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The ith Root of -1


Date: 09/16/1999 at 15:54:01
From: Carl Ford
Subject: Imaginary roots

I would like to know how the ith root of -1 (or (-1)^(1/i)) equals 
23.14069... I know that this works and why, but not how. It works by 
rearranging the equation e^(Pi*i)+1 = 0 to e^Pi = (-1)^(1/i), and 
e^Pi = 23.14069... but I can't understand how it works. It makes no 
sense to me that:
         ___
     _i /
      \/ -1   = 23.14069...

but
         ___
     _i /
      \/ 1     = 1

Thanks,
Carl


Date: 09/16/1999 at 22:23:49
From: Doctor Schwa
Subject: Re: Imaginary roots

Carl,

This is an excellent question! It certainly seems ridiculous that 
taking a root of a unit-sized number could suddenly be so big, even if 
the root is complex.

Here's the key point:

The expression e^(pi*i) is also the same as e^(-pi*i) because the pi 
here is referring to an angle in radians, so you can add or subtract 
any multiple of 2pi from it without changing the value.

Thus (-1)^(1/i) is equal to e^pi but it's ALSO equal to e^-pi and ALSO 
to e^(-3pi) and e^(-5pi) and e^(37pi) and ... 

Once you allow complex numbers into your exponents, logarithms become 
many-valued things, so roots (which can be defined in terms of 
logarithms) also have many possible answers.

I hope that helps clear things up. If you'd like to know more about 
e^(a + bi) in general, or roots or logarithms of complex numbers, 
please do write back with more questions.

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


Date: 09/22/1999 at 14:53:47
From: The Professor
Subject: Re: Imaginary roots

I don't really understand the above answer. Could you please show me 
how you would calculate the ith root of -1 if I didn't know that it 
equaled e^pi? My graphing calculator can take the ith root of -1, and 
I can't figure out how it does that. Also, how would you take the ith 
root of anything else, like the ith root of 9? (My calculator does 
that too, but I can't figure it out either.)


Date: 09/22/1999 at 15:51:33
From: Doctor Anthony
Subject: Re: Imaginary roots

We can write   -1 = cos(pi) + i.sin(pi)

               -1 = e^(i.pi)

Take ith root of both sides

       (-1)^(1/i) = [e^(i.pi)]^(1/i)

                  = e^[(i.pi)/i]

                  = e^(pi)

                  = 23.14069...

There are of course multiple answers because we could have started 
with

     -1 = e^(i.3pi)   or   e^(i.5pi)   or   ...

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

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