- 18 Jan '06 08:55

Aside from sqrt(-i)?*Originally posted by CrazyLilTing***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 15:23

No I didn't. As the google page says the answer is 0.707... - 0.707...i*Originally posted by Balla88***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 19:03-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 06:52 / 1 edit

Google is wise. Wikipedia is wiser!*Originally posted by XanthosNZ***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 07:15

Don't let it happen again!*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. - 20 Jan '06 12:17

True, but we'll assume the original poster asked for the principal value .*Originally posted by leisurelysloth***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 17:57

It's a right of passage.*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.