-i = exp{i*[(-pi)/2]}, whose square root is exp{i*[(-pi)/4]} = [sqrt(2)/2]*(1-i), as they said.

I only contributed to this thread so I could take the opportunity to be paranthetically unambiguous in as many places as possible, thus doing all I can to keep plonkers like XanthosNZ at bay.

This is also my first use of the word 'plonker' in any context, ever.

It's best to think of complex numbers geometrically as suggested in the Wikipedia page. Then it becomes clear that there are two answers to the question. Both:

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

and

- 1/sqrt(2) + i/sqrt(2)

are solutions.

EDIT: Hey Russ! How 'bout you stop fiddling around with that analysis board, and get us a decent equation editor in the forums. Jeez, you'd think this was a chess site or somethin'. ðŸ˜›

Originally posted by royalchicken -i = exp{i*[(-pi)/2]}, whose square root is exp{i*[(-pi)/4]} = [sqrt(2)/2]*(1-i), as they said.

I only contributed to this thread so I could take the opportunity to be paranthetically unambiguous in as many places as possible, thus doing all I can to keep plonkers like XanthosNZ at bay.

This is also my first use of the word 'plonker' in any context, ever.

It's best to think of complex numbers geometrically as suggested in the Wikipedia page. Then it becomes clear that there are two answers to the question. Both:

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

and

- 1 ...[text shortened]... ent equation editor in the forums. Jeez, you'd think this was a chess site or somethin'. ðŸ˜›

True, but we'll assume the original poster asked for the principal value ðŸ˜›.

Originally posted by royalchicken -i = exp{i*[(-pi)/2]}, whose square root is exp{i*[(-pi)/4]} = [sqrt(2)/2]*(1-i), as they said.

I only contributed to this thread so I could take the opportunity to be paranthetically unambiguous in as many places as possible, thus doing all I can to keep plonkers like XanthosNZ at bay.

This is also my first use of the word 'plonker' in any context, ever.