Okay so **warning** I'm not used to american mathematics notations. So i work with complex numbers a lot and i noticed a little shortcut.

So

(1+i)^2= 1 +2i-1 = 2i

So we can say that

i= ( (1+i)÷sqrt (2) ) ^2

Then

Sqrt (i) = (+or-)(1/sqrt (2) + ( 1/sqrt (2) ) i)

So sqrt (i)= -1/sqrt(2) - ( 1/sqrt(2)) i

Or sqrt (i) = 1/sqrt(2) + (1/sqrt(2)) i

Which is what mr blackpenredpen found.

Also, is this method correct? I'm still in high school so i may still be behind hahah. Also, big fan of your work.

I have a faster answer. (1+i)^2=1+2i-1=2i. So i=(1+i)^2/2. Therefore sqrt(i)=(1+i)/sqrt(2)=1/sqrt(2)+1/sqrt(2)i. I went off of your answer though, so thanks.

Great attempt! Almost the same as my answer but you didn't include the negative answer.

If a^2=4 then a=2 or a= -2

For that reason i think your answer is incomplete!

Thank you for answering!