Analytic Functions - Entire Function How do I verify it?
Verify that the formula $f(z)= \cos{(\sqrt z)}$ defines an entire function, whereas $g(z) = \sin{(\sqrt z)}$ does not.

Begin by proving that $f_x (z) = -i f_y (z) = - \sin{(\sqrt{z})} / (2 \sqrt{z})$ if $z \neq 0$, while $f_x (0) = -i f_y (0) = -1/2$. Then argue that $f_x$ and $f_y$ are continuous in $\mathbb{C}$. A possibly helpful observation is this: if a function $h: \mathbb{C} \rightarrow \mathbb{C}$ id both continuous and even, then the function $k: \mathbb{C} \rightarrow \mathbb{C}$ given by $k(z) = h(\sqrt{z})$ is continuous. 

Somebody help me please.
Thanks.
 A: Let us first discuss the meaning of $\sqrt{z}$. It is a complex number $w$ such that $w^2 = z$. For $z \ne 0$ there are exactly two choices for $w$ which differ by the factor $-1$. If we want to regard $\cos(\sqrt{z}), \sin(\sqrt{z})$ as functions defined on $\mathbb C$, we must specify which of the two possible values of $\sqrt{z}$ we want to take.
So let us say that a function $\phi : U \to \mathbb C$ defined on an open $U \subset \mathbb C$ is a root choice function if $\phi(z)^2 = z$ for all $z \in U$ Then we can consider the functions $\cos \circ \phi$ and $\sin \circ \phi$ and check under what conditions on $\phi$ they are holomorphic. Note that we do not make any assumptions on $\phi$, in particular we do not require that $\phi$ is continuous or even holomorphic.
Root choice functions exist on any $U$. Using the axiom of choice, we see that there exist uncountably many such functions. In fact, let $s : \mathbb C  \to \mathbb C, s(z) = z^2$, be the squaring function. Then $\phi : U \to \mathbb C$ is a root choice function on $U$ if and only if $\phi(z) \in s^{-1}(z)$ for all $z$. In other words, the root choice functions can be identified with the elements of $\prod_{z \in U} s^{-1}(z)$.
The most popular root choice function on $\mathbb  C$ seems to be the following. Each $z \ne 0$ has a unique representation $z = re^{it}$ with $r > 0 $ and $t \in [0,2\pi)$. Then we get the root choice function 
$$\psi : \mathbb  C \to \mathbb  C, \psi(z) = \begin{cases} 0 & z = 0 \\ \sqrt{r}e^{it/2} & z \ne 0 \end{cases}$$
where $\sqrt{r}$ denotes the positive root. It is easy to verify that $\psi$ restricts to a continuous function on $\mathbb C \setminus \mathbb R_+$, where $\mathbb R_+$ denotes the set of positive real numbers.
We need two lemmas.
Lemma 1. For each $z \ne 0$ there exists a holomorphic root choice function on some open neigborhood $U$ of $z$.
The exponential function $e^w$ has derivative $e^w \ne 0$. For $z \ne 0$ choose $w$ such that $e^w = z$. There exists an open neighborhood $V$ which is mapped by $e^w$ biholomorphically onto an open neighborhood $U$ of $z$. Let $l : U \to V$ be the holomorphic inverse of $e^w : V \to U$. Then $\phi(z') = e^{l(z')/2}$ is the desired root choice function on $U$.
Lemma 2. Let $U$ be an open neighborhood of $0$. Then there does not exist a continuous root choice function on $V  = U \setminus \{ 0\}$.
Assume there exists a continuous root choice function $\phi$ on $V$. Define $f(z) = \psi(z)/\phi(z)$. Obviously $f(z) = \pm 1$. Since $f$ is continuous on $V \setminus \mathbb R_+$, we see that $f(z) = c \in \{-1,1\}$ on $V \setminus \mathbb R_+$. Thus $\phi(z) = c\psi(z)$ on $V \setminus \mathbb R_+$. Now let $z \in V \cap \mathbb R_+$. For $n \ge N$ the points $z_n =  re^{i/n}, z'_n =  re^{i(2\pi -1/n)} = re^{-i/n}$ are contained in $V \setminus \mathbb R_+$. Both sequences converges to $z$, but we have $\phi(z_n) = c\psi(z_n) = c\sqrt{r} e^{i/2n} \to c\sqrt{r}$ and $\phi(z'_n) = c\psi(z'_n) = c\sqrt{r} e^{i(\pi -i/2n)} \to c\sqrt{r} e^{i\pi} = - c\sqrt{r}$. Hence $\phi$ cannot be continuous, a contradiction.
Theorem 1. If $h: \mathbb C \to \mathbb C$ is an even holomorphic function, then for all root choice functions $\phi : \mathbb C \to \mathbb C$ the function $h \circ \phi$ is holomorphic. In other words, the choice of $\sqrt{z}$ is irrelevant in this case.
First note that $h^* = h \circ \phi$ is the same function for all $\phi$ simply because $h(w) = h(-w)$ and the possible values of $\phi(z)$ have the form $\pm w$. Hence for all $z \ne 0$ we can choose a $\phi$ which is holomorphic on some open neigborhood of $z$. This shows that $h^*$ is holomorphic on $\mathbb C  \setminus \{ 0 \}$. The point $0$ can be regarded as a singularity. We have $\phi(z) \to 0 = \phi(0)$ as $z \to 0$, hence $h^*(z) \to h^*(0)$ as $z \to 0$. Thus $h^*$ is continuous in $0$ whence the singularity is removable. This proves that $h^*$ is an entire function.
Theorem 2. If $h: \mathbb C \to \mathbb C$ is an odd holomorphic function, then for no root choice function $\phi  : \mathbb C \to \mathbb C$ the function $h \circ \phi$ is continuous.
Since $h$ is odd, we have $h(0) = 0$. There exists $\epsilon > 0$ such that $V_\epsilon(0)  = \{ z \in \mathbb C \mid 0 < \lvert z  \rvert < \epsilon \}$ does not contain zeros of $h$ (otherwise the set of zeros of $h$ would have $0$ as an accumulation point which would imply $h = 0$ which is not odd). 
Now let $\phi  : \mathbb C \to \mathbb C$ be any root choice function. We have $\phi(V_{\epsilon^2}(0)) \subset V_\epsilon(0)$ because if $0 < \lvert z  \rvert < \epsilon^2$, then $0 < \lvert \phi(z) \rvert = \sqrt{\lvert z \rvert} < \sqrt{\epsilon^2} = \epsilon$. We know that $\phi$ is not continuous on $V = V_{\epsilon^2}(0)$. Thus there exist $z \in V$ and a sequence $(z_n)$ in $V$ converging to $z$ such that $(\phi(z_n))$ does not converge to $\phi(z)$. Since $(\phi(z_n))$ is bounded note ($\lvert \phi(z_n) \rvert = \sqrt{\lvert z_n \rvert}$), it has a convergent subsequence. W.l.o.g. assume that $\phi(z_n) \to \zeta$. Then $z_n = \phi(z_n)^2 \to \zeta^2$ which implies $\zeta^2 = z$. Thus necessarily $\zeta = - \phi(z)$. We conclude
$$h(\phi(z_n)) \to h(- \phi(z)) = -h(\phi(z) \ne h(\phi(z))$$
because $\phi(z) \in V_\epsilon(0)$. This shows that $h  \circ \phi$ is not continuous.
