More Complex Analysis-Removable Singularity

WannaBe22
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Homework Statement


Let f be analytic in the region (z:0<|z-a|<r) and isn't defined at z=a.
Prove that if there is a neighborhood of z=a where Re f(z)>0 then z=a is a removable singularity of f.


Hope you'll be able to help me
Thanks in advance

Homework Equations





The Attempt at a Solution

 
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Hint: Consider Log[f(z)].
 
thanks!
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

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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