Why $F(z) = |z|^2$ is holomorphic nowhere?

I am self-studying basic complex analysis, and am slightly confused as to how to show that $F(z) = |z|^2$ is holomorphic nowhere. A necessary and sufficient condition for the holomorphism of $F(z)$ is that $F(z)$ is independent of $\overline{z}$. That is, we require:

$$\frac{\partial F}{\partial \overline{z}}=0$$

We note that we have $F(z) = |z|^{2} = z\overline{z}$, and so we have:

$$\frac{\partial F}{\partial \overline{z}}=z$$

And so we clearly have that $F(z)$ is holomorphic everywhere, except $z = 0$, however, I don’t understand why we cannot say $F(z)$ is holomorphic at $z=0$, is it because in the neighborhood of $z = 0$ we have that $F(z)$ is nowhere holomorphic?


I think you’re fundamentally misunderstanding the condition $\frac{\partial F}{\partial \bar{z}} = 0$. After you differentiate with respect to $\bar{z}$, your resultant must be zero for the function to be complex differentiable at that point.

Your work shows that for $F(z) = |z|^2$, $\frac{\partial F}{\partial \bar{z}} = 0$ if and only if $z = 0$. (I’m not exactly sure how you came up with the statement about it being holomorphic everywhere but zero from this condition.) Hence $F$ is complex differentiable at $0$. However for a function to be holomorphic, it must be complex differentiable on an open set. Holomorphicity is a much stronger condition than just complex differentiability.

Source : Link , Question Author : Thomas Russell , Answer Author : Cameron Williams

Leave a Comment