Complex analysis - the logarithmic function

mariab89
Messages
11
Reaction score
0

Homework Statement



Show that the function Log(-z) + i(pi) is a branch of logz analytic in the domain D* consisting of all points in the plane except those on the nonnegative real axis.

Homework Equations





The Attempt at a Solution



I know that log z: = Log |z| + iArgz + i2k(pi)
I'm not sure where to start with this question, any help would be greatly appreciated!
thanks :)
 
Physics news on Phys.org
You know Log(z) is analytic except for a branch cut on the negative real axis. That means Log(-z) is analytic except for a branch cut on the positive real axis. So is Log(-z)+i*pi. Now you just have to show that it is a branch of log(z) by showing exp(Log(-z)+i*pi)=z.
 
so to show that what i did was...

exp(Log(-z) + i*pi) = exp(Log(-z))exp(i*pi) = (-z) (-1) = z

but.. I am still unclear how this shows that Log(-z) + i*pi is a branch of log z.
 
If g(z) satisfies exp(g(z))=z then it's a branch of log(z). That's what defines log(z). It's an inverse function of exp(z).
 
oh ok i see now!
Thanks a lot!
 
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