I think about on this and found a result, [tex]e^{i\pi k}=e^{-i\pi k}=-1\Rightarrow i\pi k=ln(-1)[/tex] a>1 and k=1,2,3... [tex]y=(-a)^{i\alpha }\Rightarrow lny=i\alpha ln(-a)=i\alpha (ln(-1)+lna)=i\alpha (i\pi k+lna)\Rightarrow y=e^{-\alpha \pi k}e^{i\alpha \ln{a}}[/tex] Then [tex]\alpha \ln{a}=-\pi \Rightarrow \alpha =\frac{-\pi }{\ln{a}}\Rightarrow e^{-i\pi }e^{\frac{{\pi }^2 k}{\ln{a}}}=-e^{\frac{{\pi }^2 k}{\ln{a}}} [/tex] If I choose k as infinity, I get a exponential expression as negative infinity. Is it right? Please explain to me. Thanks
Not sure what you're doing, but I'd write: [tex]a^x=-k[/tex] [tex]e^{x\log(a)}=-k[/tex] [tex]x\log(a)=\log(-k)[/tex] [tex]x=\frac{\log(-k)}{\log(a)}=\frac{\ln(k)+i(\pi+2 n \pi)}{\log(a)}[/tex] and as [itex]k\to\infty[/itex], [itex]x\to \infty+i(\pi+2n\pi)/\log(a)[/itex]