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Why z^1/2 real part is positive

Amer

Active member
Mar 1, 2012
275
as a function [tex] f(z) = z^{\frac{1}{2}} [/tex]
why the real part is positive
My work
I looked into
[tex] g(z) = z^2 [/tex] , natural domain is the complex field
we can see that
[tex]g(z) = g(-z) [/tex] , g is not 1-1
if [tex] z = r e^{i\theta}[/tex]
[tex] -z = e^{i\pi} z = re^{i(\theta + \pi)} [/tex]
so we will restrict the domain to get one-one function so we will have the inverse f
how to restrict it, or how to solve it in another way
Thanks
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,793
as a function [tex] f(z) = z^{\frac{1}{2}} [/tex]
why the real part is positive
My work
I looked into
[tex] g(z) = z^2 [/tex] , natural domain is the complex field
we can see that
[tex]g(z) = g(-z) [/tex] , g is not 1-1
if [tex] z = r e^{i\theta}[/tex]
[tex] -z = e^{i\pi} z = re^{i(\theta + \pi)} [/tex]
so we will restrict the domain to get one-one function so we will have the inverse f
how to restrict it, or how to solve it in another way
Thanks
Hi Amer! :)

The real part does not really have to be positive.
The expression $z^{\frac{1}{2}}$ represents 2 values.
Since:
$$z=re^{i\theta}=re^{i\theta + 2ik\pi}$$
$$z^{\frac 1 2} = (r e^{i\theta + 2ik\pi})^{\frac 1 2}=\sqrt r ~ e^{i \frac \theta 2 + ik\pi}$$
The two distinct values are $\sqrt r ~ e^{i \frac \theta 2}$ and $\sqrt r ~ e^{i (\frac \theta 2 + \pi)}$.

So either $\Re(z^{\frac 1 2}) = \sqrt r ~ \cos(\frac \theta 2)$ or $\Re(z^{\frac 1 2}) = \sqrt r ~ \cos(\frac \theta 2 + \pi)$.

If we pick $k=0$, and also $-\pi < \theta \le \pi$ (the principal argument), and look only at the first value, then the real part will be positive.
 
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