y to the x Power

Date: 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/