Talk:Imaginary unit
From Academic Kids
The calculation rule
root(ab) = root(a)root(b)
is only valid for real, non-negative numbers a and b.
Well, actually this equation is valid as long as you stay in the principal branch of the log function, i.e. as long as a and b don't lie on the negative real axis. But that might confuse people unnecessarily. 128.111.88.229
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What on earth is this?
(From the "i and Euler's formula" section):
You could also compute it using Bernoulli's logarithme imaginaire (imaginary logarithms):
- <math> pi/2 = 2log[(1-i^{(1/2)i} \times (1+i)^{(-1/2)i]}<math>
That expression isn't even written correctly. And it doesn't lead to anything. Were more equations supposed to follow?
I'm removing it from the main page, but I invite someone more knowledgable about imaginary logarithms than I (or the original contributor) to rework it and put it back in. --Ardonik 03:45, 2004 Aug 8 (UTC)
Damn you Wikipedia! I'll never finish this story if you keep sucking all of my time into your hideous vortex of knowledge. Damn you. DAMN YOU!
What is <math>\sqrt{-1}<math>?
Well, what is <math>\sqrt{-1}<math>???
Are you asking a philosophical question, or are you wondering what simpler terms it can be reduced to? In the latter case, it doesn't get any simpler.
What is <math>\sqrt{i}<math>?
What I meant to ask is: what is <math>\sqrt{i}<math>?
- I suppose this question is asked by someone being curious, rather than by someone being subtle (in a way beyond my comprehension).
- In elementary math, the squareroot is a single-valued function, and e.g. 25 has only the squareroot +5. In complex analysis, multi-valued functions are allowed, and 25 has two squareroots +5 and -5. This comes about in the following way:
- The squareroot of x is defined as any number y satisfying <math>y^2 = x<math>. Thus, if y is a squareroot, so is -y (since <math>(-y)\times (-y) = y\times y<math>). Thus, any number except 0 has two squareroots.
- Thus, i has two squareroots. One is <math>\frac{1+i}{\sqrt{2}}<math>; the other one is minus that number. You can verify this in the following way:
- <math>\frac{1+i}{\sqrt{2}}\times\frac{1+i}{\sqrt{2}} = \frac{(1+i)(1+i)}{\sqrt{2}\sqrt{2}} = \frac{1\times 1+1\times i+i\times 1+i\times i}{2} = \frac{1+2i-1}{2} = i<math>.
- When dealing with products, quotients, powers and roots, it's often convenient to represent complex numbers in polar form, i.e. by their distance in the complex plane from the origin, and their direction angle. i is at a distance of 1 from the origin, at an angle of 90 degrees. <math>\sqrt{i}<math> is also at a distance of 1 (i.e. at the unit circle), but at an angle of 45 degrees or 45+180 degrees.
- But I don't think any of this belongs to the article on i. Do you?
- --Niels Ø 09:05, Mar 22, 2005 (UTC)
I suppose not. I was just wondering.
This does not produce a false result.
- <math>-1 = i \cdot i = \sqrt{-1} \cdot \sqrt{-1} = \sqrt{-1 \cdot -1} = \sqrt{1} = 1<math>
Is <math>\sqrt{1} <math> not -1 also. A square root has two results.
- In standard elementary math as taught by competent teachers at all pre-university levels, the squareroot is a function, and as such cannot give two results. That's why the quadratic formula must have "plus/minus" in front of the squareroot sign to give both solutions; the squareroot itself only gives one. In this context, <math>\sqrt{-1}<math> does not exist.
- But in complex analysis, it is customary to work with multivalued functions, and the squareroot of any complex number (including 1 and -1, excluding only zero) is two-valued. So, as any paradox in math or logic, understood properly there is no contradiction. The purpose of the paradox is to raise awareness about the difficulties involved in "understanding properly".--Niels Ø 20:35, Jun 15, 2005 (UTC)
