Can a^x be negative infinity with i and e?

[coki2000]

I think about on this and found a result, $$e^{i\pi k}=e^{-i\pi k}=-1\Rightarrow i\pi k=ln(-1)$$
a>1 and k=1,2,3...

$$y=(-a)^{i\alpha }\Rightarrow lny=i\alpha ln(-a)=i\alpha (ln(-1)+lna)=i\alpha (i\pi k+lna)\Rightarrow y=e^{-\alpha \pi k}e^{i\alpha \ln{a}}$$

Then

$$\alpha \ln{a}=-\pi \Rightarrow \alpha =\frac{-\pi }{\ln{a}}\Rightarrow e^{-i\pi }e^{\frac{{\pi }^2 k}{\ln{a}}}=-e^{\frac{{\pi }^2 k}{\ln{a}}}$$

If I choose k as infinity, I get a exponential expression as negative infinity.
Is it right? Please explain to me. Thanks

 

[coki2000]

Sorry, k would be k=2n-1=1,3,5,7...

 

[jackmell]

Not sure what you're doing, but I'd write:
$$a^x=-k$$

$$e^{x\log(a)}=-k$$

$$x\log(a)=\log(-k)$$

$$x=\frac{\log(-k)}{\log(a)}=\frac{\ln(k)+i(\pi+2 n \pi)}{\log(a)}$$

and as $k\to\infty$, $x\to \infty+i(\pi+2n\pi)/\log(a)$