When does derivative of complex Log exist?

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The discussion focuses on the conditions under which the derivative of the complex logarithm exists for a function f: ℝ → ℂ. It is established that the logarithm Log(z) is differentiable in the region where |z| > 0 and Arg(z) is within (-π, π). For Log(f(x)) to be differentiable, the image of f(x) must lie within this region, and f must satisfy the Cauchy-Riemann equations. The participants confirm that these conditions are necessary for the existence of the derivative. Overall, understanding these requirements is crucial for analyzing the differentiability of complex logarithmic functions.
mnb96
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Hello,
if I have a function f:\mathbb{R}\rightarrow \mathbb{C}, what are the conditions on f for which the derivative of the Logarithm exist?

\frac{d}{dx}\mathrm{Log} (f(x)) exists?

Note that here I defined:

Log(z)=\log |z| + Arg(z), where -\pi< Arg(z) \leq \pi
 
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Try the Cauchy-Riemann equations.
 
Ok...so, given the fact that Log(z), (with z complex) as defined above, is differentiable in the region:

|z|>0
Arg(z)\in (-\pi,\pi)

In order for Log(f(x)) to be differentiable, I need to have that;

1) the image of f(x) must be in that region (the one where Log is differentiable)
2) f must satisfy the C-R equations

Is this correct?
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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