# Show No Continuous Function (Complex Analysis)

1. Nov 19, 2011

### Poopsilon

Let n ≥ 2 be a natural number. Show there is no continuous function q_n : ℂ → ℂ such that (q_n(z))^n = z for all z ∈ ℂ.

The only value of this function we can deduce is q_n(0)=0. Moreover any branch cut we take in our complex plane will touch zero. These two facts would make me a bit suspicious that z=0 is the point at which the function can't be continuous except for the fact there is a previous problem which asks the reader to prove a special case of this problem for n=2 in the punctured plane, so at least in that case there must be some other point.

Nevertheless I still have a feeling that the inability for q_n to be continuous will have something to do with having to take some branch cut. Unfortunately despite this being graduate complex analysis the explanation by my professors of the branch cut and the concept of multi-valued functions has always been a bit hand-wavy and thus I don't fully understand how these things are rigorously constructed nor do I fully understand their implications.

Some direction on this problem would be much appreciated, and maybe someone could point me to some online resources that would help me understand the concepts of branch cut and multi-valued function ( wikipedia isn't that helpful ), thanks.

2. Nov 19, 2011

### Bacle2

This is one perspective that has helped me understand the complex log, aka logz

The basis for the concept of branch and branch cut, is that the complex exponential ez is many-to-1 (countably-∞-to-1 , actually), so that ez does not have a global inverse. Still, by, e.g., the complex version of the Inverse Function Theorem, it does have local inverses. Each of these local inverses is called a branch of the function; the branch cut is the line/curve that separates different local inverses. Consider this: ez has period i2π , so that ez=ez+i2π , so that ez maps a segment [0,2π) analytically into ℂ\{0}. You can then tile the imaginary axis by segments of length 2π- , each of which will give you a bijection into ℂ-{0}. The branch cuts prevent you from jumping between consecutive branches, i.e., between the bijective map ℂ\{0}→[i2nπ, i2(n+1)π) and the branch from ℂ\{0} →[i2(n+1)π, i2(n+2)π) .

Basically, logz assigns to z its polar coordinates and ez goes in the opposite direction , assigning to a pair of polar coordinates the number z with those specific coordinates. But the periodicity of polar coordinates means that this assignment is not well-defined globally, since cosθ=cos(θ±2nπ), sinθ=sin(θ±2nπ) .

Rant of the Day (Feel free to Ignore; I haven't had my coffee yet)
I guess the handwaving is part of the wave of modern mathematics, where so much is black-boxed into commutative diagrams , functors, categories, etc. , which is sometimes helpful, but also possibly obscuring of what is going on.

3. Nov 20, 2011

### jackmell

I think learning many concepts involving multi-valuedness should begin with the complex logarithmic function and it's geometry. So understand that first: it's imaginary part consists of an infinitely-folded sheet which wraps around itself like a twisted helix shown below.

For now, just take $n=2$ so that if $(q)^2=z$ then

\begin{align} q&=z^{1/2}\\ &=e^{1/2\log(z)}\\ &=\exp\{1/2(\ln|z|+i(\Theta+2n\pi))\} \end{align}

so that if we are to have a single-valued function $myq=e^{1/2\log(z)}$, then we must necessarily excise a single-valued part of the log sheet. The largest contiguous section we can remove before it begins to overlap itself again would be an arc-section slightly less than $2\pi$ for example, a part defined as:

$$\text{mylog(z)}=\ln|z|+i\Theta,\quad 0\leq\Theta< 2\pi$$

and then write as the single-valued branch of $q$:
$$myq=e^{1/2\;\text{mylog(z)}}$$
or something similar to that using a different starting and ending region for $\Theta$. That section of the $\log$ sheet we remove is called a "branch" of the function and where we make the cut is called the branch cut. But since the imaginary part of $\text{mylog(z)}$ is equal to $\Theta$, then the function $\text{Im(mylog)}=v(r,\Theta)=\Theta,\quad 0\leq\Theta<2\pi$ or rather this surface in 3D space, is not continuous (or contiguous) along the ray $\theta=0$. You should go our of your way to draw this function and really see visually why it's not continuous; rather, there is a sharp cliff rising a height of $2\pi$ and so $\text{mylog}$ is not continuous there and so too must be $myq$. Same dif with the other n's.

