<s>[INST] What is(are) the value(s) of $\sqrt{i}+\sqrt{-i}$? When I am going to find out the value of $\sqrt{i}+\sqrt{-i}$, I stuck to evaluate $\sqrt{i}\times \sqrt{-i}$.  
Progress:  $\sqrt{i}+\sqrt{-i}=\sqrt{(\sqrt{i}+\sqrt{-i})^2}=\sqrt{2\times \sqrt{i}\times\sqrt{-i}}$.  
Now $\sqrt{i}\times\sqrt{-i}=\sqrt{i\times (-i)}=\sqrt{-i\times i}=\sqrt{-1\times i^2}=\sqrt{-1\times -1}=1$
 But again, $\sqrt{i}\times\sqrt{-i}=\sqrt{i}\times \sqrt{i}\sqrt{-1}=\sqrt{i}\times \sqrt{i}\times i=i\times i=i^2=-1$.
Which one is correct and what is the logic behind it?
and Finally what are the values of $\sqrt{i}+\sqrt{-i}$
 [/INST] I'm using your method, but I'm restraining myself from writing $\sqrt{z}$ for $z$ complex. Let's use use only $[i^2=-1]$ and $[x^2=y^2\iff x=\pm y]$
Let's have $a^2=i$ and $b^2=-i$, we are searching for the value of $(a+b)$.
$(a+b)^2=a^2+2ab+b^2=i+2ab-i=2ab$
$a^2b^2=(i)(-i)=1\iff ab=\pm 1$


*

*If $ab=1$ then $(a+b)^2=2$ and $(a+b)=\pm\sqrt{2}$

*If $ab=-1$ then $(a+b)^2=-2$ and $(a+b)=\pm i\sqrt{2}$


We found $4$ possible values, and this is normal, remember that in $\mathbb C$ each complex has $2$ possible roots so $\sqrt{z_1}+\sqrt{z_2}$ would have $4$ possible values (if $z_1\neq z_2$ of course).
Or simply notice that $z=a+b$ is solution of $z^4=(a+b)^4=(2ab)^2=4a^2b^2=4$ and $z^4-4=0$ has $4$ different roots.
</s>