# For all the math wizzards...

CrazyLilTing
Posers and Puzzles 18 Jan '06 07:37
1. 18 Jan '06 07:371 edit
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! ðŸ˜€

-J

EDIT: well, you deserve more time... it's too late in the night! ðŸ™‚
2. TheMaster37
Kupikupopo!
18 Jan '06 08:55
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! ðŸ˜€

-J

EDIT: well, you deserve more time... it's too late in the night! ðŸ™‚
Aside from sqrt(-i)?

i*sqrt(i) would work too, is the same.
3. XanthosNZ
Cancerous Bus Crash
18 Jan '06 09:10

Also 0.707106781 = sqrt(2)/2
4. 18 Jan '06 14:21
Originally posted by XanthosNZ

Also 0.707106781 = sqrt(2)/2
You forgot the (-sqrt(2)/2i)

sqrt(-i) = sqrt(2)/2 - sqrt(2)/2i
5. XanthosNZ
Cancerous Bus Crash
18 Jan '06 15:23
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.
6. 18 Jan '06 15:35
It is read incorrectly only by people that do not know math. They should be more careful then.
7. royalchicken
CHAOS GHOST!!!
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.
8. leisurelysloth
Man of Steel
20 Jan '06 06:521 edit
Originally posted by XanthosNZ

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'. ðŸ˜›
9. leisurelysloth
Man of Steel
20 Jan '06 07:15
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! ðŸ˜
10. royalchicken
CHAOS GHOST!!!
20 Jan '06 12:17
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'. ðŸ˜›
True, but we'll assume the original poster asked for the principal value ðŸ˜›.
11. ark13
Enola Straight
20 Jan '06 17:57
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.