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Raising Integers to Imaginary Powers or Exponents

Date: 07/29/2004 at 11:35:04
From: Mike
Subject: Imaginary powers of integers - Riemann hypothesis

I would like to express 2^i in the form a + bi.  This would enable me
to perform some basic calculations on the Riemann zeta function.

I know that 1 = -e^i*pi so 1^i = -e^(-pi).  But strangely, I cannot
extend this:

  2 = -2e^i*pi so
  2^i = 2^i * -e^-pi

... which suggests -e^(-pi) = 1, but it isn't!!

Date: 07/29/2004 at 12:26:33
From: Doctor Vogler
Subject: Re: Imaginary powers of integers - Riemann hypothesis

Hi Mike,

Thanks for writing to Dr Math.  There is a short answer to your 
question which really side-steps all of the important issues, but I'll
give you that first:

You wrote

  1 = -e^(i*pi),

which is correct, and then you raised both sides to the i power,

  1^i = [-e^(i*pi)]^i,

and then you INCORRECTLY pulled the negative sign out of the right

  1^i = -[e^(i*pi)]^i

But you SHOULD have gotten

  1^i = [-e^(i*pi)]^i,
  1^i = [-1]^i * [e^(i*pi)]^i,
      = (-1)^i * e^(-pi),

and then wondered what is -1 raised to the i power.

Now, the unfortunate fact is that it can equal infinitely many 
different things, depending on exactly what you mean.  See the note at
the very bottom of

  Imaginary Exponents and Euler's Equation 

This is discussed in greater detail in

  Was Euler wrong? 2*Pi=0? 

Basically, complex exponents are usually defined by Euler's Equation

  e^(ix) = cos x + i*sin x

and the formula

  a^b = (e^ln a)^b = e^(b*ln a).


  e^(2*pi*i*n) = 1

for all integers n, which throws a wrench in things.  So

  a^b = (e^(2*pi*i*n + ln a))^b
      = e^(2*pi*i*n*b + b*ln a)

for any integer n.  Well, if b is also an integer, then

  e^(2*pi*i*n*b) = 1,

and there is no confusion.  If b = 1/2, so that we want the square
root of a, then

  e^(2*pi*i*n*b) = 1 or -1,

and so we get two square roots, the negative one and the positive one.
So we have to agree that when we say "the square root of 2," that we
actually mean the positive square root of two, bearing in mind always
that there are actually two square roots.  In fact, if b is a fraction
with denominator k (in lowest terms), then


can take on a total of k different complex values, known as the
"complex k'th roots of unity," and this gives rise to k different k'th
roots of any (complex or real) number.

Now, if b is NOT a fraction, then there are INFINITELY MANY different
b'th powers of unity, since there are infinitely many values of


And if b is a pure imaginary number, then these are all different real
numbers!  Weird!  But true.

So you might say that

  1^i = (e^0)^i = e^(0*i) = e^0 = 1,

but you might also say that

  1^i = (e^(2*pi*i))^i = e^(-2*pi)

or you might say that

  1^i = (e^(2*pi*i*n))^i = e^(-2*pi*n)

for any integer n.  And that's a lot of different real numbers, and
whatever (reasonable) value you might assign to


you could also multiply it by any of those values for 1^i and argue
that it is just as reasonable a value.

That said, you wanted to know about the Riemann Zeta Function, which
is often defined as

  zeta(s) = sum n^(-s)

That's a nice sum, but when you write it down you need to bear in mind
(whenever s is not a real integer) the facts in the discussion that we
just had, and be very careful about exactly what you mean by n^(-s). 
Just as "the square root of 2" is agreed upon to mean the POSITIVE
square root of 2, so too we need to agree on a particular -s'th power
of n.  In general, this is taken as

  e^(-s*ln n),

where the ln n is the real logarithm of n.  A careful mathematician
will state this explicitly, but some mathematicians just assume you
will know this and avoid the issue entirely.  Other mathematicians
will use a different definition of the zeta function entirely.  As an
example, notice how MathWorld defines the zeta function differently at
first, and then proves that it equals the above sum when s is a
positive real integer (bigger than 1, when the series diverges).

  Riemann Zeta Function 

And yet, they get around to using the sum later on anyway.

I hope this has helped you to understand how complex powers work and
how to interpret this sum for the zeta function.  If you have any
questions about this or need more help, please write back, and I will
try to offer further suggestions.

- Doctor Vogler, The Math Forum 

Date: 08/01/2004 at 05:49:53
From: Mike
Subject: Thank you (Imaginary powers of integers - Riemann hypothesis)

Dr. Vogler, thanks very much for your very helpful information.  I 
will have to think it through a bit more first, and I may want to ask
further advice.  Thanks again.
Associated Topics:
College Imaginary/Complex Numbers

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