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
[tex]exp(z):=\sum_{k=1}^{\infty}\frac{z^k}{k!}[/tex]
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
[tex]\lim_{z\rightarrow 0}\frac{exp(z)-1}{z}=1[/tex]
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
[tex]\lim_{z\rightarrow 0}\frac{exp(z)-1}{z}=1[/tex]
[tex]\lim_{z\rightarrow 0}\frac{exp(z)-1}{z}=\lim_{y\rightarrow 0}\frac{C(y)+i S(y)-1}{i y}[/tex]
since we can let z->0 as i y->0
looking at the real part
[tex]\lim_{y\rightarrow 0}\frac{S(y)}{y}=0[/tex]
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
[tex]\lim_{y\rightarrow 0}\frac{S(y)}{y}[/tex]
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