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! 🙂

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

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

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

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

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.

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'. 😛

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.