For all the math wizzards...

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! 🙂

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! 🙂

Aside from sqrt(-i)?

i*sqrt(i) would work too, is the same.

Google knows all.

http://www.google.com/search?&q=sqrt(-i)

Also 0.707106781 = sqrt(2)/2

Originally posted by XanthosNZ
Google knows all.

http://www.google.com/search?&q=sqrt(-i)

Also 0.707106781 = sqrt(2)/2

You forgot the (-sqrt(2)/2i)

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

Originally posted by Balla88
You forgot the (-sqrt(2)/2i)

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

No I didn't. As the google page says the answer is 0.707... - 0.707...i

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.

It is read incorrectly only by people that do not know math. They should be more careful then.

-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.

Originally posted by XanthosNZ
Google knows all.

http://www.google.com/search?&q=sqrt(-i)

Also 0.707106781 = sqrt(2)/2

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/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'. 😛

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.

Don't let it happen again! 😠

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'. 😛

True, but we'll assume the original poster asked for the principal value 😛.

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 a right of passage.