there exist an holomorphic function such that ..? There exist an holomorphic function $f$ on $|z|<1$ and continuous for $|z|\le 1$ such that $ f(e^{i\theta})= \cos \theta + 2i \sin \theta$?
I have no idea what I can do in this problem. :S
 A: If $f$ was holomorphic, then so would $g(z)=f(z)-z$. However, $g$ is purely imaginary (see below) and hence must be constant. However it is not constant, which is a contradiction. Hence such an $f$ does not exist.
Note: I glibly assumed that $g$ was imaginary on the closed unit disk, when in fact, all that I know apriori is that it is imaginary on the unit circle. @PerManne below has given a nice argument showing why it is imaginary on the unit disk. Here is another proof which gives an explicit representation of $g$.
I am using Theorem 5.25 in Rudin's Real & Complex Analysis. (The set $A$ in his statement is the collection of all functions analytic in the unit ball and continuous on the closed unit disk.) Then, if $g$ is analytic on the unit ball and continuous on the closed ball, then we can write
$$g(re^{i\theta}) = \frac{1}{2\pi} \int_{-\pi}^{\pi} P_r(\theta -t) g(e^{it}) dt$$
where $P_r(\theta -t) = \frac{1-r^2}{1-2r\cos(\theta -t)+r^2}$ is the Poisson Kernel. Since the kernel is real valued, if follows that if $g$ takes purely imaginary values on the boundary then it takes purely imaginary values in the interior as well.
A: Suppose by contradiction that such a function exists. Let $g(z)=f(z)-z$.
Then 
$$g(e^{i \theta})=i \sin(\theta)$$
Then, $g$ is a holomorphic function on $|z| < 1$ whose range is imaginary.....
Added To get the contradiction, check my answer here.
Basically $h_1(z)=e^{g(z)}$ and $h_2(z)=e^{-g(z)}$ are constant on $|z|=1$, thus by applying the Maximum Modulus Principle to both, you get that $1 \leq |h_1(z) | \leq 1$. This proves that $h_1(z)$ is a constant function.
P.S. $g(z)=f(z)-2z$ satisfies the conditions of the above link, since for the proof $g$ doesn't need to be entire, just Analytic inside $|z| \leq 1$.
A: There does not exist such an $f$. Suppose, on the contrary, such an $f$ exists. Using the boundary condition of $f$, a direct computation shows that  $\int_{|z|=1}f(z)dz=-\pi i\ne 0$, which contradicts to Cauchy's integral theorem. 
An alternative way to obtain a contradiction is to consider $g(z)=\frac{3z^2-1}{2}$, which share the same boundary condition with $zf(z)$ on the uint circle. Then if $f$ is holomorpic on the unit disk and continous on the closed unit disk, $g(z)$ and $zf(z)$ must coincide with each other, i.e. $f(z)=\frac{3z^2-1}{2z}$, which has a pole at $0$, a contradiction. 
