So here is something I discovered in high school that is much more interesting than Megan.

What's the sqrt(i)? Well, let's see

sqrt(i)=i^(1/2)=i^(4/8)=(i^4)^(1/8)=1^(1/8)=1.

This logic also gives you i^1/2=i^1/10 and you can pretty much screw around with it to make any number equal any other number. Obviously, that's not true, but no one has ever been able to tell me why.

Tell me why! Tell me why!

And before you go and call me a dork, I'd like to remind you that I have not only slept with too many girls and done too many drugs for that to be true, I've also jumped out of too many airplanes to be a dork. Numerically speaking, it's statistically impossible for me to be a nerd.

It's my work, I say. I only do it for pay.

Sorry, I'll stop hijacking the blog now. Last time, I promise.

## Wednesday, August 5, 2009

### Math Time!

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## 16 comments:

It's not true from the first equivalence on; the field of real numbers is not closed and taking even roots of even powers of real numbers always returns the absolute value of the real number, i.e.,(-1)^(4/8)=[[(-1)^(4)]^(1/4)]^(1/2)=abs[-1]^(1/2)=1.

Another way to see this is to think of multiplication by i as a counterclockwise rotation of 90 degrees in the complex plane. So in polar coordinates, -1=e^pi and (-1)^(1/2)=e^(pi/2). Thus, (-1)^(4/8)=(e^(pi/8))^4, that is, take the vector 0.92+0.38i and rotate it counterclockwise pi/8 radians(22.5) degrees three more times; the ending vector will be i. Notice that in the complex plane, this operation is closed. In the line of equations you gave, the operation is not closed.

Hello,

The problem is that when you take the square root of i (or j for electrical engineers), you are only dealing with the magnitude.

Taking your example again but representing "i" using polar notation, (all complex numbers can be represented as A*e^*(i*B):)

(Where A, B are real numbers and A is the magnitude and B the phase, usually in radians)

(Note "pi" is 3.14...blah)

i=1*e^(i*pi/2)

i^(1/2)= 1*e^((i*pi/2)*1/2)

i^(1/2)= 1*e^(i*pi/4)

In rectangular form, if you like, this is:

(1/sqrt(2))*(1+i).

This is certainly not 1 as you have claimed. Rather it has a magnitude of 1. When you took the square root you only took the square root of the magnitude and you lost all the phase information.

Specifically you performed operations on "i" thinking it was real number, since real numbers have no phase information. This was your mistake.

Please tell me if I wasn't clear and I can clarify as much as I can.

(You can visually see this if you type:

sqrt(e^(i*pi/2))

in:

http://www.wolframalpha.com/

Direct Link:

http://www35.wolframalpha.com/input/?i=sqrt(e^(i*pi%2F2))

Regards,

SV

Probably because you're applying real number logic to complex numbers. Here's the exercise: draw a unit circle with i and -i on the y-axis and 1 and -1 on the x-axis. Imagine that you're standing on i, now take i*i. You'll be at -1. Multiply by i again and you'll be at -i. Etc, etc, you're moving 90° counterclockwise around the unit circle each time. Now start at 45° off the Real axis and do the same thing; instead of moving 90° CCW with each power increase, you'll move 45°. Same applies any time you have a complex number. Now apply that in reverse. i^2 rotates your starting point of 90° out to 180°; your angle has doubled in magnitude. i^1/2 rotates your starting point of 90° to 45°; your angle has halved in magnitude. i^1/4 rotates you to 22.5°, etc. So i^4/8 is actually rotating you by 4x90° = 360° then returning you to 360/8°, or 45°. Hit up "De Moivre's formula" on wikipedia for a more math-based explanation. e is such a ridiculously omnipresent number...

The explanations are awesome, in that they let me guess what field each of you are in. In order, I'm going to guess:

Math, Elec Eng, and the last one is one or the other, I think.

thanks for the explanation! I never thought of squaring as rotation around the complex plane, which is remarkable considering I use that plane a lot.

Seeing ways to get the right answer isn't the problem. I don't see which equality is wrong.

So scent, I don't from your explanation why you can't say that

sqrt(i)=i^(1/2)

Hello,

Elec Eng as you guessed correctly. I keep typing j instead of i :).

There is nothing wrong with saying sqrt(i) = i^(1/2).

(Type each of these in www.wolframalpha.com and you get the same answer).

Typically though the common association of sqrt is only applicable to positive real numbers as Scent pointed out. (Within the field of positive real numbers, the sqrt function is closed i.e. sqrt(-9) is not a real number).

(Clearly you see this when you plot y=sqrt(x)).

When you write i^4=1 you are correct, however, you have lost phase information. Whereas the number you have started with does have phase information.

In Scent's statement:

(-1)^(4/8)=[[(-1)^(4)]^(1/4)]^(1/2)=abs[-1]^(1/2)=1

Scent starts of with sign information and along the way "looses" it (handwaving at it's best). It's more clear starting at the end.

abs(-1)=1 does not imply -1=1. Also, (-1)^2=(1)^2 does not imply -1=1.

