Prove that there exist only one analytic function that holds conditions 
Let $f$ be a differentiable function in all $\Bbb{C}$ and numbers
  $$\frac{1+i}{1},\frac{2+i}{2},\frac{3+i}{3},\frac{4+i}{4},\frac{5+i}{5},
 \cdots$$  map respectively to numbers
  $$\frac{1-i}{1},\frac{1-2i}{2},\frac{1-3i}{3},\frac{1-4i}{4},\frac{1-5i}{5},
 \cdots$$ Prove that such a function is unique
  and find $f(1)+f(2i)$.

Lucky me, I found one function by myself that holds all the given conditions. It is $f(z)=-iz$ which is obviously analytic. Now I surely can find $f(1)+f(2i)=-i+2$. But it was just luck. How can I prove that such function is the only one that holds hiven conditions?

Let $g$ be a function that satisfies the same conditions. Then$$(\forall n\in\mathbb{N}):f\left(1+\frac in\right)=g\left(1+\frac in\right).$$Therefore, by the identity theorem and because the limit $\lim_{n\to\infty}1+\frac in$ exists, $g=f$.
