Raising Integers to Imaginary Powers or ExponentsDate: 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 side, 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 http://mathforum.org/dr.math/faq/faq.euler.equation.html This is discussed in greater detail in Was Euler wrong? 2*Pi=0? http://mathforum.org/library/drmath/view/51629.html 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). Unfortunately, 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 e^(2*pi*i*n*b) 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 e^(2*pi*i*n*b). 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 2^i, 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 inf zeta(s) = sum n^(-s) n=0 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 http://mathworld.wolfram.com/RiemannZetaFunction.html 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 http://mathforum.org/dr.math/ 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. |
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