Prove that this function is holomorphic

[GwtBc]

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.

 

[Ray Vickson]

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 $z$?

 

[GwtBc]

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 $a>0$

 

[Ray Vickson]

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 $\theta = \text{Arg}(z).$

 

[Dick]

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 $\sqrt{r} e^{i \frac{\theta}{2}}$.

 

[Ray Vickson]

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.

 

[GwtBc]

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

 

[GwtBc]

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.

 

[Dick]

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?

 

[GwtBc]

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 $\Omega$

 

[Dick]

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.

 

[Dick]

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.

 

[GwtBc]

I was thinking more of a proof using the difference quotient.

As in a difference quotient w.r.t. $z$ directly?

 

[Dick]

As in a difference quotient w.r.t. $z$ directly?

Sure, why not? Use the usual 'conjugate' trick.

 

[aheight]

Can we not show $f(z)^2=z$ then $f(z)=\sqrt{z}$ and thus $f(z)$ is analytic in $\Omega$?

 

[Dick]

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 $\frac{f(z+h)-f(z)}{h}$.

 

[Ray Vickson]

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 $\left. \sqrt{z}\:\right|_{z=1}$.