y to the x PowerDate: 06/10/2003 at 02:49:51 From: Bill Subject: y to the x power Computers seem able to calculate y^x when y is positive and x is any real number, but when y is negative, x must be an integer. I think that non-integer powers of negative numbers are complex numbers; (-1)^(1/2)is THE imaginary number. How can I calculate the real and imaginary parts of any non-integer powers of negative numbers? (Or do they all equal exactly sqrt(-1)?) It is interesting that you can calculate 2^2.5 but not (-2)^(2.5). Why is (-2)^x continuous on the positive side but has gaps on the negative side? I'd like to be able to visualize a 4-quadrant graph of y^x, even if imaginary numbers are involved. Date: 06/10/2003 at 15:09:39 From: Doctor Tom Subject: Re: y to the x power Hi Bill, It's a bit of a mess. As you know, if you want the square root of, say, 25, there are really two answers: 5 and -5. The square root of -25 similarly has two answers: 5i and -5i. In fact, every complex number except zero has two square roots. But there's more. Every non-zero complex number has three cube roots. For example, the cube roots of 1 are: 1, (-1 + sqrt(3))/2, (-1 - sqrt(3))/2. If you find any single cube root of a number, multiply it by each of these cube roots of one to obtain the three cube roots of that number, be that number real or complex. Similarly, every number other than zero has 4 fourth roots. The fourth roots of 1 are 1, -1, i and -i. Every non-zero number has 5 fifth roots, 6 sixth roots, and so on. Here's how to get all of them if you can get one. Multiply your one root by all the n-th roots of 1. Since e^(it) = cos(t) + i sin(t), then e^(2 pi i) = 1. The n-th roots of 1 are just: cos (2 pi k/n) + i sin(2 pi k/n) where k is an integer: k = 0, 1, 2, ..., n-1 To find the one n-th root to get started, you have to write your complex number as: e^(r + it) where r and t are real. Then one n-th root is: (e^(r/n))(cos(t/n) + i sin(t/n)) If the number whose n-th root you're taking is x + iy where x and y are real, t = arctangent(y/x) and r = sqrt(x^2 + y^2). If you take an irrational root of a non-zero number, there is an infinite number of roots. - Doctor Tom, The Math Forum http://mathforum.org/dr.math/ Date: 06/10/2003 at 15:26:58 From: Doctor Peterson Subject: Re: y to the x power Hi, Bill. A calculator that fully understands complex numbers should be able to handle (-2)^2.5; the only hard part is that there are two answers. For, say, (-2)^(1/3) there will be three answers, and it keeps getting more complicated. That's part of the reason calculators won't try to do it unless they know you want a complex answer (in every sense of the word). To find (-2)^2.5 we just do this: (-2)^(5/2) = ((-2)^(1/2))^5 = (+/- i sqrt(2))^5 = +/- i^5 (sqrt(2))^5 = +/- 4 i sqrt(2) There's a lot more to be said about fractional powers of negative numbers (or of any number, once we are talking about the complex numbers). Here are some places where we have discussed various facets of the issues you raise: Why Is (-n)^fractional Invalid ? http://mathforum.org/library/drmath/view/62979.html Decimal Exponents http://mathforum.org/library/drmath/view/55607.html i and (-1) with Multiple Powers http://mathforum.org/library/drmath/view/53881.html Square Roots of Complex Numbers http://mathforum.org/library/drmath/view/53864.html Cube Root of 1 http://mathforum.org/library/drmath/view/53842.html If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
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