Derivative of $\ln (z), z\in\mathbb{C}$

I'll use $\log$ for the complex logarithm and $\ln$ for the real-valued logarithm; you then have: $$\log z = \ln |z| + i \arg z = \ln r + i\varphi$$ where I use $r = |z|$ and $\varphi = \arg z$ for simplicity. In this form, we have written $$\log z = u(x,y)+iv(x,y)$$ with $u(x,y) = \ln r = \ln \sqrt{x^2+y^2} = \tfrac{1}{2}\ln(x^2+y^2)$ and $v(x,y) = \varphi$.

You can check the Cauchy-Riemann equations yourself and find (*) $$\frac{\partial u}{\partial x} = \frac{x}{x^2+y^2} = \frac{\partial \varphi}{\partial y} \quad ; \quad \frac{\partial u}{\partial y} = \frac{y}{x^2+y^2} = - \frac{\partial \varphi}{\partial x}$$ Use this to find the derivative directly: $$\frac{\mbox{d} \log z}{\mbox{d}z} = \frac{\partial u}{\partial x}+i\frac{\partial v}{\partial x} = \frac{x-iy}{x^2+y^2} = \frac{1}{z}$$

Remark: depending on how you define the complex logarithm, there will be different ways to find its derivative.

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(*) The derivatives for $u$ are easy. For $v$, we have:

$$\left\{ \begin{array}{ccc} x = r\cos\varphi \\ y = r\sin\varphi \end{array}\right.$$ where both $r$ and $\varphi$ are (implicit) functions of $x$ and $y$. Differentiating both equations w.r.t. $x$ gives: $$\left\{ \begin{array}{ccc} 1 = \cos\varphi\frac{\partial r}{\partial x}-r\sin\varphi\frac{\partial \varphi}{\partial x} \\ 0 = \sin\varphi\frac{\partial r}{\partial x}+r\cos\varphi\frac{\partial \varphi}{\partial x} \end{array}\right.$$ This is a linear system of two equations in the variables $\frac{\partial r}{\partial x}$ and $\frac{\partial \varphi}{\partial x}$; solve for $\frac{\partial \varphi}{\partial x}$. In the same way, take the derivative w.r.t. $y$ and solve for $\frac{\partial \varphi}{\partial y}$. Can you take it from here?

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).