Basic Complex analysis-why is z^1/3 discontinuous on real axis?

[LoA]

Homework Statement

Let $f(z) = z^{\frac{1}{3}}$ be the branch of the cube root function defined on $\left[0, 2\pi\right)$. Show that f is not continuous for $z_{0}$ where $z_{0}$ is such that Re($z_{0}$) > 0 and Im($z_{0}$) = 0.

Homework Equations

For a sequence of complex numbers, $z_{n} = x_{n} + iy_{n}$ converging to $z_{0} = x_{0} + iy_{0}$,

lim $z_{n} = z_{0}$ iff lim $x_{n} = x_{0}$ and lim $y_{n} = y_{0}$

f is continuous if:

lim f($z_{n}$) = f($z_{0}$) when lim $z_{n} = z_{0}$

The Attempt at a Solution

Let $z_{0} = a + ib$, a>0, b = 0. Let $z_{n} = x_{n} + iy_{n}$. So $lim z_{n} = z_{0}$ iff $lim x_{n} = a$ and $lim y_{n} = 0$.

$f(z_{n}) = e^{\frac{i}{3} [log(|z_{n}|) + arg(z_{n}) + 2k\pi)},$ for k = 0,1,2
$f(z_{n}) = (\sqrt{x_{n}^{2} + y_{n}^{2}})^{\frac{1}{3}}[cos(\frac{arg(x_{n}+iy_{n})+2k\pi}{3})+i sin(\frac{arg(x_{n}+iy_{n})+2k\pi}{3})]$

But $f(z_{0}) = a^{\frac{1}{3}}(cos(\frac{2k\pi}{3})+i sin(\frac{2k\pi}{3}), k = 0,1,2$ since $z_{0} = a$

I'm not sure where to go from here. When I take the limit of $f(z_{n})$, it seems to agree with the expression for $f(z_{0})$, though I'm uncertain how to handle the argument function. Is the discontinuity because of the choice of branch? But how does that give rise to a problem? Let's say I let $z_{n}$ approach a point on the real axis from above and another version approach the same point from below. Do they give the same value for the argument in the limit, or do they approach different values for the angle, say 0 for the approach from above and 2pi for the approach from below? But I thought arg(z) was defined between [-pi/2, pi/2], so wouldn't they approach the same angle, 0, from either direction? I'm asking about argument because my only clue is that somehow approaching from different positions gives different arguments for the roots, some of which may be outside the chosen branch, but I can't figure out how to show this. Any help, advice, or leading questions would be most appreciated. Thank you so much.

 

[Office_Shredder]

The problem is actually telling you that arg(z) is defined on [0,2pi) not [-pi,pi) (this is what they mean by picking a branch of the cube root function), so you should re-evaluate that approaching from above/below idea.

 

[LoA]

Ahh ok. So that clarifies my problem arg(z), but it leaves me back at step one. I'm not sure how to show this at all in that case. It seems that since $z_{n}$ approaches $z_{0}$, the function is continuous on that branch. I apologize if I'm being dense but I'm quite lost as to why this function is discontinuous when Im(z) = 0.

 

[Office_Shredder]

If you take the cube root of somebody whose argument is 2pi-epsilon, what is its new argument? Is it close to zero?

 

[LoA]

No, it's close to 2pi/3. So it's roots will be 2pi/3, 4pi/3, 6pi/3=2pi, and this last root will be outside of the selected branch of the cube root function, showing that in the limit the cube root of this guy is not the same as the cube root of a guy on the real axis, or at least doesn't exist in the chosen branch? But is it legitimate to claim that a sequence of such guys, getting closer and closer to 2pi, converges to a number on the real...oh wait I think I see the source of my confusion. Question: the choice of branch has no bearing on the domain of the function, correct? So I can argue that a sequence whose terms have arguments converging to 2pi converge to a z_0 on the axis as required by the problem. But then when we apply the function to these guys and take the limit of these functions, we get roots outside of the chosen branch. Is this the right idea? In this case, is it still true that lim f(z_n) = f(z_0), but simply that f(z_0) doesn't exist in the chosen branch?

 

[Office_Shredder]

No, it's close to 2pi/3. So it's roots will be 2pi/3, 4pi/3, 6pi/3=2pi,

No, there is only one cube root here. When they say "Take the branch defined on [0,2pi)" they are telling you how to turn z^1/3 into a single-valued function, instead of a multifunction, by picking special values. The way you decide which of 2pi/3, 4pi/3, or 6pi/3 to use for the argument of your cube root is you take the one which is one third of the original argument of your number that lay between 0 and 2pi to begin with.

The interval [0,2pi) isn't saying what interval your final answers should be in, it's saying what values of arguments you are allowed to plug into begin with. If you took the function defined on [2pi,4pi), then when you took the cube root of a guy with argument 2pi-epsilon, you would have to recalculate the argument to be 4pi-epsilon, and then take the cube root and end up with a number whose argument is about 4pi/3.

 

[jackmell]

Homework Statement

Let $f(z) = z^{\frac{1}{3}}$ be the branch of the cube root function defined on $\left[0, 2\pi\right)$

That's ill-posed. I can think of three branches of $z^{1/3}$ defined on $[0,2\pi)$. It should say, "such that $0\leq \arg(z)<2\pi$".

In my opinion, if you want to master multivalued functions, you've got to master illustrating them. So with this in mind and not wishing to interfere with others helping you, I would change the question to:

Plot the imaginary part of $z^{1/3}$, the whole $6\pi$ contraption, then excise a single-valued (non-overlapping), $2\pi$ section of it corresponding to $0\leq \arg(z)<2\pi$. Now notice the ledge. Obviously not continuous there. Get that intuitive understanding down first, then look at the algebra.

