What is the domain? Where is it differentiable? What is the derivative?

[Hodgey8806]

I'm private studying a section in Linear Algebra first dealing with complex numbers. Now, I am ok with most of the answers, but I need help particularly with where this function is differentiable.

Homework Statement

Let f(z) = ln|z| + i arg z, 0<=arg z<2pi.

What is the domain? What is it differentiable? What is the derivative.

Homework Equations

A relevant equation is that z is of the form z = x + iy

Another relevant equation might be that |z| = sqrt(x^2 + y^2)

The Attempt at a Solution

My solution attempt is to state that it is continuous for all z except z = 0.

Next, to find where it is differentiable, I use the Cauchy-Riemann equations.

du/dx = x/(x^2+y^2) , du/dy = y/(x^2 + y^2), dv/dx = 0 = dv/dy
*This of course du is the differentiation of the real parts, and dv is the differentiation of the imaginary parts of z."

However, this never satisfies the Cauchy-Riemann equations, so I would assume it's either not differentiable.

Now, it's easy to show the derivative of the function.

f'(z) = [ln sqrt(z^2)]' = 1/z . I'm assuming I'm supposed to use this somehow, but I really don't know.

Answer
Now, the answer for this solution of differentiability is:

It is differentiable everywhere except at z = 0 and the positive real axis.

I don't know what the answer means or how to show that. Please, correct my derivations if they need to be. Thanks!

 

[Dick]

Why do you think dv/dx=0? Shouldn't you find an expression for v(x,y)=arg(z) in terms of x and y before you conclude that?

 

[Hodgey8806]

I wasn't sure on what to do with that, but the section tends to suggest it is just a value or constant that goes to zero when differentiated.

Atleast, that's what I'm assuming. The text I'm studying from is "Intro to Linear Algebra and Diff Eq" by John W. Dettman. P 25, #11.

The entirety of the questions is stated above. I'm assuming it is saying that arg z takes on a specific number between 0 and 2pi, but that it is not a variable. Just a restriction.

If you have an idea of how it becomes differentiable on all z except z=0 and the entire real axis. It'd be greatly appreciated.

 

[HallsofIvy]

I wasn't sure on what to do with that, but the section tends to suggest it is just a value or constant that goes to zero when differentiated.

Atleast, that's what I'm assuming. The text I'm studying from is "Intro to Linear Algebra and Diff Eq" by John W. Dettman. P 25, #11.

The entirety of the questions is stated above. I'm assuming it is saying that arg z takes on a specific number between 0 and 2pi, but that it is not a variable. Just a restriction.

No, that's not at all what it says. "0<= arg(z)<= 2pi" means that arg z can be any number between 0 and 2 pi but does NOT say it is a constant.

If you have an idea of how it becomes differentiable on all z except z=0 and the entire real axis. It'd be greatly appreciated.

I think it would be simplest to put this function in polar form: with z= r e^{i theta}, f(z)= ln(r)+ 2i theta. r can be any non-negative number, theta any number from 0 to 2 pi.

 

[Dick]

I wasn't sure on what to do with that, but the section tends to suggest it is just a value or constant that goes to zero when differentiated.

Atleast, that's what I'm assuming. The text I'm studying from is "Intro to Linear Algebra and Diff Eq" by John W. Dettman. P 25, #11.

The entirety of the questions is stated above. I'm assuming it is saying that arg z takes on a specific number between 0 and 2pi, but that it is not a variable. Just a restriction.

If you have an idea of how it becomes differentiable on all z except z=0 and the entire real axis. It'd be greatly appreciated.

This isn't that mysterious. arg(z) is the angle measured counterclockwise from the x axis. Draw a right triangle. tan(theta)=y/x. So one expression valid for 0<=theta<=pi/2 is theta=arg(z)=arctan(y/x). Use that and check Cauchy-Riemann. Then figure out a way to define theta around the whole circle. And you should rethink the continuity answer too. What's arg(z) for a point on the positive x-axis? What are arg(z) like for a point a little above or below the point on the x-axis?

 

[Hodgey8806]

I went through the problem, and I got the answer the book has about continuity. (Every except where z=0. Which makes sense for the logarithm.

I also remember the arctan(y/x), and I checked it with Cauchy-Riemann. They satisfy the equations, but I still can't get to the point of it not working for the positive real axis.

Maybe I'm thinking the wrong way about the positive real axis. But that suggest the imaginary pieces are zero. But is it by implication that that also suggest the reals are zero too? I don't know how exactly to think of it. I really appreciate the help.

All I can see is that it can't be differentiated if x^2 + y^2 = 0 (by the Cauchy-Riemann equations). But otherwise, that doesn't really help me. Is that supposed to mean that it crosses the real axis only when x = 0 and y = 0? Again, I really appreciate the help. I hope I'm not a bother.

 

[Hodgey8806]

Ohh, do you mean because it produces a vertical slope? Thanks!

 

[micromass]

I went through the problem, and I got the answer the book has about continuity. (Every except where z=0. Which makes sense for the logarithm.

Does the book really say that, that's hard to believe. In that case, I fear the book is wrong...

 

[Hodgey8806]

The books answer says, "The domain is all z except z=0. f(z) is differentiable everywhere except at z=0 and on the positive real axis. f'(z) = 1/z."

That is the exact quote of the solution to this problem.

But it is speaking of z in terms of z = x + iy .

Firstly, I don't like the term of the derivative being 1/z. That is misleading when the the true derivative of f(z) is f'(z) = 1/|z|^2 . Thus, the imaginaries disappear.

What were you suggesting the continuity is?

 

[micromass]

The books answer says, "The domain is all z except z=0. f(z) is differentiable everywhere except at z=0 and on the positive real axis. f'(z) = 1/z."

That is the exact quote of the solution to this problem.

