Ok, so something was bothering me the other day while I was stuck in traffic going home:
I learned in high school that there are those numbers called imaginary numbers, and particularly this one number i, which when squared is -1. So, any imaginary number squared becomes a negative real number?
What happens then when you want to extract the square root of i? Is that an imaginary number or a real number? And something else…picturing the trigonometric circle where i is the segment sticking up in the air and -1 (i^2) is the segment going “straight and to the left”…where would i^1/2 be ?
Oh, the other question: is there a name for the different levels of mathematical operators-let me explain myself here:
there’s + and -, and they’re on the same level.
Then there’s * and that level is above the one above because of the distibutivity rule a* (b+c) = ac +bc
A level above that would be ^ for the same reason…
and so on and so forth…but what do you call the distinction between the levels? (Something like order of magnitude, but I know that’s not the right terminology).
Oh, and there must be a level above ^…what is it?
There’s a polar form for complex numbers: Any complex number can be written in the form R(cosx + i sinx), where R is that number’s distance from the origin in the complex plane, and x is its angle, measured counterclockwise from the positive real axis.
Complex multiplication works out very nice geometrically:
[R(cosx + i sinx)] * [S(cosy + i siny)] = RS(cos(x+y) + i sin(x+y)]
(You can verify this if you like by using the addition formulas for sin and cos).
In other words, multiplying two complex numbers together basically adds their angles together and multiplies their distances from the origin. So we’re trying to find a number whose square is i. The angle of i is pi/2, so the square root we’re looking for should have an angle half that (so that when it’s multiplied by itself, the actual angle doubles). So the angle we’re looking for is pi/4. i is one unit from the origin, so our square root should also be one unit from the origin. And so the number we want is cos(pi/4) + i sin(pi/4) = 1/sqrt(2) + i/sqrt(2).
Of course, I haven’t mentioned yet that there’s another angle whose double is pi/2; that angle is 5pi/4, so the other square root of i is cos(5pi/4) + i sin(5pi/4) = -1/sqrt(2) - i/sqrt(2) (just the negative of the previous number).
As to your second question, I’m afraid I’m not aware of any common name for that, but there may indeed be one that I’m ignorant of.
By the way, concerning your second question, I should mention that Donald Knuth invented a not uncommon notation of representing the different levels you were speaking of, arrow notation, and chained arrow notation. These notations are handy in describing Graham’s number, often described as “the largest number that has ever been put to practical use”. So maybe that’s at least a partial answer to your question.
And now that I think on it, I do seem to remember hearing the level immediately above exponentiation referred to by some name, but that name escapes me at the moment.
No, there is a nice easy way to work out sqrt(i). Here’s how you do it:
Let sqrt(i) = x + iy (i.e. some number with both real and imaginary parts)
Then (x + iy)**2 = i
Multiplying out: x2 + 2ixy - y2 = i
Equate real and imaginary parts of the two sides of the equation (the real part of the RHS must equal the real part of the LHS, ditto for the imaginary part:
**Malacandra **got it right, except of course, there’s another square root, the negative one.
Just like the square roots of 9 are 3 and -3,
sqrt(i) = (1+i)/sqrt(2) and (-1-i)/sqrt(2)
And for operator precedence, by the way, there’s no real inherent mathematical reason behind the order we use. It makes just as much sense mathematically, to start from the right or left side and go in order.
Or, to phrase it the way the OP did, sqrt(i) is up and to the right (northeast), or else down and to the left (southwest) on the unit circle. In both cases the angle is 45 degrees.
When at secondary school, operator precedence was referred to as the BODMAS rule. I haven’t heard it referred to as anything other than operator precedence in college and university maths.
I think an even easier way is to remember that i is a unit vector pointed straight up on the number axes. One of its square roots is a unit vector at 45[sup]o[/sup] which is 1/2[sup]1/2[/sup] + i/2[sup]1/2[/sup]. The other square root is a unit vector 180[sup]o[/sup] away from that or -1/2[sup]1/2[/sup] - i/2[sup]1/2[/sup].
What you’re seeing here is that there are purely real numbers, purely imaginary numbers (like i, 2i, 8.73i, etc.), and complex numbers, which have both a real and an imaginary part, like 0.707 + 0.707i, which is the square root of i.
But this same complex “plane” can also be used to represent not only square roots, but cube roots, fourth roots, logarithms, sinusoidal functions, etc. The cube root of any number on the complex plane can be expressed by another number on the complex plane, as can the log or the sine of any number there. You don’t need an extra “dimension” - just adding an axis corresponding to the square root of -1 takes care of all these.
Viewed this way, the real numbers are just a special case - the numbers that lie along the horizontal axis of the complex plane. The set of all numbers includes all of them on the complex plane.
The (r, theta) vector-in-the-complex-plane method of representing complex numbers has a few interesting properties. One is that it makes it easy to multiply them. Another is that it makes it easy to take the square root.
To multiply, you simply multiply the two r’s and add together the two theta’s. So the complex number at (5, 47deg) times the complex number at (2, 109deg) equals the complex number at (10, 156deg). Since the complex number i is at (1, 90deg), multiplication by i is like rotation by 90deg.
To take the square root, you simply take the square root of r and divide theta by two. So the square root of (9, 64deg) is (3, 32deg). And the square root of i will of course be (1, 45deg).
Actually, the whole “point” of complex numbers is that they’re algebraically complete. With the real numbers, we sometimes couldn’t represent roots as having specific values. With complex numbers, that isn’t an issue. The square root (and nth root) of a complex number will always be a complex number.
Sorry Polycarp, but that’s not correct (sorry if you were just joking). If the square root of i was -i, then we would need that (-i)[SUP]2[/SUP] = i. But, (-i)[SUP]2[/SUP] = i[SUP]2[/SUP] = -1 (not i). Several people above have pointed out methods for getting the correct answers of sqrt(i) = (+/-) (1+i)(1/sqrt(2)). As Achernar pointed out, sqrt(i[SUP]9[/SUP])=i[SUP]4.5[/SUP] (not i[SUP]3[/SUP]), and i[SUP]4.5[/SUP] = (1+i)(1/sqrt(2)), one of the correct answers.
There are a lot of things you’re not allowed to do, in the real numbers. You can’t take a log of a negative number, you can’t take the square roots of negative numbers, you can’t take the inverse sine of a number greater than 1 or less than -1, etc. But all of these problems disappear when you’re working with complex numbers. Not only can you take a log of a negative number, you can even take the log of an imaginary number, or any complex number (except zero). You can take any complex number to any power, or to any root. You can do any trig function or inverse trig function you like. And you never need to introduce any system of numbers larger than the complex numbers. That’s the great beauty of the complex numbers.
Thanks to everyone for the help- I think I see the light…
So, going back to my trig circle, the process of squaring a complex number just a matter of squaring its module and doubling its argument, right? Does this extend to any powers of a number? And conversely, extracting the nth root of a complex number will yield n solutions?
