TITLE: Derivative of $\ln (z), z\in\mathbb{C}$
QUESTION [5 upvotes]: Let $f(z) = \ln z := \ln |z| + \arg (z)i$. Then the derivative is (if it exists) by definition:
$$\lim_{h\to 0}\frac{\ln (z+h)-\ln (z)}{h}=\lim_{h\to 0}\frac{\ln |z+h| +\arg(z+h)i-\ln |z| -\arg(z)i }{h}$$

REPLY [6 votes]: Let $ U =\mathbb C \setminus (-\infty,0].$ Then $f(z)=\ln |z| + i\arg z$ is continuous on $U$ and we have $e^{f(z)} = z$ there. This shows $f(z)$ is injective on $U.$ Fix $z\in U.$ Then for small nonzero $h$ we have
$$1=\frac{e^{f(z+h)}-e^{f(z)}}{h} = \frac{e^{f(z+h)}-e^{f(z)}}{f(z+h) - f(z)}\frac{f(z+h) - f(z)}{h}.$$
The injectivity of $f$ shows that $f(z+h) - f(z)\ne 0,$ so we're OK dividing by it above. As $h\to 0, f(z+h) \to f(z)$ by the continuity of $f.$ So the first difference quotient on the right tends to the derivative of $e^w$ at $w=e^{f(z)},$ which is $e^{f(z)} = z \ne 0.$ Knowing all of this, we can now write
$$\tag 1 \frac{f(z+h) - f(z)}{e^{f(z+h)}-e^{f(z)}} = \frac{f(z+h) - f(z)}{h}.$$
Since the left side of $(1) \to 1/z,$ we get $f'(z) = 1/z$ (as expected).