Complex root

$\begin{aligned} \sqrt{i} & = \frac{1}{\sqrt{a}} + \frac{i}{\sqrt{b}} \quad \quad \quad \quad \space \small \color{#3D99F6}{\text{Squaring both sides...}} \\ i & = \frac{1}{a} - \frac{1}{b} + \frac{2}{\sqrt{ab}}i \quad \quad \small \color{#3D99F6}{\text{Equating the imaginary part...}} \\ \Rightarrow 1 & = \frac{2}{\sqrt{ab}} \\ \sqrt{ab} & = 2 \\ \Rightarrow ab & = \boxed{4} \end{aligned}$

Since $i = cos(90°)+isin(90°)$ , we use the reverse De Moivre's Theorem to find that $\sqrt{i} = cos(\frac{90°}{2})+isin(\frac{90°}{2}) = cos(45°)+isin(45°) = \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i$ Other value of it is $cos(45°+\frac{360°}{2})+isin(45°+\frac{360°}{2}) = cos(225°)+isin(225°) = \frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i$ , but we won't use this other value. As $a = b = 2$ , $ab = 2\times2 = 4$

$\large \sqrt{\iota} = \left(e^{\iota\pi/2}\right)^{\frac{1}{2}} = e^{\iota\pi/4}$ $\large e^{\iota\pi/4} = \cos\left(\dfrac{\pi}{4}\right) + \iota \sin\left(\dfrac{\pi}{4}\right)$ $\large = \pm\dfrac{1}{\sqrt{2}}( 1 + \iota)$