Saturday, 26 November 2011

Imaginary Exponents: x^i

To learn how we do this I first need to explain a very, very useful mathematical formula. It is called "Euler's formula", and this formula gives us a way to find the value of the imaginary exponential function (e^ix) using methods that we already have well defined and are easy to deal with. Euler's formula is: e^ix = Cos(x)+iSin(x), where x is an angle in radians. I may do a post on the actual maths behind why this is the case, but for the purpose of this post it has no relevance.

Now, we can find e^ix, but what use is this if we want to find 2^i, i^i or just anything raised to the power of i, let's call this a^i. So, we want to find a^x where x = i, so we need to try and remember a^x as something involving e raised to the power of something. So this means we have, a^x = e^y.
This then means that a^x = e^[xln(a)]. So we now have a^x in a form involving e raised to a power. So now we can input when x = i. Now by simply placing this into the equation we get, a^i = e^[i*ln(a)]. We can then turn this into something we can solve using Euler's formula, e^[i*ln(a)] = Cos[ln(a)]+iSin[ln(a)].
So now to actually input some numbers to this. Let's say I want to find 2^i, so from our previously defined formula we now have that: e^[i*ln(2)] = Cos[ln(2)]+iSin[ln(2)]. Using our calculators we will find that this is roughly Cos(0.693147)+iSin(0.693147), which then equates to roughly 0.76924+0.63896i. So 2^i ≈ 0.76924+0.63896i

As you can see, this is a complex number and it will be a lot of the time when we deal with imaginary exponents, but (as you may have thought) there are times when the solution to a^i will be a real answer. This is when iSin(x) = 0, and this will happen at Sin(x) = 0 and if you know your Sine curves you will know that this is at Sin(kπ), where k is any integer. Using our formula derived from Euler's, e^[i*ln(a)] = Cos[ln(a)]+iSin[ln(a)], we can see that if a^i is a real number, ln(a) = kπ. If we make both sides to the power of e, we can clear our logarithm to get: a = e^kπ. This should also then mean that the solution that is real (where a = e^kπ) should be equivalent to Cos(kπ).

Therefore if this is correct, then (e^3π)^i should produce a real value. (e^3π)^i = Cos[ln(e^3π)]+iSin[ln(e^3π)]. And when you do work this out, low and behold you get the answer of -1 (which incidentally is the same as Cos(3π)).

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