18 Jan 06
Originally posted by CrazyLilTingAside from sqrt(-i)?
You have at most an hour to reply:
Which is the square root of -i ?
P.S.: I dont know the answer. but could intenti to find it! And this isn't a college duty! 😀
-J
EDIT: well, you deserve more time... it's too late in the night! 🙂
i*sqrt(i) would work too, is the same.
18 Jan 06
Originally posted by Balla88No I didn't. As the google page says the answer is 0.707... - 0.707...i
You forgot the (-sqrt(2)/2i)
sqrt(-i) = sqrt(2)/2 - sqrt(2)/2i
I also pointed out that 0.707... = sqrt(2)/2 as the squareroot form is much more useful (and precise). I did not state that the answer was 0.707...
Also be careful of your brackets. sqrt(2)/2i is easily read incorrectly. Better would be sqrt(2)/2 * i.
18 Jan 06
-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.
20 Jan 06
1 edit
Originally posted by XanthosNZGoogle is wise. Wikipedia is wiser!
Google knows all.
http://www.google.com/search?&q=sqrt(-i)
Also 0.707106781 = sqrt(2)/2
http://en.wikipedia.org/wiki/Complex_numbers#Geometric_interpretation_of_the_operations_on_complex_numbers
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'. 😛
20 Jan 06
Originally posted by royalchickenDon't let it happen again! ðŸ˜
-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.
20 Jan 06
Originally posted by leisurelyslothTrue, but we'll assume the original poster asked for the principal value 😛.
Google is wise. Wikipedia is wiser!
http://en.wikipedia.org/wiki/Complex_numbers#Geometric_interpretation_of_the_operations_on_complex_numbers
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'. 😛
20 Jan 06
Originally posted by royalchickenIt's a right of passage.
-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.