Unable to prove that $\sqrt{i} + \sqrt{-i}$ is a real number.

Question

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 ?

Answer 1

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

Answer 2

Using the the polar representation for complex numbers, there holds that $$ \sqrt{i}=e^{i\frac{\pi}{4}}~~~\mbox{and}~~~\sqrt{-i}=e^{-i\frac{\pi}{4}}$$ so that $$ \sqrt{i}+\sqrt{-i}=e^{i\frac{\pi}{4}}+e^{-i\frac{\pi}{4}}=2\cos\left(\frac{\pi}{4}\right)=\sqrt{2}.$$

Answer 3

Note that $i=e^{(\pi/2) i}$ and $-i=e^{(-\pi/2) i}$

Thus $\sqrt i = e^{(\pi/4) i}$ and $\sqrt -i = e^{(-\pi/4) i}$

Adding the two we get $$\sqrt i+\sqrt -i =e^{(\pi/4) i}+e^{(-\pi/4) i}=2 \cos (\pi/4)=\sqrt 2$$ Which is real.