Challenge IV: Complex Square Roots, solved by jgens

[micromass]

This is a well-known result in complex analysis. But let's see what people come up with anyway:

Challenge:
Prove that there is no continuous function $f:\mathbb{C}\rightarrow \mathbb{C}$ such that $(f(x))^2 = x$ for each $x\in \mathbb{C}$.

 

[jgens]

Suppose such a function existed and consider the map g:S¹→S¹ given by restriction. Then the degree of g² is a multiple of two but the degree of id_S¹ is one. This contradiction completes the proof.

 

[Infrared]

Suppose a function f did exist. For a complex number z=Re^{i\theta}, the only two square roots are \pm \sqrt{R}e^{\frac{i\theta}{2}}. Call the function with the plus sign f_1 and the function with the minus sign f_2

First let's show that a function that takes on the value f(z)=\sqrt{R}e^{\frac{i\theta}{2}} for some z's and f(z)=-\sqrt{R}e^{\frac{i\theta}{2}} for others is discontinuous.

For nonzero z_1=R_1e^{i\theta_1} and z_2=R_2e^{i\theta_2} assume that f(z_1)=\sqrt{R_1}e^{\frac{i\theta_1}{2}} and f(z_2)=-\sqrt{R_2}e^{\frac{i\theta_2}{2}}.

Let \gamma:[0,1] \to ℂ such that \gamma(0)=z_1 and \gamma(1)=z_2 be an injection and a parameterization of a path from z_1 to z_2 that does not pass through the origin . Let t' be the infimum over all t \in [0,1] such that f(\gamma(t))= f_2(\gamma(t)). It is easy to show that f is discontinuous at \gamma(t')

Now, we just have to deal with the case f=f_1 or f=f_2.

If f(z)=\sqrt{R}e^{\frac{i\theta}{2}}, where z=Re^{i\theta}, then we discover that f isn't really a function because Re^{i\theta}=Re^{i\theta+2\pi} but f(Re^{i\theta+2\pi})=\sqrt{R}e^{\frac{i\theta}{2}+\pi} \neq \sqrt{R}e^{\frac{i\theta}{2}}=f(Re^{i\theta}) The case for f_2 is similar.

 

[micromass]

OK, so that was definitely too easy for the smart crowd of people here

A big congratulations for jgens for his topological solution and for beating HS-Scientist in just a minute. But the latter did give a nice elementary solution. There are other solutions though, so if somebody finds them, please do post!

 

[jostpuur]

what does a degree of a function g:S^1->S^1 mean?

 

[micromass]

what does a degree of a function g:S^1->S^2 mean?

It's a notion of algebraic topology. See http://en.wikipedia.org/wiki/Winding_number

 

[jostpuur]

Anyway, my choice of words would be this: We define an operation f\mapsto g by setting

<br /> g:[0,2\pi[\to\mathbb{C},\quad g(\theta)=f(e^{i\theta})<br />

The property f^2=\textrm{id} implies

<br /> g(\theta)=\epsilon(\theta)e^{i\theta/2}<br />

with some function \epsilon:[0,2\pi[\to \{-1,1\}.

If f is continuous (as anti-thesis), g must be continuous too, which implies \epsilon(\theta) is a constant. Then we find that \lim_{z\to 1}f(z) does not exist.

In other words, I would prefer forcing the discontinuity somewhere with the assumptions.

 

[micromass]

I might as well post my own solution. Fix $z\neq 0$, consider the function

G(w) = \frac{f(z)f(w)}{f(zw)}

This function is well defined on $\mathbb{C}\setminus \{0\}$. By squaring, we see that

G(w) = \pm 1

By continuity of $G$ and since $\mathbb{C}\setminus \{0\}$ is connected, we see that $G$ is constant, so we have

f(z)f(w) = f(zw)~\text{or}~f(z)f(w) = f(zw)

By (if necessary) multiplying $f$ with $-1$, we can assume that the first identity is satisfied. But then

f(1) = f(-1)f(-1) = (f(-1))^2 = -1

but also

f(1) = f(1)f(1) = (f(1))^2 = 1

Contradiction.

 

[jgens]

Another solution: With a little voodoo we can actually show that f is holomorphic. This means that differentiating at zero tells us that 1 = 2f(0)f'(0) = 0 and then we are done.

 

[Mandelbroth]

This is a well-known result in complex analysis. But let's see what people come up with anyway:

Challenge:
Prove that there is no continuous function $f:\mathbb{C}\rightarrow \mathbb{C}$ such that $(f(x))^2 = x$ for each $x\in \mathbb{C}$.

I've done this proof before. I did it using the definition of continuity of complex functions by the continuity of their real and imaginary parts, id est, given $x=a+bi$ and $f(x)=u(a,b)+iv(a,b)$ for real numbers $a$ and $b$ and real valued functions $u$ and $v$, $f$ is continuous at $x_0=a_0+ib_0$ if and only if $$\lim_{(a,b)\rightarrow(a_0,b_0)}[u+iv]=u(a_0,b_0)+iv(a_0,b_0)=f(x_0).$$ I can see if I can find my proof later if anyone is interested, but it essentially came down to a statement like "if $u$ is continuous, $v$ is not."

If you find this too easy, then there is this generalization:
Given $n$ points on a plane. If they do not all lie on a straight line, then there is a straight line in the plane that contains exactly two of the points.

Suppose $n=1$.