Answer
We need to prove $$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)^2=i$$ first.
By doing so, it is true that $$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)=\sqrt i$$
Work Step by Step
$$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)=\sqrt i$$
To prove this, we can prove that
$$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)^2=i$$
We come from the left side first:
$$A=\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)^2$$
$$A=\Big(\frac{\sqrt2}{2}\Big)^2+\Big(\frac{\sqrt 2}{2}i\Big)^2+2\times\frac{\sqrt 2}{2}\times\frac{\sqrt 2}{2}i$$
$$A=\frac{1}{2}+\frac{1}{2}i^2+\frac{\sqrt 2\times\sqrt 2}{2}i$$
$$A=\frac{1}{2}+\frac{1}{2}(-1)+\frac{2}{2}i$$
$$A=\frac{1}{2}-\frac{1}{2}+i$$
$$A=i$$
Therefore, it is true that $$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)^2=i$$
Thus, $$\Big(\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\Big)=\sqrt i$$
