a. Since [tex]i=e^{i\pi/2}[/tex], we have
[tex]i^i=(e^{i\pi/2})^i=e^{i^2\pi/2}=e^{-\pi/2}[/tex]
b. Since [tex]1=e^0[/tex], we have
[tex]-1^{-i}=-(e^0)^{-i}=-(e^0)=-1[/tex]
c. Yes, for the reason illustrated in part b. [tex]1=e^0[/tex], and raising this to any power [tex]z\in\mathbb C[/tex] results in [tex]e^{0z}=e^0=1[/tex].
