# 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?

• $|z|$ is NOT analytic. – Crostul Mar 2 '15 at 17:57
• Note that "$u + iv$ form" means splitting the value into real and imaginary parts. Here you have $u(x, y) = \sqrt{x^{2} + y^{2}}$ and $v(x, y) = 0$. – Andrew D. Hwang Mar 2 '15 at 19:25
• No complex function taking only real values is analytic, unless it is constant. – lhf Mar 2 '15 at 19:27
• I had a feeling it wasn't, i just was not sure if i shold have treated the magnitude portion exckusively as the real portion of the complex value – dc3rd Mar 2 '15 at 20:51

$f(z) = |z|$ is not analytic. So you are being asked to prove something that is false.

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.

Recall that

1. 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 $V$.

2. 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.

Is the function $f(z)=|z|^2=z\bar{z}$ analytic?

Let $z=x+iy$. $f(z)=z\bar{z}=(x+iy)(x-iy)=x^2-y^2$

If $$f=u+iv$$ C-R imply

$\dfrac{\partial u}{\partial x}=2x\neq 0=\dfrac{\partial v}{\partial y}$

$\dfrac{\partial u}{\partial y}=-2y \neq 0=-\dfrac{\partial v}{\partial x}$

Then $f$ is not analytic

• oops: $|z| \neq z\bar{z}$ it is the square root of that. – Mark Fischler Mar 2 '15 at 19:10