# Part of a proof in Spivak's Calculus, 4th ed.

## Homework Statement

I'm following along in Spivak's Calculus, 4th ed. In the chapter on complex functions (page 547 in my copy), Spivak outlines a proof that there is no "square root function" on the complex numbers; he leaves one step to the reader, claiming that it is a standard least-upper-bound argument. I've got a question about that step, but I'll outline the rest of the proof as well for context:

Claim: There is no continuous function f defined on the complex numbers such that for all complex numbers z, $(f(z))^2=z$. In fact, there is no such function even for just the complex numbers z with $\vert z \vert = 1$.

Proof:
Suppose, aiming for contradiction, that such f exists. Without loss of generality, take f(1)=1, since f can always be replaced with -f. This takes care of both cases of $(f(1))^2=1$.

***This is the step left to the reader. Spivak says the argument is "a standard type of least upper bound argument.") ***
We claim that for all $\theta$ with $0\leq \theta < 2\pi$, $f(\cos\theta + i\sin\theta) = \cos(\frac{\theta}{2})+i\sin\theta(\frac{\theta}{2})$.

With the equality established, taking the limit as $\theta$ goes to $2\pi$, f goes to -1. Taking this limit is equivalent to taking the limit of f(z) as z goes to 1, so $\lim_{z\rightarrow 1}f(z)=-1$. But since f(1)=1, f is not continuous at 1, a contradiction.

My small issue with the proof: I did the step left to the reader completely algebraically; I'll show what I did below. What I don't get is the need for any sort of least-upper-bound argument.

## The Attempt at a Solution

My solution to the claim left to the reader:

Since by assumption $(f(z))^2=z$, we have $(f(\cos\theta+i\sin\theta))^2=\cos\theta+i\sin \theta$. Applying the trigonometric identities $\cos 2\theta = \cos^2\theta-\sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$, we get
$$(f(\cos\theta+i\sin\theta))^2=\cos\theta+i\sin \theta$$
$$=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos \frac{\theta}{2}$$
$$=(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})^2$$.

Taking square roots then proves the claim.

Seeing as I was able to prove the claim just via some trigonometric identities, I'm not sure why Spivak says the claim follows from a least-upper-bound argument. I'm not disputing that there is one, but is there a reason why Spivak mentions it specifically when such an easy algebraic argument exists? It seems to go against his style throughout the rest of the book. Maybe the l.u.b. argument is even simpler? Or am I just overlooking something?

## Homework Statement

I'm following along in Spivak's Calculus, 4th ed. In the chapter on complex functions (page 547 in my copy), Spivak outlines a proof that there is no "square root function" on the complex numbers; he leaves one step to the reader, claiming that it is a standard least-upper-bound argument. I've got a question about that step, but I'll outline the rest of the proof as well for context:

Claim: There is no continuous function f defined on the complex numbers such that for all complex numbers z, $(f(z))^2=z$. In fact, there is no such function even for just the complex numbers z with $\vert z \vert = 1$.

Proof:
Suppose, aiming for contradiction, that such f exists. Without loss of generality, take f(1)=1, since f can always be replaced with -f. This takes care of both cases of $(f(1))^2=1$.

***This is the step left to the reader. Spivak says the argument is "a standard type of least upper bound argument.") ***
We claim that for all $\theta$ with $0\leq \theta < 2\pi$, $f(\cos\theta + i\sin\theta) = \cos(\frac{\theta}{2})+i\sin\theta(\frac{\theta}{2})$.

With the equality established, taking the limit as $\theta$ goes to $2\pi$, f goes to -1. Taking this limit is equivalent to taking the limit of f(z) as z goes to 1, so $\lim_{z\rightarrow 1}f(z)=-1$. But since f(1)=1, f is not continuous at 1, a contradiction.

My small issue with the proof: I did the step left to the reader completely algebraically; I'll show what I did below. What I don't get is the need for any sort of least-upper-bound argument.

## The Attempt at a Solution

My solution to the claim left to the reader:

Since by assumption $(f(z))^2=z$, we have $(f(\cos\theta+i\sin\theta))^2=\cos\theta+i\sin \theta$. Applying the trigonometric identities $\cos 2\theta = \cos^2\theta-\sin^2\theta$ and $\sin 2\theta = 2\sin\theta\cos\theta$, we get
$$(f(\cos\theta+i\sin\theta))^2=\cos\theta+i\sin \theta$$
$$=\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}+2i\sin\frac{\theta}{2}\cos \frac{\theta}{2}$$
$$=(\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})^2$$.

Taking square roots then proves the claim.

OK, but you can't take square roots just like that. For example, you know that if $x^2=y^2$, then $x=y$ or $x=-y$.

In this case, you know that

$$(f(\cos\theta+i\sin\theta))^2= (\cos\frac{\theta}{2}+i\sin\frac{\theta}{2})^2$$

this is fine. But this only implies that

$$f(\cos\theta+i\sin\theta)= \cos\frac{\theta}{2}+i\sin\frac{\theta}{2}$$

OR

$$f(\cos\theta+i\sin\theta)= -\cos\frac{\theta}{2}-i\sin\frac{\theta}{2}$$

You must still prove that the last possibility cannot occur. This is where the lower bound argument comes in (i.e. let $\theta$ be the smallest number such that the last possibility occurs, ...)