A function differentiable only at $0$ and for $|z|=1$ I need to find a polynomial function  that is differentiable at the origin where $f'(0)=1$ and at every point $|z|=1$ but at no other point in the complex plane. I just have no clue how to solve something like this. Any ideas? Thanks
 A: It is easier if we write the polynomial as a polynomial in $z$ and $\overline{z}$ and use the Wirtinger calculus than to use the real coordinates.
The circle $\lvert z\rvert = 1$ is the zero locus of the polynomial (in $z$ and $\overline{z}$)
$$z\cdot\overline{z}-1,$$
so a polynomial $p$ (in $z$ and $\overline{z}$) is complex differentiable on the circle $\lvert z\rvert = 1$ if $\frac{\partial p}{\partial\overline{z}}(z)$ is a multiple of $z\cdot\overline{z}-1$. If we directly took $\frac{\partial p}{\partial\overline{z}}(z) = z\cdot\overline{z}-1$, we would get $p(z) = \frac{1}{2}z\cdot\overline{z}^2-\overline{z} + h(z)$ with a holomorphic polynomial $h$. That would however not be complex differentiable at $0$. So we multiply $z\cdot\overline{z}-1$ with a holomorphic polynomial $g$ that has a zero at $0$ and no zeros $\zeta$ with $0 < \lvert\zeta\rvert\neq 1$ to obtain something complex differentiable at $0$. Taking $g(z) = z$ fits the bill,
$$\frac{\partial p}{\partial \overline{z}}(z) = (z\cdot\overline{z}-1)\cdot z\tag{1}$$
leads to
$$p(z) = \frac{1}{2}z^2\overline{z}^2 - z\overline{z} + h(z)$$
with a holomorphic polynomial $h$. From $(1)$ it follows that $\frac{\partial p}{\partial\overline{z}}(z) = 0$ if and only if $z = 0$ or $\lvert z\rvert = 1$, so such a $p$ satisfies that part of the requirements, and it only remains to arrange it so that
$$\frac{\partial p}{\partial z}(z) = z\overline{z}^2-\overline{z} + h'(z)$$
attains the value $1$ at $z = 0$. We could for example take $h(z) = z$.
