Suppose $f$ is holomorphic, and is written as $f=u+iv$ with $u,v$ real-valued. Why is the partial derivative $u_x(z)$ equal to $\operatorname{Re}(f'(z))$?
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This fact is used in the proof linked below, at the line that starts with "By assumption, $f'(z)=0$...)
Recall the definition of complex derivative: $$ f'(z) = \lim_{h\to 0} \frac{f(z+h)-f(z)}{h} \tag1 $$ In the definition, $h$ can be any complex number. Let's focus on real $h$ only: $h=t$, $t\in\mathbb R$ $$ f'(z) = \lim_{t\to 0} \frac{f(z+t)-f(z)}{t} $$ Plug $f=u+iv$ into this limit and separate the real and imaginary parts (this is where it helps that $t$ is real): $$ f'(z) = \lim_{t\to 0 } \frac{u(z+t)-u(z)}{t} + i \lim_{t\to 0 } \frac{v(z+t)-v(z)}{t} $$ The limits are just the partial derivatives of $u$ and $v$ with respect to $x$ variable. So,
$$ f' = u_x + iv_x $$ from where the claim follows.
For completeness, let's now take purely imaginary values of $h$, that is $h=it$, $t\in\mathbb R$. Then (1) becomes $$ f'(z) = \lim_{t\to 0} \frac{f(z+it)-f(z)}{it} $$ Again, separate real and imaginary parts: $$ f'(z) = -i \lim_{t\to 0 } \frac{u(z+it)-u(z)}{t} + \lim_{t\to 0 } \frac{v(z+it)-v(z)}{t} $$ and so $$f' = -iu_y+v_y$$