Branch Cut for (lnz)^2: Determining Branch Points and Range

CrimsonFlash
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Homework Statement


It is simply the same as the one for lnz i.e. does it go from 0 to ∞?
Also, is there any proper way to figure out branch points of a function?

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CrimsonFlash said:

Homework Statement


It is simply the same as the one for lnz i.e. does it go from 0 to ∞?
Also, is there any proper way to figure out branch points of a function?
Do you mean, "Does z go from 0 to ∞?" Your question is unclear, which is why no one will answer.

Generally, branch cuts can be made fairly arbitrarily as long as they make sense. Though, I personally prefer to use Riemann surfaces because they are more fun for me.
 
Certainly if you have a domain where ln(z) is defined holomorphically, then ln(z)2 is also holomorphic on the same domain
 
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Ok thanks.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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