# Let $S$ be the open unit disk and $f: S\to \Bbb C$ be a real-valued analytic function with $f(0)=1$

I was thinking about the problem that says:

Let $S$ be the open unit disk and $f:S\to \Bbb C$ be a real-valued analytic function with $f(0)=1$.Then which of the following option is correct?

The set $\{z \in S:f(z) \neq 1\}$ is:

(a) empty,

(b) non-empty finite,

(c) countably infinite,

(d) uncountable.

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You seem to be very confused. I'm not sure why you give a sequence of functions when you're asked about a single function, but let's look at your $f_1$. You have $f_1(z)=z+1$, so $f$ is not real-valued: $f(i/2)=i/2+1$ for example. In addition, $\{ z \in S:f_1(z)\neq 1\}$ is not countably infinite: it is $S \setminus \{0\}$, which is uncountable. –  Chris Eagle Dec 8 '12 at 13:35

Hint: Since $f$ is analytic it's also holomorphic. Using the Riemann-Cauchy equations one can show that a real-valued holomorphic function (on a domain) has to be constant.
Well, it's enough that one of the options is correct, isn't it? What do you mean by form of $f$? In general you would have $f=f_1 + \imath \, f_2$ where $f_1,f_2$ are real-valued functions. In this case you know $f_2=0$ (since $f$ is real-valued) and therefore in particular $\partial_1 f_2=\partial_2 f_2 = 0$. –  saz Dec 9 '12 at 8:24
(a) is correct: $f$ is analytic on the & real-valued on the domain $S\implies f$ is constant on $S\implies f=0$ on $S\implies$ $\{z \in S:f(z) \neq 1\}=\emptyset.$