Apply Cauchy-Riemann equations on $f(z)=z+|z|$? I am trying to check if the function $f(z)=z+|z|$ is analytic by using the Cauchy-Riemann equation.
I made 
$z = x +jy$ 
and therefore 
$$f(z)= (x + jy) + \sqrt{x^2 + y^2}$$
put into $f(z) = u+ jv$ form:
$$f(z)= x + \sqrt{x^2 + y^2} + jy$$
where  
$u = x + \sqrt{x^2 + y^2}$
and that
$v = y$
Now I need to apply the Cauchy-Riemann equation, but don't know how would I go about doing that.
Any help would be much appreciated.
 A: The Cauchy-Riemann equations are
\begin{align}
\dfrac{\partial u}{\partial x} & = \dfrac{\partial v}{\partial y}\\
\dfrac{\partial v}{\partial x} &= -\dfrac{\partial u}{\partial y}
\end{align}
In your case, $u(x,y) = x + \sqrt{x^2+y^2}$ and $v(x,y) = y$. Assuming $(x,y) \neq (0,0)$, the partial derivatives are
\begin{align}
\dfrac{\partial u}{\partial x} & = 1 + \dfrac{x}{\sqrt{x^2+y^2}}\\
\dfrac{\partial v}{\partial x} & = 0\\
\dfrac{\partial u}{\partial y} & = \dfrac{y}{\sqrt{x^2+y^2}}\\
\dfrac{\partial v}{\partial y} & = 1
\end{align}
Hence, from the Cauchy-Riemann equations, we get that
$$1 + \dfrac{x}{\sqrt{x^2+y^2}} = 1 \implies \dfrac{x}{\sqrt{x^2+y^2}} = 0$$
$$\dfrac{y}{\sqrt{x^2+y^2}} = 0$$
This has no solutions since $(x,y) \neq (0,0)$. Hence, the function is not differentiable on $\mathbb{C} \backslash \{(0,0)\}$. The only point we need to check whether it is differentiable is $(0,0)$. At this point, we can check for differentiability directly from the definition. You will find that it is also not differentiable at $(0,0)$. Hence, the function is nowhere analytic.
A: In order for your function to be analytic, it must satisfy the Cauchy-Riemann equations (right?  it's good to think about why this is true).  So, what are the equations?
Well, du/dx = dv/dy.
Does this hold?  Or you could consider du/dy = -dv/dx.
If either of these equations do not hold, then the function is not analytic.
