In general: \(e^{i\theta}=\cos \theta+i\sin\theta\)
You already know that \(\sqrt{i}=e^{i\pi /4}\)
Put these together to get: \(\sqrt{i}=\cos{(\pi /4)+i\sin{(\pi /4)}}\)
Since we have: \(\cos{(\pi /4)}=\frac{1}{\sqrt{2}} \space \text{and }\sin{(\pi /4)}=\frac{1}{\sqrt{2}}\)
we can see that: \(\sqrt{i}=\frac{1+i}{\sqrt{2}}\)
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