Question

How do I evaluate irrational/complex exponents without a calculator?

I feel like I should know this, because it seems like basic algebra... but I can't recall having learned it.

Answer 1

do you have an example?

Answer 2

No, I'd like to deal with this on a general basis.

Evaluate an "exact" expression for x^n for n that is a complex or transcendental number.

Answer 3

Well, if it's complex, you can use the more general form of Euler's identity. In particular, \[re^{i \theta} = r(\cos(\theta) +i \sin(\theta))\]And this should also work for non complex numbers, i.e., \(i\)=0. As for evaluating the sines and cosines, I don't know.

Answer 4

For a complex number \(z=a+bi\), we have
\(e^z=e^a\cdot e^{ib}=e^a(\cos(b)+i\sin(b)).\) This is a very famous identity.
I think, however, for most irrational numbers it's hard to find "exact" value. Euler's identity should be mentioned here, that is
\(e^{i\pi}+1=0\).
This identity contains, probably, the most important numbers in Mathematics (e, i, pi, 1 and 0).

Answer 5

But then getting any exponent into that form is not necessarily easy without a calculator.

Answer 6

Well, I think just being able to express a complex power in terms of sines and cosines is a pretty powerful tool. I was just hoping that I'd missed a "better" method due to the weird direction of my pre-college education.

As for Euler's identity... well, I could subtract one then take the i'th root of both sides...? I'm not sure how to isolate the transcendental power.

Answer 7

What would you do that for? I mean which "number" do you want to solve for, if that makes sense?

Answer 8

I'd like to isolate e^pi somehow without having to take the i'th root of both sides, because that seems a bit weird.

Answer 9

you just end up taking the sine and cosine of it in the more general form. There's no need to take the i-th root of anything as long as you can get your \[x^n = re^{i \theta}\]for some \(r\). (which may be a power in itself)

Answer 10

Well... that makes sense? I'd like to try writing one out, though. Give me a second.

Answer 11

Well we have \(e^{i\pi}=-1\), if that's what you're looking for. You can also find using the formula above that
\(\large e^{i\frac{\pi}{2}}=i\).
\(\large e^{-i\pi/2}=-i\).

Answer 12

I've got to leave. Good luck!

Answer 13

Thanks!

Answer 14

Thanks again for the help!