Does there exist a continuous function $f$ on $\Bbb C\setminus \{0\}$ such that $f(z)^2=z$ 
Does there exist a continuous function $f$ on $\Bbb C\setminus \{0\}$ such that $f(z)^2=z$?

I tried to do using contradiction.But that does not help out also.Please give some way to resolve this.
 A: Suppose $f$ exists.
Consider $c:[0,1) \to \mathbb{C}$ given by $c(t)=e^{2\pi i t}$.
WLOG (that this is the case will be clear later), we can assume $f(1)=1.$
Let $s:=\sup \{t \in [0,1) \mid f(c(t))=e^{\pi it}\}$.
Clearly the set above is non-empty. Also, supposing $s<1$, since the two complex numbers $z_1$ and $z_2$ in the circle such that $z_1^2=z_2^2=z$ are distant from one another (in fact, $d(z_1,z_2)=2$), continuity assures us that there exists $t>s$ such that $t$ is in the set, contradicting the fact that $s=\sup$. Therefore, $s=1$.
It follows that $f (c(t))=e^{\pi it }$ for all $t \in [0,1)$. But since $c(t) \to_{t \to 1}1$, we would have $f(1)=e^{\pi i}=-1$ by continuity, a contradiction.
A: Let $D$ be the complex plane minus the negative real axis.
Observe that the principal square root function $g(z) = \sqrt{z}$ is continuous on $D$, and also satisfies $g(z)^2 = z$.
It's not hard to show that, at every point in $a \in D$, $f(a)/g(a)$ is either $1$ or $-1$.
Since $f(z) / g(z)$ is continuous, it is either everywhere $1$ or everywhere $-1$.
Thus, we either have $f(z) = \sqrt{z}$ or $f(z) = -\sqrt{z}$.
However, neither possibility extends continuously to $\mathbb{C} \setminus \{ 0 \}$, so the claimed function cannot exist.
A: Hint: find the image of $\mathbb{C}\setminus\{0\}$ by $z\mapsto f(z^2)/z$.
