# Problems with the complex square root

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?

• What do you mean by script? Feb 24, 2021 at 15:24

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

• 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 $-1^{\frac{1}{2}}=-i$?
– Nick
Feb 24, 2021 at 15:57
• 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 $-\sqrt{|z|}e^{i\operatorname{arg}(z)/2}$. Feb 24, 2021 at 19:08
• @ 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.
– Nick
Feb 24, 2021 at 19:36
• I think that you missed the minus sign in the definition of $z^{1/2}$. Feb 24, 2021 at 19:41
• 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.
– Nick
Feb 24, 2021 at 20:02