showing the function |z| is analytic So i need to shwo that f(z) = |z| is analytic. All i really have avavailable to me really are the cauchy riemann equations. So with that being the case i guess the assumption that the partials are continuous is present.
But how do i show analyticity for this function?
Following the advice from some earlier posts i saw, i should convert the function into u+iv form, but all all that gives me is (u^2 + v^2)^1/2. 
What if i treated the modulus as a positive and negative case?
 A: $f(z) = |z|$ is not analytic. So you are being asked to prove something that is false.
A: A function $F(z)$  is analytic if $\dfrac{\partial F(z)}{\partial \bar{z}}=0.$ We have $f(z)=|z|=\sqrt{z \bar{z}}$. It is clear that $\dfrac{\partial f(z)}{\partial \bar{z}} \neq 0,$ so  $|z|$ is not analytic.
A: 
Recall that 
  
  
*
  
*A complex function $f=u+iv:\Bbb C\to \Bbb C$ is analytic at a point $z_0=x_0+iy_0$ if there is a neighborhood $V=B(z_0,r)$ (say) of $z_0$ such that $f$ is differentiable (in the complex sense)  at every point $z$ of $V$.
  
*A necessary and sufficient condition that a complex function $f=u+iv:\Bbb C\to \Bbb C$ is differentiable at $w=a+ib$ is that 
($i$)  all the partial derivatives $u_x,u_y,v_x$and $v_y$ exist and continuous at $(a,b)\in\Bbb R^2$
($ii$) the Cauchy-Riemann equations $u_x=v_y$ and $u_y=-v_x$ must hold at $(a,b)\in \Bbb R^2$

For your problem $f(z)=|z|$ and so $u(x,y)=\sqrt{x^2+y^2}$ and $v(x,y)=0$. Clearly it can be observed that the condition $2(i)$ is failed to hold at $(0,0)$ for $u$. So $f$ is not analytic at any point of the complex plane.
