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?

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

1 Answer 1


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

  • $\begingroup$ 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

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