 

[LoA]

The interval [0,2pi) isn't saying what interval your final answers should be in, it's saying what values of arguments you are allowed to plug into begin with. If you took the function defined on [2pi,4pi), then when you took the cube root of a guy with argument 2pi-epsilon, you would have to recalculate the argument to be 4pi-epsilon, and then take the cube root and end up with a number whose argument is about 4pi/3.

Wow ok that helps a lot, thanks! So the branch determines what arg(z) is for the log that appears in complex powers? So the multivalued function f(z) = z^1/3 gives three distinct values all the way up to 6pi, and THEN the values begin to repeat? This was a lot simpler when we were just computing "all" the roots with de Moivre's formula. Somewhere in there I got really confused about this multi-valued stuff.

Ok, so as it's been given to me, the function is single valued. In particular, for this choice of branch, anything with argument ≥ 2pi (question: just how could this even happen algebraically? Isn't arg(z) = arctan(y/x)? So how do ever get something with arg > 2pi?) must have it's argument adjusted down by at least 2pi. And vice versa; if the branch were say [2pi,4pi), we would have to adjust a number with arg < 2pi UP by that much. Then the argument of the output will be 1/3 of the recalculated argument. Then when we compute the actual resulting complex number by evaluating the cos and sin of the recalculated argument, it's imaginary part will have jumped from where it looked like it was going before the recalculation. Am I on the right track?

 

[Office_Shredder]

That sounds right, although I don't know what you mean by the imaginary part jumping - the number will have a different imaginary and real part.

 

[LoA]

Great, I see what you mean now. I've worked out a rough sketch but I'm still a little uncertain of the algebra. Let me run my sketch past you and then ask how I can make it more concrete.

Let $z_{0}$ s.t. $|z_{0}| > 0$ and $arg(z_{0}) = 2\pi$.

Then $f(z_{0}) = |z_{0}|^{\frac{1}{3}}[cos(\frac{2\pi}{3} + \frac{2k\pi}{3}) + isin(\frac{2\pi}{3} + \frac{2k\pi}{3})]$ for $k = -1$

i.e. $f(z_{0}) = |z_{0}|^{\frac{1}{3}}[cos(\frac{2\pi}{3} - \frac{2\pi}{3}) + isin(\frac{2\pi}{3} - \frac{2\pi}{3})]$

So $f(z_{0}) = |z_{0}|^{\frac{1}{3}}$

Now let $z_{n}$ s.t. $\forall n$ $|z_{n}| > 0$ and $0 ≤ arg(z_{n}) < 2\pi$ and lim $z_{n} = z_{0}$.

Then $f(z_{n}) = |z_{n}|^{\frac{1}{3}}[cos(\frac{arg(z_{n})}{3} + \frac{2k\pi}{3}) + isin(\frac{arg(z_{n})}{3} + \frac{2k\pi}{3})]$ for $k = 0$

i.e. $f(z_{n}) = |z_{n}|^{\frac{1}{3}}[cos(\frac{arg(z_{n})}{3}) + isin(\frac{arg(z_{n})}{3})]$

But then lim $f(z_{n}) =$ lim $|z_{n}|^{\frac{1}{3}}[cos(\frac{arg(z_{n})}{3}) + isin(\frac{arg(z_{n})}{3})]$ goes to

$= |z_{0}|^{\frac{1}{3}}[cos(\frac{2\pi}{3}) + isin(\frac{2\pi}{3})]$
$= |z_{0}|^{\frac{1}{3}}(-\frac{1}{2} + \frac{\sqrt{3}}{2}i)$

i.e. lim $f(z_{n}) \neq f(z_{0}).$ So f is not continuous here.

Some questions:

a) Have I set up my sequence in an acceptable manner? Can I show that z_n can meet the constraints I put on it while still converging to the z_0 I gave?

b) How does arg(z) interact with limits? I know that the cube root of the modulus is continuous, and cos and sin are continuous, so I can pass the limit through those functions. But what is lim[arg(z_n)]? Is it really just arg(z_0)? Is the trick really just that the function gets "set up" differently for different arguments, which then may be altered by doing things to the function like taking limits, which in turn produces the disagreement between the outputs in this case?

c) Where can I make this more "algebraic"? While I think I'm starting to get the intuitive point here, I'm still very uncertain of how to write it out.

I know I'm being a little dense here but this material is both interesting and important to me and I very much want to master it. I am very grateful to you for your help Office_Shredder.

 

[Office_Shredder]

Scrap that arg(z₀) = 2pi stuff. Let z₀ be a positive real number. Then f(z₀) is a positive real number as well, and there's nothing more you have to say about it. Other than that, I think that your argument is a correct one .

You never said that the argument of these z_n converges to 2pi, which means that your argument is not correct - if you include that part though then I think that you're 100% fine (and note that once you state that you can pass limits through cos and sin because they are continuous functions). If you don't include that then your sequence could have been guys whose arguments are close to 0 and of course that wouldn't be a sequence that gives a counterexample. If you don't feel comfortable with the level of abstraction, all you need to do is pick a single sequence converging to z₀ to show discontinuity, not all of them. Let
$$z_n = z_0 e^{(2\pi - 1/n) i}$$
and show that f(z_n) converges to something whose argument is 2pi/3, along with noting that z_n converges to z₀ if z₀ is a positive real number.