Find an entire and not constant function such that $f(\lvert z\rvert=1) = i$ I had an exam and there was a question:  

find, if exist, an entire and not constant function $f(z)$, so that for every $\lvert k\rvert=1$, $f(k) = i$.
  if not exist - explain why.

I had no direction here.. How do you solve it?
 A: METHODOLOGY $1$:
If $f(z)=u(x,y)+iv(x,y)$ is analytic inside the unit disk, then we have $\nabla^2 u(x,y)=\nabla^2 v(x,y)=0$ for $x^2+y^2<1$.  
Note that if in addition, $f(z)=i$ on the unit disk, then the problem is equivalent to the pair of interior Dirichlet Problems
$$\begin{align}
\nabla^2 u(x,y)&=0, \,\,x^2+y^2<1\\\\
u(x,y)&=0,\,\,x^2+y^2=1 \tag 1
\end{align}$$ 
and
$$\begin{align}
\nabla^2 v(x,y)&=0, \,\,x^2+y^2<1\\\\
v(x,y)&=1,\,\,x^2+y^2=1 \tag 2
\end{align}$$ 
The unique solutions to $(1)$ and $(2)$ are trivially $u(x,y)=0$ and $v(x,y)=1$
Therefore, repeating this line for the exterior Dircihlet problem, the only entire function $f(z)$ with $f(z)=i$ on the unit circle is $f(z)=i$ for all $z$.

METHODOLOGY $2$:
Another way forward as suggested by @user1952009 is to apply Cauchy's Integral Formula
$$f(z)=\frac{1}{2\pi i}\oint_C \frac{f(z')}{z'-z}\,dz' \tag 3$$
for $z$ inside the region for which $C$ is the boundary.
If $C$ is the unit circle on which $f(z)=i$, then for $|z|<1$ we find
$$\begin{align}
f(z)&=\frac{1}{2\pi i}\oint_C \frac{i}{z'-z}\,dz' \\\\
&=i 
\end{align}$$
Note that Cauchy's Integral Theorem guarantees that for $|z|>1$
$$\frac{1}{2\pi i}\oint_C \frac{f(z')}{z'-z}\,dz'=0 \tag 4$$
Applying $(3)$ to a contour $C_R=\{z': |z'|=R>1\}$ and for $z$ such that $1<|z|<R$, and exploiting $(4)$, we find
$$\frac{1}{2\pi i}\oint_{C_R} \frac{f(z')}{z'-z}\,dz'=\frac{1}{2\pi i}\oint_C \frac{f(z')}{z'-z}\,dz'=i$$
Inasmuch as $R>1$ is arbitrary, we conclude immediately that $f(i)=i$ for all $z$.  And we are done! 
