Imaginary axis of a complex analytic function has zero measure?

It is established in [1] that the zero set of a non-constant real analytic function on $$\mathbb{R}^d$$ has measure zero. To me, this result should intuitively extend to the entire imaginary axis of a non-constant complex analytic function since this is a measure zero set with respect to $$\mathbb{C}$$. But, I'm having a hard time showing it, particularly since the imaginary axis is an uncountably infinite set. Any suggestions here or is this not necessarily true for some obvious reason?

• The vanishing set has locally measure zero, thus it has globally measure zero. In dimension one this is the isolated zero theorem. In dimension $n$ : at a vanishing point, there is a direction where the function is $f(z_0+t v) = C t^k + O(t^{k+1}), C \ne 0$ thus the vanishing set has dimension $n-1$ – reuns Jul 30 at 1:02

1 Answer

You can use the canonical identification $$\mathbb C = \mathbb R^2$$, which happens to preserve measure, together with the fact that the imaginary axis is the zero set of the nonconstant real analytic function $$f(x,y)=x$$.