Complex analysis is different

Let $F = \mathbb C^D$ be the set of functions that fit the description in (a).

a) There are infinitely many functions with the desired property.

If $a = e^{i\phi} \in D$ , then the equation $z^2 = a$ has two solutions, namely $z = \pm e^{i\phi/2}$ . For each of the uncountably many values of $a$ one can choose either value of $z$ ; thus there is a bijection between $\{-1,+1\}^{[0,2\pi\rangle} \sim 2^{\aleph_0}$ and $F$ , given by $(g:\ [0,2\pi\rangle \to \{-1,+1\}) \mapsto (f:D \to \mathbb C;\ \ f(e^{i\phi}) = g(\phi)e^{i\phi/2}).$

b) There are no such functions that are continuous.

Starting at point $a_0 = 1$ , choose a sign: $f(a_0) = \pm 1$ . To keep this continuous on all of $e^{i[0,2\pi\rangle}$ , we must have $f(e^{i\phi}) = \pm e^{i\phi/2}$ with the same sign. Then continuity requires

$\lim_{\phi \to 2\pi} f(e^{i\phi}) = \pm e^{i\pi} = \mp 1.$

But for a continuous function $f$ , this means $f(a_0) = \lim_{a\to a_0} f(a) = \mp 1$ , while we started with $f(a_0) = \pm 1$ . This is contradiction. Therefore no such continuous function exists.

The answer, then, is $\tfrac 1 \infty + 0 = 0 + 0 = \boxed{0}$ .

a) $A = 2^{\aleph_1} = \aleph_2 = \infty$ , because the cardinal of $[0, 2 \pi)$ is equal to cardinal of $\mathbb{R}$ . By contradiction, let's suppose that there are a finite number $\{f_1, ... , f_n\}$ such that given an $a \in D$ , $f(a)$ is one and only one of the roots of the equation $z^2 = a$ . Let's choose $n$ points $a_1, ... , a_n \in D$ and a function $g$ satisfying $g(a_i) = - f_i (a_i), \forall \space i = 1, 2, ... , n$ and for the rest of points in $D$ one and only one of the roots of the equation $z^2 = a$ . Then $g \neq f_i, \forall \space i = 1, 2, ... , n$ and $g$ fulfills the requisites.

b) By contradiction. Let's suppose that there exists a continuous function $f$ . WLOG $f(1) = 1$ . Because of the image of a point belonging to the first quadrant is a point belonging to the first quadrant or to the third quadrant, the image by $f$ of a point belonging to the first quadrant is a point of the first quadrant, in fact, of the first octant, because if there was a point $z$ belonging to the first quadrant such that $f(z)$ belongs to the third quadrant, the function $f$ would "jump" from the first quadrant to the third quadrant and this is a contradiction with the continuity of $f$ . Hence, $f(i) = e^{i \frac{\pi}{4}}$ .

With the same reasoning the image of a point belonging to the second quadrant is a point belonging to the second octant or to the sith octant, so the image by $f$ of a point belonging to the second quadrant is a point of the first quadrant,in fact, of the second octant, because if there was a point $z$ belonging to the second quadrant such that $f(z)$ belongs to the third quadrant or sixth octant, the function $f$ would "jump" from the first quadrant to the third quadrant and this is a contradiction with the continuity of $f$ . Hence, $f(-1) = e^{i \frac{\pi}{2}} = i$ .

With the same reasoning the image of a point belonging to the fourth quadrant is a point belonging to the fourth quadrant and the image of a point of the third quadrant is a point of the fourth quadrant and hence, $f(-1) = e^{-i \frac{\pi}{2}} = -i$ (Contradiction).