TITLE: Prove that $f(z) = |z|^2$ is differentiable only at the origin
QUESTION [7 upvotes]: Question:
I need to show that $f(z) = |z|^2$ is differentiable only at the origin
Sufficient condition for differentiability: The sufficient condition for a function $f(z) = u + iv$ to be analytic at all the points in region $R$ are
$\displaystyle (a)\; \frac{\partial u}{\partial x}  = \frac{\partial v}{\partial y} ,\frac{\partial u}{\partial y}  = -\frac{\partial v}{\partial x} $ 
$\displaystyle (b)\; \frac{\partial u}{\partial x}  , \frac{\partial v}{\partial y} ,\frac{\partial u}{\partial y}  ,\frac{\partial v}{\partial x} $ are continuous functions of $x$ and $y$ in region $R$ .


My answer:
$f(z) = |z|^2$ Let, $z =  x + iy$

$f(z) = x^2 + y^2$
But $f(z) = u + iv$
$u = x^2 + y^2 , v = 0$
$\displaystyle \frac{\partial u}{\partial x} = 2x$ , 
$\displaystyle \frac{\partial v}{\partial y} = 0$
$\displaystyle \frac{\partial u}{\partial y} = 2y$ , 
$\displaystyle \frac{\partial v}{\partial x} = 0$
if $f(z) $ is differentiable then,
$\displaystyle \frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ or, $2x = 0$ or $x = 0$

& $\displaystyle \frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$ or, $2y = 0$ or $y = 0$
cauchy riemann equation is satisfied at $(0,0)$
For the second part of the condition (checking for continuity of the partial derivatives ) I see that $\displaystyle \frac{\partial u}{\partial x} = 2x$ which is continuous for any value of $x$ & $\displaystyle \frac{\partial v}{\partial x} = 0$ regardless of the value of x, so it's continuous.The other two are also similar.

I am confused about the second part of the proof.Have I understood it properly?

REPLY [2 votes]: What you've done is correct. Another way to see that $f$ is complex differentiable only at $0$ is after writing $f(x+iy) = x^2+y^2$ and noticing that $f$ is differentiable as function on $\mathbb R^2$, we can calculate$$\frac{\partial f}{\partial\bar z} = \frac{\partial}{\partial\bar z}(z\bar z) = z$$ which is $0$ only at $z = 0$.