- #1
Bacle
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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?
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?