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Are Exponents Associative?


Date: 02/05/2002 at 07:43:41
From: Lucas
Subject: Complex numbers

I have a question that I couldn't answer:

   How much is 2^(i) or X^(i) ?

I asked my teacher, and he said that

2^(i) = 2^(-1)^(0.5)
      = 0.5^0.5~0,707106781186547524400844362104849

I think a formula exists to find y and z:

2^(i) = y and y^(i) = 0.5
2^(-i) = z and y^(-i) = 0.5
2^(-i) = z and y^(i) = 2

Is this correct? If not, why not?
Thanks.


Date: 02/05/2002 at 11:37:37
From: Doctor Peterson
Subject: Re: Complex numbers

Hi, Lucas.

It looks as if your teacher accidentally assumed that exponents are 
associative, that is, that

    (a^b)^c = a^(b^c)

We are used to this being true for addition and multiplication, but it 
is NOT TRUE of exponents; for that reason, it is important to include 
parentheses to clarify this, though by convention we take a^b^c to 
mean a^(b^c). Note that, in reality,

    (a^b)^c = a^(b*c)

(Your own facts are correct, if I fix one typo, and they make use of 
this fact. Unfortunately, I don't see that they go anywhere useful.)

Here is what your teacher claimed, with the wrong move stated 
explicitly:

    2^i = 2^[(-1)^0.5] = [2^-1]^0.5 = 0.5^0.5 = 0.707
                      NO!

To solve this correctly, you need to use Euler's formula, which you 
can read about in the Dr. Math FAQ:

   Imaginary Exponents and Euler's Equation
   http://mathforum.org/dr.math/faq/faq.euler.equation.html   

This says that

    e^(xi) = cos(x) + sin(x) i 

We can transform your power of 2 to a power of e this way:

    2^i = (e^ln(2))^i = e^(ln(2) i)

Now apply Euler to this:

        = cos(ln(2)) + sin(ln(2)) i

        = 0.769 + 0.639 i

I know that sounds weird, but this definition of imaginary powers 
makes all the rules for exponents work, so it is what we use. You can 
find more about this in our archives.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Exponents
High School Imaginary/Complex Numbers

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