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### 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

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/
```
Associated Topics:
College Exponents
College Imaginary/Complex Numbers
High School Exponents
High School Imaginary/Complex Numbers

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