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### i and (-1) with Multiple Powers

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Date: 08/09/2001 at 22:32:03
From: Kenneth Martin
Subject: Imaginary numbers (the order in which i and (-1) should be
raised by when using multiple powers)

My question is about the order in which i and (-1) should be raised
when using multiple powers. The problem that I have run into is that
when I change the order of powers I can obtain many different values.
This is just one of the many examples I have:

(-1)^(3/2)^(1/3) equals .8660254038-.5i, but when it is simplified
to (-1)^(1/2) it of course equals i. Stranger still, when I switch the
order of the powers to (-1)^(1/3)^(3/2), the value equals (-i).

I have many more examples of this, but they're generally the same. Why
is it possible to obtain so many different values? I am quite aware
that i's have a nature all their own that negates rules in math that
are followed when dealing with real numbers. So what are the rules we
MUST follow when dealing with i?
```

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Date: 08/11/2001 at 23:15:50
From: Doctor Peterson
Subject: Re: Imaginary numbers (the order in which i and (-1) should
be raised by when using multiple powers)

Hi, Kenneth.

It sounds as if you are finding powers on a calculator without
stopping to think about what they mean. Let's think about that a
moment.

We're working with complex numbers here. Recall that x^(1/2), the
square root, has _two_ values, not just one; in real numbers we can
choose one as the primary root, but we can't even do that in complex
numbers. Further, there are _three_ cube roots, _four_ fourth roots,
and so on. So it really shouldn't surprise you at all that you can get
several different values for the same expression; they are all
correct! What should be surprising is that the calculator would tell
you that one of them is "the" answer. That's the real cause of the
difficulty; whatever rule they use to decide which answer to give
produces different results when you do it in a different order. It
would be better if it told you somehow that this is only one of a
certain number of values.

A side issue to consider is the meaning of x^y^z. We usually take
this to mean x^(y^z); you appear to mean (x^y)^z, which is the same
as x^(yz). I'll add brackets to avoid confusion.

So let's try doing the calculation as if we were the calculator, doing
just what we're told. The expression

[(-1)^(3/2)]^(1/3)

tells the calculator to raise -1 to the 3/2 power, which it probably
does by taking "the" square root, i, and cubing it, to get -i. (Of
course, this is actually the other square root of -1.) It would be
equally valid to take -i as the square root and cube it to get i; but
let's forget that for now. Now the calculator will take "the" cube
root of this, which might be done either by thinking of it as having
angle 270 degrees and dividing by 3 to get 90 degrees (giving the cube
root as i); or it might think of the angle as -90 degrees and take the
cube root with angle -30 degrees. The latter agrees with the answer
you give.

Since there are two square roots and three cube roots, there will be
six possible values for this expression. You've found ways to get
three of them. See if you can do the same thing I just did to explain

[(-1)^(1/3)]^(3/2)

I'd get i, myself; but we don't know for sure what rules the
calculator is using. You might want to read the manual to see if they
tell you, or keep it a secret.

You can also read these answers from the Dr. Math archives for some

Multiplying Radicals of Negative Numbers
http://mathforum.org/dr.math/problems/amy.7.12.00.html

Cube Roots of Numbers
http://mathforum.org/dr.math/problems/martin11.5.97.html

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 08/12/2001 at 17:54:11
From: Kenneth Martin
Subject: Thank you

I recently submitted a question to Dr. Math about a problem I was
having with i. Within a couple of days I received a response that
helped with the problem that I had. Dr. Math explained enough for me
to understand the solution, yet left enough unexplained to make me
rethink the problem. You have managed to answer more for me in two
days then my teachers could in two years! I just wanted to thank you
and I hope you will continue helping students like myself for a long
time to come.
```
Associated Topics:
High School Exponents
High School Imaginary/Complex Numbers

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