The book is correct. The domain is indeed everything except 0. However, the book doesn't say anything about continuity! The function is not continuous on the positive real axis, can you see why?

Firstly, I don't like the term of the derivative being 1/z. That is misleading when the the true derivative of f(z) is f'(z) = 1/|z|^2 . Thus, the imaginaries disappear.

Why would the true derivative be 1/|z|²?? This is not true. How did you find this?

 

[Hodgey8806]

Oh, goodness I'm sorry about that.

f(z) = ln|z| = ln sqrt(z^2)
f'(z) = 1/sqrt(z^2)*1/2sqrt(z^2)*2z = z/z^2 = 1/z. But where does the imaginary part go. Because isn't this translated into f'(z) = 1/(x^2+y^2) .

I now realize my error in that interpretation. But if the derivative is 1/z. Doesn't that suggest that it is 1/(x+iy). Where does the imaginary part go? Thanks!

 

[micromass]

Oh, goodness I'm sorry about that.

f(z) = ln|z| = ln sqrt(z^2)

But it simply isn't true that $|z|=\sqrt{z^2}$. For example, for z=i, then it isn't true that

$$1=|i|=\sqrt{i^2}=\sqrt{-1}$$

whatever that right-hand side mught mean.

I highly suggest you take the partial derivatives of the function and apply Cauchy-Riemann...

 

[Hodgey8806]

Also, I really can't see why it's not continuous on the positive real axis.

May I have a hint? Is it the arctangent or is it the ln|z|? Or am I just missing it.

 

[micromass]

Also, I really can't see why it's not continuous on the positive real axis.

Consider the sequences:

$$x_n=e^{i(2\pi-\frac{1}{n})}$$

What does it converge to? What does Log(x_n) converge to?

 

[Hodgey8806]

I took the partial derivatives and found the Cauchy-Riemann equations are satisfied. I don't see how I can say that it is 1/z using them.

I think I can see it. If I say:

f(z) = ln|z| + arctan(y/x) = ln(sqrt(x^2+y^2)) + arctan(y/x)
f'(z)=1/z

and each partial is either U' or V' with respect to either y or x. Is that somewhat correct?

 

[Hodgey8806]

At the risk of making a fool of myself. I didn't cover imaginary sequences in Calculus. But, from the section I'm studying, isn't it safe to say:

e^i(2pi-1/n) can be written as cos(2pi-1/n) + isin(2pi-1/n). Which as n approaches infinity, it converges to cos(2pi)+isin(2pi) = 1.

And thus the log(1) = 0. So log(xn) converges to 0. Is that correct?

 

[micromass]

So, you've found the partial derivatives, you say? Well, then the derivative is easy: if f=u+iv, then

$$f^\prime(z)=\frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x}$$.

At the risk of making a fool of myself. I didn't cover imaginary sequences in Calculus. But, from the section I'm studying, isn't it safe to say:

Dpn't worry, you'll find complex sequences behaving a lot like real sequences...

e^i(2pi-1/n) can be written as cos(2pi-1/n) + isin(2pi-1/n). Which as n approaches infinity, it converges to cos(2pi)+isin(2pi) = 1.

And thus the log(1) = 0.

So far so good.

So log(xn) converges to 0. Is that correct?

This is not correct, you assumed continuity of the log here. It's not because x_n converges to x, that f(x_n) converges to f(x). This is only true for continuous functions, but continuity is exactly what we're trying to show here!

So, you'll need to calculate log(x_n) directly...

 

[Dick]

Maybe I'm thinking the wrong way about the positive real axis. But that suggest the imaginary pieces are zero. But is it by implication that that also suggest the reals are zero too? I don't know how exactly to think of it. I really appreciate the help.

Concentrate on the arg(z) part of log(z). Along the positive real axis arg(z) as you've defined it is 0. As you rotate counterclockwise at fixed radius from the positive x-axis it increases until it's almost 2pi at the end of going through the fourth quadrant. Then it has to snap back to 0 again on the positive x-axis. Doesn't sound continuous to me.

 

[Hodgey8806]

So, for both responses:

It's because we need to calculate the ln|z| directly and that's why it's not continuous? Is that because when you reach 2pi it goes back to 0?

I'm really trying to completely wrap my head around it. And I really appreciate the help! Have a great Memorial Weekend.

 

[Dick]

So, for both responses:

It's because we need to calculate the ln|z| directly and that's why it's not continuous? Is that because when you reach 2pi it goes back to 0?

I'm really trying to completely wrap my head around it. And I really appreciate the help! Have a great Memorial Weekend.

Yes, it's because the arg(z) part is discontinuous on the x-axis because, as you said, it has to go from 2pi to 0. Can't be continuous. You can move the discontinuity around by redefining where it has to shift by 2pi, but you can't get rid of it altogether.

 

[Hodgey8806]

Alright, great! Is this always ok to assume for any arg z that is bounded between 0<=argz<2pi?

 

[Dick]

Alright, great! Is this always ok to assume for any arg z that is bounded between 0<=argz<2pi?

Well, no, depends on how you define arg(z). I'll define my 'arg(z)' to be pi/2 along the y-axis and equal to pi/2+(the angle measured counterclockwise from the y axis) until you hit the y-axis again. Nothing wrong with that, right? Except the discontinuity is now along the y-axis. And it still has the property that exp(|z|+i*'arg(z)')=z.

 

[Hodgey8806]

Excellent. Thanks for your help, both of you! If either of you has time, I have posted another thread in the other forum:

https://www.physicsforums.com/showthread.php?t=502082

And if you would help, it'd be much appreciated. The last post, I just posted the problem as the help I received didn't make sense for what the problem was asking. But no worries if you cant. I won't be offended ;)

Again, I really do appreciate the help!