the set where a harmonic function's gradient is zero If $u$ is a harmonic function in a region  $\Omega$ what can you say about the  set of points at which the grradient of $u$ is zero.
 A: If $u$ is complex valued, the vanishing set of the gradient of $u$ is the intersection of the vanishing sets of the gradients of $\operatorname{Re} u$ resp. $\operatorname{Im} u$. As we will see, that is a set of exactly the same type as the vanishing set of $\operatorname{grad} (\operatorname{Re} u)$, so let us suppose that $u$ is real-valued.
Every point $z\in \Omega$ has a connected neighbourhood $W$ on which $u$ is the real part of a holomorphic function, that is, on $W$ there exists a real-valued harmonic function $v$ such that $f(z) = u(z) + iv(z)$ is holomorphic.
The Cauchy-Riemann equations
$$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}; \qquad \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$$
tell us that the gradient of $u$ vanishes in exactly the same points as the gradient of $v$ (they tell us more, that $\lVert \operatorname{grad} u(z)\rVert = \lVert \operatorname{grad} v(z)\rVert$ for all $z\in W$, but the vanishing is all we need), and those points are the zeros of $f'$.
Now, as the derivative of a holomorphic function, $f'$ is also holomorphic, and we know that the zeros of a non-constant holomorphic function on a domain (connected open set) $W$ form a discrete closed subset of $W$. So either $f' \equiv 0$, or $Z(u,W) = \{ z\in W : f'(z) = 0\} = \{ z \in W : \operatorname{grad} u(z) = 0\}$ is a discrete closed subset of $W$.
By the identity theorem, the set of points $\{z \in \Omega : \operatorname{grad} u(z) \equiv 0 \text{ in a neighbourhood of } z\}$ is closed. It is open per definition, so it is a union of connected components of $\Omega$.
Thus the vanishing set of the gradient of a harmonic function $u$ on $\Omega$ is the union of some components of $\Omega$ and a closed discrete subset of $\Omega$. (If by "region" you mean "connected open set" - the terminology is not uniform, some authors require regions to be connected, others don't - then it simplifies to "the vanishing set is either all of $\Omega$, or a closed and discrete subset of $\Omega$".)
