Finding Analyticity Region of a Function - Matt

  • Thread starter Thread starter ultramat
  • Start date Start date
  • Tags Tags
    Function
ultramat
Messages
1
Reaction score
0
Hi, I have some questions regarding how to find the analytisity region of a funtion.
I'm a little confuse after I studied the definition of analytic function: which it saids
[if a function f is differentiable at every z in A, then f is analytic on A]

eg. Log z is analytic on the entire complex plane EXCEPT the -ve real axis.
Which make sense to me since Log z is undefind when x<=0 & y=0 , for z=x+iy

Log z^2 is analytic on the entire complex plane again EXCEPT z=0, and exclude the
Imaginary axis. Is that right?

I'm wondering if there's a way to actually compute/calculate the region instead of doing it in the head?
Since Log z & Log z^2 is kinda basic, it'll be hard to do if it is comething like Log (1+2/z)

Thanks in advance

Matt
 
Physics news on Phys.org
log(-1) CAN be defined since e^(i*pi)=-1. The upper case 'L' in Log is not is not just ornamental. It's a branch of log that's undefined for negative reals precisely so it can be uniquely defined for all other complex numbers. You can't compute why a function like that is undefined. Where it's undefined is a matter of convention. Look up "analytic continuation".
 
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...
Back
Top