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\(z-\sqrt{-2i},z=-\sqrt{2i},z=\sqrt{-2i},\text{ and }z=\sqrt{2i}.\) since we quad root to get \(\pm\sqrt{\pm2i}\)

This question has been asked multiple times...you could look for it as well Guest.

Actually the previously posted anserers found

z =  1 +i       1 - i      -1 +i    and   -1 - i