Principal branch of the logarithm

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Homework Statement



Define an analytic branch f(z) of w, such that f(z)=0 for the limit of z->\infty

Now what is f(1)?

Homework Equations



w=\frac{z+i}{z-i}

The Attempt at a Solution



The branch cut of the logarithm is: (-\infty,0)
All branches of the logarithm are:
f(z)=Log(z)+iArg(z)=Log(z)+2i\pik

But then f(1)=0, which is wrong.
 
Last edited:
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There was a typo in my first post:

mxmt said:
w=\frac{z+i}{z-i}

Of course this should be w=log(\frac{z+i}{z-i})

Anybody who understands it now?
 
A guess is that the line segment of z=(i,-i) is mapped onto the normal branch cut of the logarithm (-inf,0). Therefore, f(1)=exp(iPi/2) because this is where i is located in the complex plane.
 
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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