Question-122854

Question Number 122854 by Study last updated on 20/Nov/20

Answered by mathmax by abdo last updated on 20/Nov/20

{(- 2)}^{\sqrt{2}} = e^{\sqrt{2} \ln (- 2)} = e^{\sqrt{2} (\ln (- 1) + \ln (2))} = e^{\sqrt{2} (i π + \ln 2)}

= e^{\sqrt{2} \ln (2) + i \sqrt{2} π} = e^{\sqrt{2} \ln (2)} {\cos (π \sqrt{2}) + isin (π \sqrt{2})}