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-- Square root of the square root of -1

Square root of the square root of -1

also, sqrt(i)

try that one, ladies.

1?

sqrt(i) = 0.707106781 + 0.707106781 i

quote:

Originally posted by reveal sqrt(i) = 0.707106781 + 0.707106781 i

gah! cheater.

quote:

Originally posted by reveal sqrt(i) = 0.707106781 + 0.707106781 i

0wn3d

Now my question is: What are you in life by knowing this?

(been asking myself that question for many years now, still no answer )

quote:

Originally posted by reveal sqrt(i) = 0.707106781 + 0.707106781 i

quote:

Originally posted by Zenchowdah gah! cheater.

congrats reveal.

sqrt ( i ) = sqrt 2 + sqrt 2 i ?

quote:

Originally posted by zayd sqrt ( i ) = sqrt 2 + sqrt 2 i ?

Actually, that's 1/sqrt(2), not sqrt(2).

It's easier if you figure it as i = e^i(90�). When you take the square root, halve the angle to get e^i(45�), then use the identity e^(ix) = cos x + i sin x. You could use this to find pretty much any real or complex power of i.

Anyway, complex numbers have very real meaning, you just don't learn it in high school. They're as fundamental to signal processing as arithmetic is to your high school courses.

quote:

Originally posted by Mister_Michel Now my question is: What are you in life by knowing this? (been asking myself that question for many years now, still no answer )

Your avatar is hot, got any pics of this girl?

quote:

Originally posted by Scorchio Your avatar is hot, got any pics of this girl?

Yes, and while you're at it, why don't you start another fucking "post pics of hot girls" thread dedicated to her!

quote:

Originally posted by DigiNut Yes, and while you're at it, why don't you start another fucking "post pics of hot girls" thread dedicated to her!

Actually you'd have to start another thread about "POST PICS OF HOT MEN" because that is actually a male's ass in his avatar.

Just a technicality but j^1/2 has two solutions actually...

take j^1/2 = (1^1/2)*e(j*pi/2)*1/2 = e(j*pi/4) and -e(j*pi/4)

On the unit circle this corresponds to [rho,phi] = [1,pi/4] and [-1,pi/4].

Remember kids, the nth root of a number has n solutions.





oh god...

quote:

Originally posted by Mister_Michel Now my question is: What are you in life by knowing this?

a knob.

quote:

Originally posted by DigiNut Anyway, complex numbers have very real meaning, you just don't learn it in high school. They're as fundamental to signal processing as arithmetic is to your high school courses.

from my understanding, they are often used only to avoid having to solve differential equations though.

quote:

Originally posted by Mister_Michel Now my question is: What are you in life by knowing this? (been asking myself that question for many years now, still no answer )

you are someone who knows the answer to this. If you know and understand the answer then it shows a little bit more of your reasoning ability.

Actually...

Actually, Zarathustra, just a little technicality...

You're wrong.

The square root of any complex number (including the real set) is a one-to-one mapping. For example, the square root of 4 is only 2, not {2,-2}. Otherwise you wouldn't be able to distinguish -(4^0.5) from +(4^0.5). If this confuses you, let me know....I'll try to explain it differently :P

--HyprLogik

Re: Actually...

quote:

Originally posted by HyprLogik Actually, Zarathustra, just a little technicality... You're wrong. The square root of any complex number (including the real set) is a one-to-one mapping. For example, the square root of 4 is only 2, not {2,-2}. Otherwise you wouldn't be able to distinguish -(4^0.5) from +(4^0.5). If this confuses you, let me know....I'll try to explain it differently :P --HyprLogik

What are you talking about? This is absolutely wrong. If my memory of my complex numbers course serves me correctly, it can actually be shown that the nth root of any complex number actually has n values, not 1.

When you write a complex number Z as Ae^ix, which is actually Ay corresponds to (A^y)e^(ixy). The value of A (magnitude) is raised to the power y as with real arithmetic, and the angle is multiplied by y.

But neither the original angle x nor the resulting angle xy are unique. Since we're dealing with polar coordinates, x = arg(z) = arg(z) + 2*pi*k, where k is any integer value including zero. So when you multiply your angle by 2 - say it's 45� (pi/4) in the context of the original question - you end up with 90� (pi/2). Simple right? But if your original angle was 225� (5*pi/4), you end up with 450� (5*pi/2). 5*pi/2 is the same angle as pi/2, and your angle is only defined from (-pi,pi] so it must be written as pi/2.

And like I said, you can extend it to any nth root. If you take a 60� (pi/3) angle and take its cube to get a 180� (pi) value, you don't know whether the original angle was 60�, 180�, or -60�.

What you've stated here:

quote:

Otherwise you wouldn't be able to distinguish -(4^0.5) from +(4^0.5).

is simply not correct. When you raise a complex number to the power 0.5, you are by definition multiplying the angle by 0.5. To put it another way, when you use this operation on a real number, you are actually by definition taking the positive square root.

Mathematically, x^0.5 = |sqrt(x)|. Therefore, -(x^0.5) = -|sqrt(x)|. The former must be positive, the latter must be negative, assuming x is positive and real. There is no discrepancy here, it's simply a misunderstanding on your part of the difference between sqrt(x) and x^0.5.

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