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i and (-1) with Multiple Powers


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?


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 
additional insights:

   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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