For a function $f(z)=z^2$ why aren't $u=0$ and $v=0$ orthogonal at $(0,0)$? Here $u=x^2-y^2=0$ and $v=2xy=0$ do not come out to be orthogonal despite the fact that $u=c_1$ and $v=c_2$ are orthogonal at points of intersection for an analytical function $f=u+iv$ in its domain.
 A: As Daniel Fischer wrote, 

The reason is that $f'(0) = 0$. Since $f'(0) = 0 \neq f''(0)$, angles at $0$ are multiplied with $2$. Generally, if a holomorphic function attains the value $w_0$ with multiplicity $n$ at $z_0$, then angles at $z_0$ are multiplied with $n$.

More specifically, suppose the derivatives $f',f'',\dots,f^{(n-1)}$ vanish at $z_0$, while $f^{(n-1)}(z_0)\ne0$. Then $f$ can be written as $f(z) = w_0+ g(z)^n$ in a neighborhood of $z_0$, where $g(z_0)=0$ and $g'(z_0)\ne 0$. (Reference). The function $g$, having nonzero derivative, preserves all angles. But 
the set $$\{ z : \operatorname{Re} (f(z)-f(z_0))=0\}$$ is the inverse image under $g$ of the set 
$$\{w: \operatorname{Re} (w^n) =0\}\tag{1}$$ which is the union of $n$ lines crossing at $0$ at angles $\pi/n$. 
Similarly, the set
$$\{ z : \operatorname{Im} (f(z)-f(z_0))=0\}$$ is the inverse image under $g$ of the set 
$$\{w: \operatorname{Im} (w^n) =0\}\tag{2}$$ which is just like the set $(1)$  but rotated by $\pi/(2n)$.
