Complex: Defining Sqr(-1) in Main Branch of logz

[Bacle]

Complex: Defining Sqr(-1) using Main Branch Logz
Posted: Jul 3, 2011 12:46 PM Plain Text Reply


Hi, All:

I am having trouble understanding how/if sqr(-1) can be defined when we use the standard branch Logz of logz:

As far as I know, we define the complex exponential z^b , in a region R, for z,b both complex, we first define (if possible), a branch of logz in R, after which we _define_:


z^b:=e^(b.logz)

Right?

So, say we want to define a log in the plane using the branch Logz of logz, i.e, the branch given by removing [0,oo), where the points in the real axis have argument 0.

But, once we removed [0,oo), how can we define Sqr(-1), given that Logz is not defined there, and Sqr is defined in terms of Logz by:

z^(1/2):= e^(Logz/2)? ($) And then removing one half of the remaining plane to avoid z^(1/2) being a multi-function. Still: Logz is not defined on the negative real axis, where -1 is, so how can we then define (-1)^(-1/2) using ($) above? Or is the existence of a log sufficient but not necessary for defining a square root?

 

[Feldoh]

Are you having trouble because you think Log(z) is undefined along the negative real axis? We can extend logarithms into the complex plane quite easily actually: http://en.wikipedia.org/wiki/Complex_logarithm

 

[Bacle]

Feldoh: I am not sure I get your point:

The log cannot be defined globally on the whole of C, and I don't know
how one would define it if we were to work on a Riemann surface. Of course, we
can find a branch for which log is defined on the negative real axis, but then we must
cut-off another part of the plane to be able to define a branch. It all comes from the
fact that e^z is many-to-one ; actually oo->1 , so that it does not have a global
inverse ( on C-{0} ), but instead it only has local inverses when we define it in
a strip (t,t+2Pi), where it is 1-1.