Revisiting the original statement and in particular i^4=1. If instead we write,

i^4=e^(i*(pi/2))*4

=e^(2*i*pi)

And then

(i^4)^(1/8)=e^((2*i*pi)*(1/8))

(i^4)^(1/8)=e^((i*(pi/4))

Which gives the desired result.

Hopefully this is clearer now.

Regards,

Sanjay

Similarly if we main

So it seems to me that problem is that imaginary components are analogous to negatives, in that they get lost in some multiplications. It's basiucally the same reason you can't say sqrt(-1)=-1^2/4=1.

So for real numbers, the problem lies in the fact that even roots have two answers, that is, sqrt isn't monotonic. That's not the case for imaginary numbers, though. -i*-i=1, not -1.

I guess that that's resolved by remembering that powers need to use complex conjugates, which are equal to each other for real numbers?

...and that's why Megan McArdle should be fired.

QED, bitches.

Hello,

I am not sure if I am helping but,

-i*-i=-1.

Technically we should write,

-i*-i=e^(-i*(pi/2))*e^(-i*(pi/2))

-i*-i=e^( (-i*(pi/2)) +(-i*(pi/2)) )

-i*-i=e^(-i*pi).

This way we preserve phase information.

Very technically you can't say

sqrt(-1). As Scent says, the field of real numbers is not closed on sqrt.

That is sqrt(-1) is not a real number thus real number manipulations on this technically can't be done. (You can nevertheless do so but you have to be careful otherwise your results can be wacky).

(A little aside, complex numbers came about not because people looked at the equation x^2=-1 and said what gives but rather because they were trying to get solutions to cubic equations. The formula was derived by two different men Gerolamo Cardano and Scipione del Ferro. But today, it's known as Cardano's method of solving cubics. More at http://en.wikipedia.org/wiki/Cubic_function. Anyway, the solution sometimes required that one take the square root of a negative number as a intermediate step yet the final answer was real. They kept aside this sqrt(-1) as separate variable and usually the formula would require that this variable be multiplied by itself so everything was nice and clean. Complex numbers were born from further study of these solutions).

I am not sure what you mean by "remembering that powers need to use complex conjugates".

However, taking the magnitude of complex number requires taking that number and multiplying by it's complex conjugate). I.e.

|5+6i|=sqrt( (5+6i)x(5+6i)*)

where x is multiply and * is conjugate.

Conjugate of a real is itself. So you're right there.

Hopefully that clears stuff up.

Regards,

SV

-i*-i=1, you proveit yourself by saying it = e^-ipi

What I meant by the power statement is that a+bi^2=a+bi*a-bi

If it's worth mentioning at this point, your last step, 1^(1/8), has eight solutions: 1, -1, i, -i, sqrt(2)*(1+i)/2, sqrt(2)*(1-i)/2, -sqrt(2)*(1+i)/2, and -sqrt(2)*(1-i)/2. The two correct solutions to your first step are among them, but you picked one of the six that (according to previous commenters who understand more about it than I do) involve the loss of phase information.

Hello,

Err...You might want to check that.

e^(-i*pi)=-1

or even type -i*-i into

www.wolframalpha.com.

Also (a+bi)^2=(a+bi)x(a+bi). No conjugate needed.

Magnitude, or Absolute Value of (a+bi):

|a+bi|=sqrt((a+bi)x(a+bi)*)

where (a+bi)* is the conjugate of (a+bi).

Again you can type:

(a+b*i)^2 and |a+b*i| at wolframalpha.com to see more clearly.

Regards,

SV

No way in hell a braggardly, insecure attention-whore who equates taking drugs and un-substantiated boasts of sexual conquests with being teh cool could EVAH be a nerd.

That unpossible.

OK, I concede to SV cause I don't really know my complex stuff that well.

Also, last Nony, love your grasp of irony and self deprecation. Have you been on the internet long?

So scent, I don't from your explanation why you can't say thatsqrt(i)=i^(1/2)

Assuming you're still reading this: it's perfectly okay to say that Sqrt[i]=i^(1/2). The best way of looking at this is by using polar coordinates.. For example, the complex number 1+Sqrt[3]*i has a modulus of 2, and can be written as 2*e^(i*pi/3). Similarly, i=O+1*i, which has a modulus of 1, and can be written as 1*e^(i*pi/2). So Sqrt[i]=i^(1/2)=[1*e^(i*pi/2)]^(1/2)=1^(1/2)*[e^(i*pi/2)]^(1/2)=1*e^(i*pi/4).

Here's the deal now,

anynon-negative real number r can be written as r=r+0*i. So the modulus of r is just r, and the polar representation is r*e^(i*pi*0). But 0 is very special; this reduces to r=r*e^(0)! So taking, say, the square root of r gives us: Sqrt[r]=r^(1/2)=[r*e^(0)]^(1/2)=r^(1/2)*[e^(0)]^(1/2)=r^(1/2)*e^(0*1/2)=r^(1/2)*e^(0). Iow, taking roots of non-negative real numbers results in a real number, and this is because of the exponent 0. Taking the roots of anything else will give you a complex number.Hope that clears things up.

Yup, thanks!

[Unable to type, as both jaw & fingers have fallen to the floor.]

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