I let $f(z) = r$ where $r$ is a constant. Then I want to see if the limit of $$\frac{f(z+h) – f(z)}{h}$$
exists as $h$ goes to $0$:
$$\frac{f(z+h) – f(z)}{h}$$
$$ = \frac{r-r}{h}$$
$$ = 0$$
so this goes to $0$ as $h$ goes to $0$. is this correct?
Now to prove that a real valued holomorphic function $f= u+iv$ on complex plane is constant, I used Cauchy Riemann equations and got
$$u_x = v_y = u_y = v_x = 0.$$
To show that $f$ is constant, I want to show that $f' = 0$.
But I am confused about what $f'$ actually means here. Do I need to show that $f_x = u_x + iv_x = 0$ and $f_y = u_y + iv_y = 0$ ?
or is $f' = u_x + v_y$ ?
Best Answer
If the real valued holomorphic function is defined on the complex plane, you could also define $g(z) = \exp(if(z))$ which would be a bounded entire function, hence constant, so that $f$ must be constant.
Your approach works too, however, and one can actually do it in this case by showing that $$0 = f'(z) := \lim_{h \to 0}\frac{f(z+h) - f(z)}{h}$$ for all $z$. The merit of this is that it works in any connected set, not just the entire plane. You say you already know that for $f = u + iv$ and for all $z = x+iy$ $$u_x(z) = u_y(z) = v_x(z) = v_y(z) = 0$$ from the Cauchy-Riemann equations. The missing piece of information is that for all $z$ and $w = w_1 + iw_2$ $$f'(z)\cdot w = \begin{pmatrix} u_x(z) & u_y(z)\\v_x(z) & v_y(z)\end{pmatrix}\begin{pmatrix} w_1\\w_2\end{pmatrix}$$ where we have used the natural identification of the complex number $w$ with the vector $(w_1,w_2)$. Since the left side is zero, so is the right, and since it holds for all $w$ we conclude that $f'(z) =0$.
Notice that the matrix on the right side is the total derivative (Jacobian matrix) of $f$ regarded as a mapping $\mathbb{R}^2 \to \mathbb{R}^2$. The proof of the relation between complex and total derivative is simple: complex multiplication by $f'(z)$ induces a linear transformation that clearly satisfies the definition of total derivative.