0
$\begingroup$

I have a problem understanding the following procedure. ( It's from a script)

Consider the domain C[0,$\infty$) and the branch of the logarithm given by

$ log(z)=ln(|z|)+i \cdot arg(z)$ ,with $arg(z) \in(0,2\pi)$.

Then we find: $z^{\frac{1}{2}} = \sqrt{|z|} e^{i arg(z)/2}$, where $arg(z^{\frac{1}{2}}) \in (0,\pi)$ The square root of any complex number that is not a positive real then lies in the upper halfplane and in this case we find $(-1)^{\frac{1}{2}}=i$

I don't understand why the last one follows. Why applies $(-1)^{\frac{1}{2}}=i$ ?

I know I can write: $-1^{\frac{1}{2}}= e^{\frac{1}{2}ln(|-1|)+arg(-1)}=e^{\frac{1}{2}arg(-1)}$ and arg(-1) ist $\pi$ and with that the upper equation would be correct....BUT that's actually wrong, since $\pi$ is not in the domain of arg. Can someone explain that to me?

$\endgroup$
1
  • $\begingroup$ What do you mean by script? $\endgroup$
    – TonyK
    Feb 24, 2021 at 15:24

1 Answer 1

0
$\begingroup$

The only argument of $-1$ in $(0,2\pi)$ is $\pi$. Therefore, by that definition we have$$z^{1/2}=\sqrt1e^{i\pi/2}=i.$$

$\endgroup$
9
  • $\begingroup$ okay i guess i understand. But what about the following: Can i find a branch, for which $1^{\frac{1}{2}}$ and $-1^{\frac{1}{2}}$ are defined and $-1^{\frac{1}{2}}=-i$? $\endgroup$
    – Nick
    Feb 24, 2021 at 15:57
  • $\begingroup$ Yes: for each $z\in\Bbb C\setminus\{\lambda i\mid\lambda\in(-\infty,0]\}$ let $\operatorname{arg}(z)$ be the only argument of $z$ in $\left(-\frac\pi2,\frac{3\pi}2\right)$. Then define $z^{1/2}$ as $-\sqrt{|z|}e^{i\operatorname{arg}(z)/2}$. $\endgroup$ Feb 24, 2021 at 19:08
  • $\begingroup$ @ José Carlos Santos Unfortunately I don't understand that. In this case we would be have arg(-1) =! - $\pi$ and in the first case we had arg(-1) = $\pi$. I do not understand how the definition area changes the sign of $\pi$. Sorry I'm a little confused. $\endgroup$
    – Nick
    Feb 24, 2021 at 19:36
  • $\begingroup$ I think that you missed the minus sign in the definition of $z^{1/2}$. $\endgroup$ Feb 24, 2021 at 19:41
  • $\begingroup$ Yeah I didn't see that. But why can I just define a minus there? And why do I choose exactly this domain? I think my problem is still the same and only in a different place. $\endgroup$
    – Nick
    Feb 24, 2021 at 20:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.