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Last edited: Nov 20, 2011
4. Nov 20, 2011

### Poopsilon

Ok this mostly makes sense, especially when you say "there is a sharp cliff rising a height of 2π" as that is how I have always envisioned it.

One thing I can't figure out is that I thought this problem of multi-valuedness with period 2π was a problem for all complex functions, what is so special about my q_n functions? I mean I understand that for example the square-root function is already a multi-valued function, even if we just consider it as a function from ℝ to ℝ so I guess in my brain I have the root functions which seem intrinsically multi-valued, where one must specify which root one wants produced for each z in the domain, and then on the other hand one has the problem of multi-valuedness for all complex functions because of the period 2π thing, and this somehow feels more extrinsic, sort of forced on the function by the nature of the complex plane.

The way I see the q_n function is that it is some function that takes every z to exactly one of its nth roots, you just don't know which. And thinking of it like that it is not at all obvious to me why such a function would have problems being continuous unless ALL functions in the complex plane would have a similar problem being continuous, which brings me back to my feeling that this branch-cut issue is somehow being forced on the function by the nature of the complex plane, and is not really a problem with the function itself.

So you're saying that helicoid you attached has its x and y coordinates as the real and imaginary input and then just outputs the angle and that's the z coordinate. That makes sense. I mean I understand conceptually most of what you're saying but I still can't figure out what makes q_n so special in not being continuous.

5. Nov 20, 2011

### jackmell

The origin of multivaluedness lies in the complex logarithm function. Functions which do not include them either explicitly or implicitly, are not multivalued (pretty sure about that but maybe there are some that are. Mostly anyway then). So polynomials, the exponential function, the trig functions, are all non-multivalued although you could make them multivalued by supplying multivalued arguments such as $\sin(\log(z))$. The inverse trig functions are multivalued because their definitions include the log function. Roots are multivalued because again, they include in their definitions, the complex log function. Other non-obvious functions are multivalued because somewhere within them lies a multivalued component.

q_n is being implicitly defined as a single-valued "function" by saying "a function maps C to C." That implicitly means "single-valued" so that's why we define it as we did.

The single-valued root function is discontinuous along it's branch cut but the multivalued root function $\sqrt{z}$ is differentiable (locally) everwhere except at zero. Yeah, I know, the single-valued and multi-valued one look the same.

Branch-cuts are a consequence of the complex log function and the requirements for functions to have single-values for many applications in math.

I should have written $\log(z)=ln|r|+i\arg(t)$ and for any function f(z) we can write it's real and complex component as :

$$f(z)=u(r,\theta)+iv(r,\theta)$$

then the imaginary component of log(z) is $u(r,\theta)=\theta$. Ok, if you plot $u(r,\theta)$as a function of the two variables r and theta, you'll get the helicoid.

Last edited: Nov 20, 2011
6. Nov 20, 2011

### Poopsilon

Ahhhh I see. Ok I see what you mean about the log function being implicit within the nth root functions. And I also understand that the log function in the complex plane is the inverse of the exponential function which is not injective (since it gives the same output for every 2π increase/decrease in period). But even the the identity function f(z)=r*exp(iθ) gives the same output for every 2π increase/decrease in its period, thus I still can't figure out what's so special about the inverse of exp(z).

7. Nov 21, 2011

### jackmell

8. Nov 21, 2011

### Bacle2

I think the multivaluedness here also has to see with the periodicity of the sin and cos functions, and the polar form z=rcosθ+isinθ=rcos(θ+2nπ)+isin(θ+2nπ) , which is what the logz gives you, so that you get a countably-infinite-to-1 representation of a complex number
every time you use a log, or a polar representation. But , while you cannot get a global inverse, you can still get many local inverses (if f'(z)±0 ) called logθ , and each of these inverses is a branch, where you have a unique, consistent, polar representation in a range [θ,θ+2π) , with the openness on the right to prevent periodicity from creeping-in .

Notice too, that because of the fundamental theorem of algebra, all polynomials of degree higher-than-one are many-to-one; only entire biholomorphic maps are of the type az+b.