# Is $f(z) = |z|$ holomorphic on $\mathbb{C}$?

I am trying to refine my complex analysis skills and have come across a curious problem. It is as follows:

Determine whether the function $f(z) = |z|$ is holomorphic on $\mathbb{C}$.

I have tried to use the Cauchy-Riemann equations but am not sure how to proceed. Any help is appreciated.

• I have tried to use the Cauchy-Riemann equations but am not sure how to proceed. How hard did you try? Did you write down the real and imaginary parts of $f$ in terms of the real and imaginary parts of $z$? Because there is only one thing left to do if you want to use the Cauchy-Riemann equations. Apr 13, 2018 at 13:59
• It isn't even real differentiable Apr 13, 2018 at 13:59
• Not holomorphic because it is not analytic Apr 13, 2018 at 14:00

To use the CR equations, we need to set up the functions $u, v:\Bbb R^2\to \Bbb R$ such that $$f(x + iy) = u(x, y) + iv(x, y)$$ for real $x, y$. This isn't that difficult, as we already know that $$|x + iy| = \sqrt{x^2 + y^2}$$ which means that $$u(x, y) = \sqrt{x^2 + y^2}\\ v(x, y) = 0$$ Now insert into the CR equations and see that you don't get equality.
More generally, if $f: \mathbb C \to \mathbb C$ is a holomorphic function such that $f(\mathbb C)\subseteq \mathbb R$, then $f$ is constant.
Indeed, the imaginary part of $f$ is zero and so the Cauchy–Riemann equations imply that the real part of $f$ has zero partial derivatives and so is constant.
Since $z \mapsto |z|$ is real but not constant, it cannot be holomorphic.
Your function isn't even differentiable at $$0$$: if you restrict $$z$$ to the non-negative reals, you get $$\lim_{z\to0}\frac{|z|}z=1$$ and if you restrict $$z$$ to the non-positive reals, you get $$\lim_{z\to0}\frac{|z|}z=-1$$. It turns out that it is differentiable nowhere.
No, because $$|z|=\sqrt{z\bar z}$$ and a function $f(z)$ of complex variable $z$ is holomorphic only if $\frac{\partial f}{\partial \bar z}=0$