GravitatisVis said:
It's a really interesting equation. I'm not sure how it works out though. On a similar note, when I raise a whole number to an imaginary non whole number, say 2^2.5i, sine and cosine become involved for some reason. I'm not sure why that happens. Can someone explain this? I finished precalc this year; on to calculus I when school starts in a month. Thanks.
So what you need is a definition of e^z=exp(z) that works for complex numbers z. The best motivation for such is analytic continuation, but that requires to much theory, so one of these is often employed.
1) Infinite series
exp(z):=\sum_{k=1}^{\infty}\frac{z^k}{k!}
2)differential equation
exp'(z)=exp(z) for all complex z
exp(0)=1
3) functional relation
exp(a+b)=exp(a)exp(b) all complex a and b
\lim_{z\rightarrow 0}\frac{exp(z)-1}{z}=1
4) unmotivated definition
exp(z)=exp(Re(z))cos(Im(z))+i exp(Re(z))sin(Im(z))
5) various ad hoc definitions that state what properties of exp are wanted and define a unique function.
I like 3) best
consider
z=x+i y
exp(z)=exp(x+i y) for x and y real
then
exp(z)=exp(x)exp(i y)
We already know what to do with exp(x) so we turn attention to exp(i y)
exp((x+y)i)=exp(x i)exp(y i)
now for convienance we will define
exp(i x)=C(x)+i S(x)
exp((x+y)i)=[C(x)C(y)-S(x)S(y)]+i[S(x)C(y)+C(x)S(y)]
C(x+y)=C(x)C(y)-S(x)S(y)
S(x+y)=S(x)C(y)+C(x)S(y)
now if x=x y=-x
1=exp(0)=exp(i(x-x))=exp(i x)exp(-i x)
so exp(-i x)=1/exp(i x)=1/(C(x)+i S(x))
=(C(x)-i S(x))/(C(x)^2+S(x)^2)=exp(-i x)/(C(x)^2+S(x)^2)
hence
1=C(x)^2+S(x)^2 for all real x
Now consider
\lim_{z\rightarrow 0}\frac{exp(z)-1}{z}=1
\lim_{z\rightarrow 0}\frac{exp(z)-1}{z}=\lim_{y\rightarrow 0}\frac{C(y)+i S(y)-1}{i y}
since we can let z->0 as i y->0
looking at the real part
\lim_{y\rightarrow 0}\frac{S(y)}{y}=0
Now collecting what we know
C(x+y)=C(x)C(y)-S(x)S(y) for all real x and y
S(x+y)=S(x)C(y)+C(x)S(y) for all real x and y
1=C(x)^2+S(x)^2 for all real x
\lim_{y\rightarrow 0}\frac{S(y)}{y}
We recognize this can only be if C(x) and S(x) are our beloved functions
C(x)=cos(x)
S(x)=sin(x)
so
exp(x+i y)=exp(x)(cos(x)+i sin(y))
exp(x+i y)=exp(x)cos(x)+i exp(x)sin(y)
for complex exponetials we define
u^v=exp(v log(u))
where the log used is the natural log so log(e)=1
thus the sin and cos in complex exponentials
In your example
2^2.5i=exp(2.5 i log(2))
2^2.5i=cos(2.5 log(2))+i sin(2.5 log(2))
2^2.5i~-.161363+.986895i