Square root of -i

Ersby

Fortean
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Sep 12, 2001
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What is it? Perhaps this is a stupid question, but it's been bugging me all week. And any links to pages expaining the answer would be nice. Thanks.
 
z = (-i)^0.5
=> z^2 = -i
=> (a + b*i)(a + b*i) = -i
=> a^2 - b^2 = 0 and 2ab = -1

This is two equations with two unknowns. A bit of manipulation results in the two solutions:

z = 2^-0.5(-1 + i) and
z = 2^-0.5(1 - i)
 
Another way to get this:

-i = i sin 3π/2 = e<sup>3π/2</sup>

(using e<sup>i θ</sup> = cos θ + i sin θ )

sqrt(-i) = e<sup>3π/4</sup> = cos 3π/4 + i sin 3π/4 = sqrt(2) - sqrt(2) i.

Of course in that first expression for i, you can add an arbitrary multiple of 2π to the argument and still get an equation for i. That means you can add an arbitrary multiple of π to the argument in the expression for sqrt(i); that gives you the other root.
 
when taken the square root of a complex number, the result has two solutions with a magnitude that is the square root of the magnitude of the original. The angle or the root half the original, with the second solution being 180 degrees off the first solution.

So -i has magnitude 1 and angle of 270 (or -90) degrees.
so sqrt(-i) has magnidute of sqrt(1) = 1, and angle of 270/2 = 135 degrees. A second solution exists at -45 degrees.

So sqrt(i) = -0.707 + 0.707i or 0.707 - 0.707i

You can double check by multiplying either of those by itself.

Walt

P.S. to take the nth root of any number, change its format to its magnitude and angle in the complex plane. The solution then has magnitude of root n of original and an angle of angle of the original over n, and additional solutions spaced at 360/n degrees around the complex plane.

So take the fourth root of 1, express one as magnitude / angle (1 / 0 degrees) then take forth root of magnitude 1 which is 1, and divide the angle by 4 (0/4 = 0). So one solution to the fourth root of 1 is 1, but there are additional ones of similar magnitude spaced 90 (360/4) degrees. Magnitude 1 at 180 degrees corresponds to -1 and magnitude 1 at 90 and -90 degrees corresponds to i and -i. So our four solutions are

1^1/4 = 1, i , -1 or -i


Edited to add: yeah, what zombified said.
 
Ersby said:
What is it? Perhaps this is a stupid question, but it's been bugging me all week. And any links to pages expaining the answer would be nice. Thanks.
Click to expand...

Ersby,

My post from this thread from 9/28/04 should help answer it.

I'll repost here:


Note that we can write 2i as: (1+i)(1+i)

So sqrt(2i) = 1+i

So sqrt(i) = 1/sqrt(2) + i/sqrt(2)
Click to expand...


To change it to the sqrt(-i) case,

note that we can write -2i as: (1+i)*-(1+i)

So sqrt(-2i) = 1-i

So sqrt(-i) = 1/sqrt(2) - i/sqrt(2)
 
Zombified said:
Another way to get this:

-i = i sin 3π/2 = e<sup>3π/2</sup>

(using e<sup>i θ</sup> = cos θ + i sin θ )

sqrt(-i) = e<sup>3π/4</sup> = cos 3π/4 + i sin 3π/4 = sqrt(2) - sqrt(2) i.

Of course in that first expression for i, you can add an arbitrary multiple of 2π to the argument and still get an equation for i. That means you can add an arbitrary multiple of π to the argument in the expression for sqrt(i); that gives you the other root.
Click to expand...

My highlighting. You may want to look up your cosine and sine tables again. ;)
 
Gaaah, you're right, I'm off by a sign. :( But now it's too late to edit my post.
 
Zombified said:
Gaaah, you're right, I'm off by a sign. :( But now it's too late to edit my post.
Click to expand...
I hate those days when I try to correct something ... and the correction is wrong as well. :p

1/sqrt(2) - i 1/sqrt(2), :D
 
#@*&$%@()*&#)(@*&%@#)(*&)(@#*&^&#@$%#@

Dammit.

I've obviously spent far too much time standing on my head trying to figure out Kumar.
 
(a cis b)^n = (a^n cis bn).

So for -i:

(1 cis 3π/2)^(1/2) = (1 cis 3π/4)

= -sqrt(1/2) + sqrt(1/2)i.

Finding the cube root of -i is much easier.
 
phildonnia said:
(a cis b)^n = (a^n cis bn).

So for -i:

(1 cis 3π/2)^(1/2) = (1 cis 3π/4)

= -sqrt(1/2) + sqrt(1/2)i.

Finding the cube root of -i is much easier.
Click to expand...

i is one, what are the other two?

You gotta love the fundamental theorem of algebra

(complex analysis is so cool)
 
phildonnia said:
I get sqrt(3)/2 - i/2 and -sqrt(3)/2 - i/2. But i is the "principal" cube root of -i.
Click to expand...

I don't know what the "principal" root means. If it means the "easy" one, then I agree. However, in complex analysis, there is no reason that one is any more relevent than the other. In the complex world, for any polynomial of order n, there are n solutions, all as equally valid as the next. Some may be more simple than others, but that is pretty much the only way to distinguish them.

If you were talking about the cube root of -1 it would be different, because then you could distinguish between the real and complex solutions. But for -i, it is all complex.
 
pgwenthold said:
I don't know what the "principal" root means. If it means the "easy" one, then I agree. However, in complex analysis, there is no reason that one is any more relevent than the other.
Click to expand...

After some research, I am informed that the "principal root" only applies for even roots of real numbers, and means the positive real root.

I had understood it to mean "the root with the smallest angle", which it apparently does not mean.
 
I guess if you want to annoy your friends, you can ask "What's the fourth root of 16?" If they say "2", say "No, 2i". When you ask for the nth root, it's assumed you mean "the positive root". ("The" fourth root of 16 is 2, but -2, 2i, and -2i are all roots). Once you leave the real numbers, "positive" has no meaning, so there is no such thing as "the" root.

Now, there is something called "the principle root": http://mathworld.wolfram.com/PrincipalRootofUnity.html "Informally, the term 'principal root' is often used to refer to the root of unity having smallest positive complex argument. "
 
Ersby said:
What is it? Perhaps this is a stupid question, but it's been bugging me all week. And any links to pages expaining the answer would be nice. Thanks.
Click to expand...

The various methods used here will work, of course. To get the answer quickly (and in your head), though, consider the problem using vectors in the complex plane.

Imagine two perpendicular axes (much like the "x" and "y" axis normally seen), where the horizontal axis is real and the vertical axis is imaginary. -i is represented by a vector pointing straight down (i.e. (0,-1)), and 1 is the vector pointing to the right (i.e. (1,0)).

When you square a complex number, its angle (as measured from (1,0)) is doubled. So the question becomes "which angles, when doubled, give you the same angle as (0,-1) (namely, 3*pi/2 rads)?" The answers will be the angles halfway to (1,0) from (0,-1) on both sides--namely (-sqrt2/2, sqrt2/2) and (sqrt2/2, -sqrt2/2) or -sqrt2/2 + sqrt2/2*i and sqrt2/2 - sqrt2/2*i.

Diagrams probably would have helped, and it takes a few tries to get used to, but really... it's fast and easy ;)
 
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