I see $i$

$\begin{aligned} \left(\frac{\sqrt{i} + i\sqrt{i}}{i}\right)^2 & = \left(\frac{\color{#D61F06}{\sqrt{i}} + i\color{#D61F06}{\sqrt{i}}}{\color{#D61F06}{\sqrt{i}} \sqrt{i}} \right)^2 \quad \quad \small \color{#D61F06}{\sqrt{i} \text{ in the nominator and denominator cancel off.}} \\ & = \left(\frac{1 + i}{\sqrt{i}}\right)^2 \\ & = \frac{2i}{i} \\ & = \boxed{2} \end{aligned}$

Rewrite it into the form of $(\frac{1}{\sqrt{i}}+\sqrt{i})^2 = i+\frac{1}{i}+2 = i-i+2 = 2$

$\Large \left( \frac{\sqrt{i}+i\sqrt{i}}{i}\right)^2 = \dfrac{i+i^3+2i(\sqrt{i})^2}{i^2} = \dfrac{i-i+2i^2}{i^2} =2$

$i = e^{\frac{i\pi}{2}}$ ; $(\frac{\sqrt{i} + i\sqrt{i}}{i} )^2 = ( \frac{e^{\frac{i\pi}{4}} + e^{\frac{3i\pi}{4}}}{e^{\frac{i\pi}{2}}})^2 = (\frac{e^{\frac{i\pi}{2}} + 1}{e^{\frac{i\pi}{4}}})^2 = \frac{2i}{i} = 2$ or $( \frac{e^{\frac{i\pi}{4}} + e^{\frac{3i\pi}{4}}}{e^{\frac{i\pi}{2}}})^2 = ( \frac{e^{\frac{-i\pi}{2}} \cdot (e^{\frac{i\pi}{4}} + e^{\frac{3i\pi}{4}})}{e^{\frac{-i\pi}{2}} \cdot e^{\frac{i\pi}{2}}})^2 = (e^{\frac{i\pi}{4}} + e^{\frac{-i\pi}{4}})^2 = (2 \cos \frac{\pi}{4})^2 = 2$

$\Rightarrow \left( \frac{\sqrt{i}+i\sqrt{i}}{i} \right)^2$

$\left(\frac{\sqrt{i}+\sqrt{i^3}}{i} \right)^2$

$\left(\frac{(\sqrt{i})^2+(\sqrt{i^3})^2+2×\sqrt{i^4}}{(i)^2}\right)$

$\left(\frac{i-i+2×(i)^2}{-1}\right)$

$\frac{-(2)}{-(1)}=\boxed{2}$