Unable to prove that $\sqrt{i} + \sqrt{-i}$ is a real number. I did this :-
$$
\sqrt{i} = \sqrt{\frac{1}{2}.2i} = \sqrt{\frac{1}{2}.(1 + 2i - 1)} = \sqrt{\frac{1}{2}.(1 + 2i + i^2)} = \sqrt{\frac{1}{2}.(1+i)^2} = \frac{1}{\sqrt{2}}(1+i) \\= \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \mbox{ --------- 
1} \\ \mbox{Also, }
\sqrt{-i} = \sqrt{-1.i} = \sqrt{-1}.\sqrt{i} = i\sqrt{i} = i(\frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}}) \mbox{ --------from 1} \\ = \frac{1}{\sqrt{2}}i - \frac{1}{\sqrt{2}} \\ = -\frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \mbox{ --------- 2} \\
\mbox{Adding 1 and 2, we get} \\ \sqrt{i} + \sqrt{-i} \\ = \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \\= i.\frac{2}{\sqrt{2}} \\= i.\frac{\sqrt{2}.\sqrt{2}}{\sqrt{2}} \\= i\sqrt{2}
$$
This is a complex number, not a real number. 
What am I doing wrong here? Is there another way to prove this ?

so to sum up. the question is incorrect. there are 4 solutions for this because root of any number has 2 solutions. in this question  there are 2 real solutions and 2 solutions are on the imaginary axis
