
#1
Sep1610, 12:57 PM

P: 91

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 



#2
Sep1710, 05:06 AM

P: 91

Sorry, k would be k=2n1=1,3,5,7...




#3
Sep1710, 07:56 AM

P: 1,666

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] 


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