Is function $f(z)=|z-9|$ differentiable? 
Let $z \in \mathbb{C}$. Is function $f(z)=|z-9|$ differentiable?

For a function to be differentiable in $\mathbb{C}$, it must satisfy the Cauchy-Riemann equations:
$$\begin{cases}u_x=v_y \\ u_y = - v_x \end{cases}$$
Let $f(z)=|x+iy-9|=\sqrt{(x-9)^2+y^2}$. This means that $u(x,y)=\sqrt{(x-9)^2+y^2}$ and $v(x,y)=0$. Then
$u_x = \frac{x-9}{(x-9)^2+y^2}$ and $u_y=\frac{y}{\sqrt{(x-9)^2+y^2}}$. $v_x=v_y=0$ for $z \ne 9$
I am not sure what to do next. Also i don't know what to do about point $z=9$.
 A: There is no need to use the Cauchy-Riemann equations here. The function $f$ is not differentiable at $9$ because the limit$$\lim_{z\to9}\frac{f(z)-f(9)}{z-9}=\lim_{z\to9}\frac{|z-9|}{z-9}$$doesn't exist. It is equal to $1$ if $z\in 9+\Bbb R$  and it is equal to $-i$ if $z\in9+i\Bbb R$.
A: If you move from a point $z$ in a direction at a right angle to the line from $9$ to $z,$ you see that the rate of change of $|z-9|$ is $0.$ But if you move directly along that line, the rate of change is $1.$ Since $0$ and $1$ are not both the same number, this function is not differentiable.
A: $f(z)=|z-9|=\sqrt{(x-9)^2 +y^2} $
$$u(x, y) =\sqrt{(x-9)^2 +y^2}$$
$$v(x, y) =0$$
$\frac{\partial u}{\partial x}=\frac{1}{ \sqrt{(x-9)^2 +y^2} } (x -9)$
$\frac{\partial u}{\partial y}=\frac{1}{ \sqrt{(x-9)^2 +y^2} } y$
$\frac{\partial u}{\partial x},\frac{\partial u}{\partial y}$ doesn't exists at the point $z=(9, 0)$, hence the function $f$ doesn't satisfy C-R equation at $z=9$ and so $f$ not Holomorphic at $z=9$.
$f:D\to \mathbb{C} $ is Holomorphic on $D\subset {\mathbb{C}} $ open set. Then $f$ satisfy C-R equations.
By Contrapositive argument, if $f$ doesn't satisfy C-R equations at $z_0$ implies $f$ is not Holomorphic at $z_0 \in D$
A: The slickest argument here is based on the following fact:
If a function $\mathbb C\to\mathbb C$ always takes real values and is differentiable, then it is constant.
Since $|z-9|$ always takes real values and is not constant, it cannot be differentiable.
You can prove the italicized fact directly from the Cauchy-Riemann equations with less calculation than it would take to argue from the particular definition of $|z-9|$. Since you're assuming the function takes only real values, two of $u_x, u_y, v_x, v_y$ (which ones?) will be zero. Then Cauch and Riemann claim the two others are zero too, and two applications of the familiar real mean value theorem now tells you that the function value must be the same at any two complex points.
