Why $\dfrac{\partial v}{\partial y}=0$ which also gives us $\dfrac{\partial v}{\partial x}=0$?

Question

I passed through this answer here and I don't understand the following. We are at the moment:

If $u^2+v^2 \neq 0$, then $\dfrac{\partial v}{\partial y}=0$ which also gives us $\dfrac{\partial v}{\partial x}=0$. Now by Cauchy-Riemann, we also get that $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial y}=0$. Hence, $f'(z) = 0$ which gives us that $f$ is constant.

Why we get "$\dfrac{\partial v}{\partial y}=0$ which also gives us $\dfrac{\partial v}{\partial x}=0$? I don't see this...

Another way to see this is that if $v$ harmonic and independent of $x$ then $v$ is a linear polynomial in $y$ since $v_{yy}=0$ and then the original hypothesis precludes that — Conrad, Jan 17 '21 at 15:03

score 1 · Answer 1 · answered Jan 17 '21 at 13:51

There is a comment in the answer to the question you linked that answers your question. The original question was about proving that for a holomorphic function $f$, $|f|$ constant implies $f$ constant. Now if $|f(z)|$ is constant, so is $|f(iz)|$, and running the same argument that showed $\frac{\partial v}{\partial y}=0$ on this rotated function gives $\frac{\partial v}{\partial x}=0$.

Why $\dfrac{\partial v}{\partial y}=0$ which also gives us $\dfrac{\partial v}{\partial x}=0$?

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