I don't think so. So ##z## can be any number in the complex plane but the negative reals or zero. This still means the function has to be differentiable at any point ##(a,0)## where ##a \in \mathbb{R}## and ##a>0##
 
  • #4
GwtBc said:
I don't think so. So ##z## can be any number in the complex plane but the negative reals or zero. This still means the function has to be differentiable at any point ##(a,0)## where ##a \in \mathbb{R}## and ##a>0##

If ##z = r e^{i \theta},## then ##|arg(z)| < \pi## means ## 0 \leq \theta < \pi##. Together with ##z \neq 0##, this ensures that you will never have ##y = 0## when ##x \leq 0##. Certainly we have ##y > 0## in the interior of the region ##\Omega##. Is that enough? How do your course notes and/or textbook deal with the concept of differentiability at a boundary point of a region? (Back in the Stone Age when I took complex variables we did not let "holomorhpic" look at boundary points.)

Note added in edit: If you are using the convention that ##\theta \in (-\pi,pi]##, then, of course, the condition ##|\theta| < \pi## just eliminates the negative real axis, nothing more. However, if you use the convention that ##\theta \in [0,2 \pi)## then ##|\theta| < \pi## is the upper half-plane with the negative real axis removed.

Check your textbook or course notes to see which convention is assumed for ##\theta = \text{Arg}(z).##
 
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  • #5
Ray Vickson said:
Check your textbook or course notes to see which convention is assumed for ##\theta = \text{Arg}(z).##

Since they specified ##|arg(z)|## I'm pretty sure they intend to cover negative values of ##arg(z)##. Also ##sgn(y) |y|## might not look differentiable, but it is. An easy way to show it's holomorphic is to show that if ##z = r e^{i \theta}## what you have there is a formula for ##\sqrt{r} e^{i \frac{\theta}{2}}##.
 
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  • #6
Dick said:
Since they specified ##|arg(z)|## I'm pretty sure they intend to cover negative values of ##arg(z)##. Also ##sgn(y) |y|## might not look differentiable, but it is. An easy way to show it's holomorphic is to show that if ##z = r e^{i \theta}## what you have there is a formula for ##\sqrt{r} e^{i \frac{\theta}{2}}##.

I guess you are right on that score. Anyway, ##\text{sgn}(y) |y| = y,## so of course it is smooth.
 
  • #7
Ray Vickson said:
If ##z = r e^{i \theta},## then ##|arg(z)| < \pi## means ## 0 \leq \theta < \pi##. Together with ##z \neq 0##, this ensures that you will never have ##y = 0## when ##x \leq 0##. Certainly we have ##y > 0## in the interior of the region ##\Omega##. Is that enough? How do your course notes and/or textbook deal with the concept of differentiability at a boundary point of a region? (Back in the Stone Age when I took complex variables we did not let "holomorhpic" look at boundary points.)

Note added in edit: If you are using the convention that ##\theta \in (-\pi,pi]##, then, of course, the condition ##|\theta| < \pi## just eliminates the negative real axis, nothing more. However, if you use the convention that ##\theta \in [0,2 \pi)## then ##|\theta| < \pi## is the upper half-plane with the negative real axis removed.

Check your textbook or course notes to see which convention is assumed for ##\theta = \text{Arg}(z).##

If ##f(z)## is defined in a nbd of ##z_{0}## we say that ##f(z)## is differentiable at ##z_{0}## if we have that ##f(z) = f(z_{0})+L(z-z_{0})+\eta## where the complex number ##L## is independent of ##z## and ##\frac{\eta}{\left |z-z_{0} \right |} \rightarrow 0## as ##z \rightarrow z_{0}##. I take this to mean that a function cannot ever be differentiable at a boundary point i.e. it cannot be differentiable on a close set, but that's ok cause this one is open.

edit: Still not sure why I can't differentiate at ##(a,0)##. Mathematica also outputs indeterminate for the limit defining the derivate at such points. My expression for it was ##\lim_{h \to 0}\frac{\sqrt{\sqrt{x^{2}+h^{2}}-x}}{h}##

edit #2: I was wrong. The equality where ## y \neq 0## is pretty much immediately obvious
 
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  • #8
Dick said:
Since they specified ##|arg(z)|## I'm pretty sure they intend to cover negative values of ##arg(z)##. Also ##sgn(y) |y|## might not look differentiable, but it is. An easy way to show it's holomorphic is to show that if ##z = r e^{i \theta}## what you have there is a formula for ##\sqrt{r} e^{i \frac{\theta}{2}}##.
I did show this and it pretty much solves the second half of the question, but I don't see how it helps with the first.
 
  • #9
GwtBc said:
I did show this and it pretty much solves the second half of the question, but I don't see how it helps with the first.

