Called Complex For A Reason - iii

Calculating the Square Root of $i$ :

$(a+bi)^2 = i$

$a^2 + 2abi + (bi)^2 = (a^2-b^2) + 2abi$

Therefore, first the real part, $(a^2-b^2),$ must equal $0$ (aka, $a^2 =b^2 ).$ And also, the imaginary part, $2abi,$ must equal $i$ , yielding a second equation: $2ab = 1$ .

Solving the two simultaneous equations:

$a = \frac{1}{2b}$
$a^2=b^2 \rightarrow \left(\frac{1}{2b}\right)^2 = b^2 \rightarrow \frac{1}{4b^2} = b^2 \rightarrow \frac{1}{4} = b^4 \rightarrow b = \pm \frac{1}{\sqrt{2}}$ and, therefore $a$ is also equal to $\pm \frac{1}{\sqrt{2}},$ suggesting two potential values for $\sqrt{i}:$

$\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$ and $-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i.$

However, only $\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i$ is a correct evaluation of $\sqrt{i}$ since the square root function is specified to have the range of only complex values with positive real components.

One can similarly calculate that $\sqrt{-i} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i.$

Therefore, $\sqrt{-i} + \sqrt{i} = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}i + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}i = \frac{2}{\sqrt{2}} = \sqrt{2}$

Zandra herself wrote a standard solution, so now I will present another. Using the reverse De Moivre's identity, we will get that $\sqrt{i} = cos(\frac{0.5\pi}{2})+isin(\frac{0.5\pi}{2}) = cos(\frac{\pi}{4})+isin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}(1+i)$ and $\sqrt{-i} = cos(\frac{\pi}{4})-isin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}(1-i)$ Summing them gives $\sqrt{-i}+\sqrt{i} = \frac{1}{\sqrt{2}}(2) = \sqrt{2}$ Here, we just ignore another root of $\sqrt{\pm i}$ (of $arg(z) = \frac{5\pi}{4})$ as it will yield negative results, which is in contrast to the sign of roots (we are looking for principal, positive roots)

$\begin{aligned} \sqrt{-i} + \sqrt i & = \sqrt{\left(\sqrt{-i} + \sqrt i\right)^2} \\ & = \sqrt{-i + 2\sqrt{-i \cdot i} + i} \\ & = \boxed{\sqrt 2} \end{aligned}$

let sqrt(-i) + sqrt(i)=x (-i) + (i) +2 sqrt(-i i)=x^2 ...squaring both sides 2*(1)=x^2 x=+sqrt(2) or -sqrt(2) ignoring negative one we get sqrt(2) as answer bold text