# Challenge IV: Complex Square Roots, solved by jgens

1. Aug 6, 2013

### micromass

Staff Emeritus
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 $x\in \mathbb{C}$.

Last edited by a moderator: Aug 8, 2013
2. Aug 6, 2013

### 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 idS1 is one. This contradiction completes the proof.

3. Aug 6, 2013

### 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.

Last edited: Aug 6, 2013
4. Aug 6, 2013

### micromass

Staff Emeritus
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!

Last edited: Aug 8, 2013
5. Aug 6, 2013

### jostpuur

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

Last edited: Aug 6, 2013
6. Aug 6, 2013

### micromass

Staff Emeritus
7. Aug 6, 2013

### jostpuur

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

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

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

$$g(\theta)=\epsilon(\theta)e^{i\theta/2}$$

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.

Last edited: Aug 6, 2013
8. Aug 6, 2013

### micromass

Staff Emeritus
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

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

9. Aug 6, 2013

### 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.

10. Aug 6, 2013

### Mandelbroth

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."

Suppose $n=1$.

Last edited: Aug 6, 2013