If you write ##\sqrt{r} e^{i \frac{\theta}{2}}## as ##z^\frac{1}{2}## it's pretty clearly holomorphic, isn't it?
 
  • #10
Dick said:
If you write ##\sqrt{r} e^{i \frac{\theta}{2}}## as ##z^\frac{1}{2}## it's pretty clearly holomorphic, isn't it?
Well you can write ##z^{1/2}## as ##\frac{\mathrm{d} }{\mathrm{d} z}(\frac{2}{3}z^{3/2})## which I do believe means it must be holomorphic, but then my question is, is this true ##\forall z \in \mathbb{C}## ? so Why would they specify the set ##\Omega##?

edit: Actually then you would have to show that ##z^{3/2}## is holomorphic (either way a proper proof is required for this question) but again I can't understand why this would only be true on ##\Omega##
 
  • #11
GwtBc said:
Well you can write ##z^{1/2}## as ##\frac{\mathrm{d} }{\mathrm{d} z}(\frac{2}{3}z^{3/2})## which I do believe means it must be holomorphic, but then my question is, is this true ##\forall z \in \mathbb{C}## ? so Why would they specify the set ##\Omega##?

edit: Actually then you would have to show that ##z^{3/2}## is holomorphic (either way a proper proof is required for this question) but again I can't understand why this would only be true on ##\Omega##

Because ##z^\frac{1}{2}## isn't well defined over all of ##\mathbb{C}##. Every number has two different square roots. You need to pick a continuous branch of the function to say it's holomorphic. You can do that by omitting the negative real axis.
 
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  • #12
GwtBc said:
Well you can write ##z^{1/2}## as ##\frac{\mathrm{d} }{\mathrm{d} z}(\frac{2}{3}z^{3/2})## which I do believe means it must be holomorphic

I was thinking more of a proof using the difference quotient.
 
  • #13
Dick said:
I was thinking more of a proof using the difference quotient.
As in a difference quotient w.r.t. ##z## directly?
 
  • #14
GwtBc said:
As in a difference quotient w.r.t. ##z## directly?

Sure, why not? Use the usual 'conjugate' trick.
 
  • #15
Can we not show ##f(z)^2=z## then ##f(z)=\sqrt{z}## and thus ##f(z)## is analytic in ##\Omega##?
 
  • #16
aheight said:
Can we not show ##f(z)^2=z## then ##f(z)=\sqrt{z}## and thus ##f(z)## is analytic in ##\Omega##?

Basically, yes. Except that ##\sqrt{z}## is not particularly well defined. I would just evaluate ##\frac{f(z+h)-f(z)}{h}##.
 
  • #17
Dick said:
Basically, yes. Except that ##\sqrt{z}## is not particularly well defined. I would just evaluate ##\frac{f(z+h)-f(z)}{h}##.

In the domain ##\Omega## specified in the problem statement, ##\sqrt{z}## is perfectly well-defined and is definitely holomorphic, at least if one specifies the principle branch. Furthermore, one can guarantee that the principle branch is the correct one to use in this problem just by evaluating ##f(1 + 0 i)## and comparing that with the principle value of ##\left. \sqrt{z}\:\right|_{z=1}##.
 

1. What does it mean for a function to be holomorphic?

Holomorphic functions are complex-valued functions that are differentiable and analytic on a domain in the complex plane. This means that they can be represented by a power series and have a well-defined derivative at every point within their domain.

2. How can you prove that a function is holomorphic?

To prove that a function is holomorphic, you must show that it satisfies the Cauchy-Riemann equations, which state that the partial derivatives of the function with respect to its real and imaginary components must exist and be continuous at each point within its domain. Additionally, the function must also be differentiable and have a continuous derivative at each point within its domain.

3. What is the importance of holomorphic functions in mathematics?

Holomorphic functions are important in complex analysis and other areas of mathematics because they have many useful properties, such as being analytic and having a well-defined derivative at every point. This makes them useful for solving problems involving complex numbers and for understanding the behavior of functions in the complex plane.

4. Can a function be holomorphic on a non-continuous domain?

No, a function must have a continuous domain in order to be holomorphic. This is because the Cauchy-Riemann equations require the partial derivatives to exist and be continuous at each point within the domain. If the function is not continuous, then it cannot satisfy these equations and therefore cannot be considered holomorphic.

5. Are there any other conditions that a function must meet in order to be considered holomorphic?

In addition to being differentiable and satisfying the Cauchy-Riemann equations, a function must also be analytic within its domain in order to be considered holomorphic. This means that it can be represented by a convergent power series, and its derivatives can also be represented by power series. If a function meets all of these criteria, then it is considered holomorphic.

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