Exponents with i? How does that work?

Question

Prove that (e^π)^i = -1
What does 2^i equal?

Answer 1

2^i=0.769238901 + 0.638961276 i
2^i=cos(ln2) + Sin (ln2)i
What you have there (actually should be e^(πi)=-1 is Euler's equation- one of my favorites! It has the five most basic mathematical constants in one equation! The beauty of it! The source I posted has a great explanation that isn't on too high of a level. The second requires a little bit of calculus knowledge.

Answer 2

I know you know about Taylor series, so I will just skip talking about how to find it.

e^x=1+x+x^2/2+x^3/3!+ . . .

(e^x)^i=e^(xi) = 1+xi+(xi)^2/2+ . . . = 1-x^2/2+x^4/4!-x^6/6! + i(x-x^3/3!+x^5/5!-x^7/7!) = cos(x)+i•sin(x).

Thus (e^π)^i = cos(π)+i•sin(π)=-1.

2^i= e^(ln(2^i))= e^(ln(2)•i)= cos(ln(2))+i•sin(ln(2)).

Answer 3

e^(i*theta) = cos(theta) + i*sin(theta)

You can prove the above by taking the taylor series for e^x, sin(x), and cos(x), and showing that these are equal.

e^(i*pi) = -1
Well if you substitute pi for theta you get...

cos(pi) + i*sin(pi) = -1 + i*0 = -1

Calculatin 2^i requires will look like:

2^i = cos(ln(2)) + i*sin(ln(2))
approximately equal to: 0.76924 + 0.63896i

Answer 4

I just wanted to compliment you on your icon picture of the Andromeda. 2^i = -2?

Answer 5

im pretty sure that i = root -1...i dunno if that helps u at all but i tired...