Ivars Wrote:I wondered what results can be obtained if we calculate 2^I , 2^(2^I) , etc.
\( 2^I = e^{I*\ln2} \)
\( 2^{(2^I)} = 2^{e^{I*\ln2}} \)
\( e^{I*\ln2}= cos (I*\ln2)+I*sin(I*\ln2) \)
There is a little error here:
\( e^{i\varphi} = \cos(\varphi) + i\sin(\varphi) \)
There is no i contained inside the cos and sin.
Otherwise I dont know, looks pretty chaotic to me, so why not taking a fractal explorer (for example chaos pro) putting in the formula z_0=I, z=c^z (c is the to be colored point in the plane) and show us the results? So you can see how it behaves at all points in the plane not only your favourite values.
